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IPE SECOND YEAR ` KUKATPALLY CENTRE

FIITJEE KUKATPALLY CENTRE: # 22-97, Plot No.1, Opp. Patel Kunta Huda Park, Vijaynagar Colony, Hyderabad - 500 072. Ph.: 040-64601123 Regd. Off.: 29A, ICES House, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110 016. Ph: 011 - 2651 5949, 2656 9493, Fax: 2651 3942

INORGANIC CHEMISTRY VA GROUP ELEMENTS

1. Discuss the structures of the oxides of 2N

A: Nitrogen forms several oxides 2 2 3 2 2 4 2 5N O, NO, N O , NO or N O and N O . The structures of these oxides

are as follows:

* Nitrous oxides ( )2N O is a linear molecule with the following resonance structures.

N N – O N = N = O

* Nitric oxide ( )NO is an odd electron molecule containing unpaired electron. It is a resonance hybrid of

the following structures.

N = O N = O N = O

* Dinitrogen trioxide ( )2 3N O may have one of the following two structures.

O

O = N N = O

(Symmetric)

N – N

O O

O (Unsymmetric)

* Nitrogen dioxide ( )2NO is an odd electron molecule and paramagnetic in nature in gaseous state due

to the presence of unpaired electron. It is a resonance hybrid of the following structures. N

O O

N

O O

N

O O

N

O O In the solid state it exist as dimmer 2 4N O . It is diamagnetic in nature.

N N O O

O O N N

O

O O

O

* Nitrogen Pentoxide ( )2 5N O : Its structure is

N N

O O

O O O

2. Write an essay on the preparation and structures of any three oxides of Nitrogen.

A: * Nitrous oxide ( )2N O : It can be prepared by heating ammonium nitrate.

4 3 2 2NH NO N O 2H OD¾¾® +

Properties: a) It is a colourless, neutral gas b) When inhaled it produces laughing hysteria so it is called laughing gas. Structure:

N N – O N = N = O

* Nitric oxide or Nitrogen oxide ( )NO : It is prepared by the catalytic oxidation of ammonia.

Catalyst

3 2 24NH 5O 4NO 6H O+ ¾¾¾¾® +


Properties: a) It is a colourless, neutral gas. b) It is an odd electron molecule with paramagnetic character in gaseous state. c) It is diamagnetic in nature in solid state. Structure:

N = O N = O N = O

* Nitrogen dioxide ( )2NO : It is prepared by the thermal decomposition of lead nitrate.

( )2 2 222Pb NO 2PbO 4NO OD¾¾® + +

Properties: a) It is an odd electron molecule. b) It is brown coloured gas.

On cooling it converts into solid state. In the solid state it is colourless and diamagnetic. In solid state it do not contain unpaired electron

Structure:

N

O O

N

O O

N

O O

N

O O In the solid state it exist as dimmer 2 4N O . It is diamagnetic in nature.

N N O O

O O N N

O

O O

O

3. Write an essay on the structural aspects of the phosphorous series of acids. A: Phosphorous forms three oxyacids in phosphorous series of acids.

* Orthophosphorous acid 3 3H PO

* Metaphosphorous acid 2HPO

* Hypophosphorous acid 3 2H PO

a) In all oxyacids, phosphorous is involved in 3sp hybridization and is surrounded by the atoms or groups tetrahedrally.

b) In all these oxyacids atleast one –OH group is in bond with phosphorous. c) The hydrogen atoms in –OH groups are ionizable and are responsible for the acidic nature. d) The basicity of an oxyacid is equal to the number of OH groups in that acid.

e) The phosphorous series of acids contain P H- bonds and P OH- bonds. The reducing property of these acids is due to this P H- bonds.

Structures:

* Orthophosphorous acid 3 3H PO : Its basicity is two. It contains two –OH groups and one P – H bond.

P

O

H OH

OH

* Metaphosphorous acid 2HPO : Its basicity is one. Generally it exists as cyclic compound. The structure

of monomer is HO P O- =

* Hypophosphorous acid 3 2H PO : Its basicity is one. It contains one –OH group and two P – H bonds.

P

H

H

OH

OH


4. Write an essay on the preparation and properties of (a) Hypophosphorous acid (b) Hypophosphoric acid (c) Metaphosphoric acid

A: (a) Hypophosphorous acid ( )3 2H PO :

Preparation: It is prepared in the laboratory by heating white phosphorous with dilute solution of

( )2Ba OH

( ) ( )4 2 2 2 32 22P 3Ba OH 6H O 3Ba H PO 2PH+ + ® +

From this barium salt 3 2H PO is obtained.

Properties: It is a mono basic acid. It is very strong reducing agent in basic medium and is oxidized to

3 3H PO .

(b) Hypophosphoric acid ( )4 2 6H P O :

Preparation: When phosphorous oxide is exposed to moist air hypophosphoric acid is formed. Properties: It is a tetra basic acid. Oxidation number of phosphorous in hypophosphoric acid is +4

(c) Metaphosphoric acid ( )3HPO :

Preparation: It is prepared by heating orthophosphoric acid at 870 K.

870 K

3 3 3 2H PO HPO H O¾¾¾® +

Properties: It gives a glassy transparent solid. So it is called glacial phosphoric acid. It is a monobasic acid. Its salts are called metaphosphates. Its salts exist as cyclic metaphosphates.

5. Describe Haber’s process for the preparation of 3NH . Give a Neat diagram of the plant. Lable it.

A: Ammonia can be manufactured by Haber’s process. In this method ammonia is manufactured by the direct reaction between nitrogen and hydrogen.

2 2 3N 3H 2NH ; H 93.63 kJ+ ® D = -

* The above reaction is reversible reaction * The forward reaction is exothermic reaction. * When the reactants converts into products the volume decreases, as the number of product molecules

are less than the total number of reactant molecules. Synthesis of NH3 by Haber’s Process:

or At ordinary temperature the reaction occurs slowly. Accoridng Le Chatelier’s principle the following conditions are more favourable for more yield of ammonia. (i) Low temperature (ii) High pressure (iii) Catalyst Ammonia is manufactured at about 725 – 775 K; using a pressure of 200 to 300 atmospheres. Iron powder mixed with molybdenum is used as catalyst. Molybdenum acts as promoter to iron catalyst. A mixture of

potassium and aluminium oxides ( )2 2 3K O and Al O also act as promoter to the iron catalyst.

Pure and dry 2N and 2H mixture in 1 : 3 ratio by volume are made to react at a pressure of 200 to 300

atmospheres. Then it is passed into a catalytic chamber. These gases react at a pressure of 200 to 300 atmosphere. Then it is passed into a catalytic chamber. These gases react in the presence of catalyst at 725-775 K. Ammonia is formed to the extent of about 10%. It is separated from the unreacted 2N and 2H

by condensation.

6. Detail the Ostwald’s process for the manufacture of 3HNO . Give balanced equations wherever possible.

A: Ostwald’s process for the manufacture of 3NH : Ammonia mixed with air in 1 : 7 or 1:8 ratio is passed over

hot platinum gauze catalyst. Then ammonia is oxidized to NO.

Pt. gauze

3 2 21155K4NH 5O 4NO 6H O+ ¾¾¾¾® +


or Manufacture of HNO3 – Ostwald’s process:

The NO formed in the above reaction is cooled and mixed with oxygen to convert into 2NO . The 2NO is

then passed into warm water under pressure in the presence of excess of air to give 3HNO

2 22NO O 2NO+ ®

2 2 2 34NO 2H O O 4HNO+ + ®

The acid formed is 61% concentrated. This is further concentrated in three states.

Stage 1: 61% 3HNO is distilled until 68% 3HNO is obtained.

Stage 2: 68% 3HNO is mixed with conc. 2 4H SO and distilled to get 98% acid.

Stage 3: 98% 3HNO is cooled in a freezing mixture. Crystals of pure 3HNO separate out.

7. Describe an industrial method for the preparation of superphosphate of lime. Why is it converted into “triple phosphate of lime”?

A: Superphosphate of lime is a mixture of calcium dihydrogen phosphate ( )2 4 2Ca H POé ùë û and Gypsum

( )4 2CaSO .2H O

It is manufactured by treating the powdered phosphate rock with calculated amount of sulphuric acid.

( ) ( ) ( )3 4 2 4 2 2 4 4 22 2Ca PO 2H SO 4H O Ca H PO 2 CaSO .2H O+ + ® +

Manufacture of Superphosphate of lime:

The phosphate rock is powdered. It is taken in a cast iron mixer. Calculated amount of sulphuric acid is added and mixed well. This reaction mixture is dumped in one of the dens 1D and 2D through the values

1 2V and V . The reaction is allowed to take place for about 24 – 36 hours. The impurities like carbonate and

fluoride are removed in the form of 2CO and HF gases. The hard mass formed is powdered and sold as

superphosphate of lime.


The 4CaSO in the superphosphate of lime is an insoluble waste product and is of no value to the plants. To

avoid this waste product formation, it is converted into triple superphosphate by using phosphoric acids.

( ) ( )3 4 2 4 2 42 2Ca PO 4H PO 3Ca H PO+ ®

The triple phosphate completely dissolves in water i.e., no waste product formed. 8. What are the similar features of ammonia and phosphine? Write a comparative note on them. A: Both ammonia and phosphine can be prepared by similar methods. They can be prepared by the hydrolysis

of their binary compounds either with water or dilute acid.

Ex: ( )3 2 2 32Mg N 6H O 3Mg OH 2NH+ ® ¯ +

( )3 2 2 32Ca P 6H O 3Ca OH 2PH+ ® +

Comparative notes on their general properties:

* Stability of 3NH is greater than 3PH because N – H bond is stronger than P – H bond.

* Both contain one lone pair on central atom, so they can act as Lewis bases. But 3NH is stronger Lewis

base than 3PH

* Both can act as reducing agents but 3NH is weaker reducing agent then 3PH because of more

stability.

* Both 3NH and 3PH are soluble in water but the solubility of 3NH in water is more than 3PH .

* 3NH and 3PH are volatile and colourless gases but 3NH is less volatile because it can form inter

molecular hydrogen bonds.

* The hydrogen atoms in 3NH and 3PH can be substituted by alkyl group like 3CH and halogens like

Cl. The ease of substitution decrease from 3NH to 3PH

* 3NH can form salts with any acid forming salts like 4 3NH Cl.PH can react with very strong acids like

HI.

* Both 3NH and 3PH are pyramidal in shape. In 3NH nitrogen is involved in 3sp hybridization but in

3PH the pure p-orbitals participate in bonding. The bond angle in 3PH ( 93 36 '° ) is less than in

( )3NH 107°

9. In what way the structures of trioxides and pentoxides of 2N and 4P differ. Draw their structures and show.

A: Trioxides and Pentoxides of nitrogen and phosphorous are chemically similar but differ in their structures. The oxides of nitrogen exists as monomers 2 3N O and 2 5N O while the oxides of phosphorous 4 6P O and

4 10P O exists as dimers. Their structures are given below.

Sl.No. Nitrogentrioxide structure Nitrogen pentoxide structure

1.

O N N

O

O asymmetrical form

N N

O

O O O

O

O O = N N = O

(Symmetric)

N N

O O

O O O

2. Phosphoric trioxide structure Phosphoric pentoxide structure


P

P

P

P

O O

O O

O O

(dotted line does not represent bond)

P

P

P

P

O O

O O

O O

Phosphorous Oxygen

O

O

O

O

O

10. Write balanced equations for the reactions of 2H O with

a) 4 6P O b) 4 10P O c) 3PCl d) 5PCl

A: ( )4 6 2 3 3P O 6H O 4H PO Phosphorous acid+ ®

( )4 10 2 3 4P O 6H O 4H PO Orthophosphoric acid+ ®

( )4 10 2 3little

P O 2H O 4HPO Metaphosphoric acid+ ®

( )3 2 3 3PCl 3H O 3HCl H PO Phosphorous acid+ ® +

( )5 2 3 4PCl 4H O 5HCl H PO Orthophosphoric acid+ ® +

11. Write a note on the Cyanamide process. Mention any three important uses of 3NH .

A: Cyanamide process: Ammonia can be manufactured synthetically by cyanamide process. In this method the calcium is made to react with nitrogen gas in an electric furnace at 1273 – 1373 K. In this reaction a mixture of

2CaCl and 2CaF acts as catalyst. Then a mixture of calcium cyanamide and graphite known as nitrolium is

produced. When super heated steam is passed over calcium cyanamide at 453 K ammonia is produced.

1273 1373 K

2 2 2GraphiteCalcium

Cyanamide

CaCl N CaCN C-+ ¾¾¾¾¾® +

453 K

2 2 3 3CaCN 3H O CaCO 2NH+ ¾¾¾® +

Uses of Ammonia: * Ammonia is used in the manufacture of fertilizers like ammonium sulphate, urea, calcium ammonium

nitrate etc. * Ammonia is used as refrigerant. * Ammonia is used in the manufacture of sodium carbonate by Solvay process. * Ammonia is used in the manufacture of nitric acid and explosives like ammonium nitrate.

12. Discuss the principle underlying the manufacture of 3HNO

A: Nitric acid can be manufactured by two methods: * Birkland and Eyde Method – Principle:

a) An electric arc converts 2N and 2O of air into NO.

2 2N O 2NO+ ®

b) NO is oxidized to 2NO by air.

2 22NO O 2NO+ ®

c) 2NO is dissolved in water in the presence of oxygen to get 3HNO

2 2 2 34NO 2H O O 4HNO+ + ®

* Ostwald’s process-Principle

a) Ammonia is oxidized by air when 3NH and air in 1 : 7 or 1 : 8 ratio passed over platinum gauze

catalyst. Pt. guaze

3 2 21155 K4NH 5O 4NO 6H O+ ¾¾¾¾® +


b) NO is then oxidized to 2NO by air.

2 22NO O 2NO+ ®

c) 2NO is dissolved in water in the presence of oxygen to get 3HNO

2 2 2 34NO 2H O O 4HNO+ + ®

13. What makes the difference between the two following reactions? Phosphate rock treated with a) Chambers acid b) excess of phosphoric acid write the equation.

A: a) When phosphate rock reacts with chambers acid superphosphate of lime is formed.

( ) ( )3 4 2 4 2 2 4 4 22 2Ca PO 2H SO 4H O Ca H PO 2CaSO .2H O Heat+ + ® + +

Superphosphate of lime contains insoluble waste product 4CaSO . So the percentage of phosphorous in

superphosphate of lime become less. b) To avoid the waste product gypsum in superphosphate of lime the phosphate rock is treated with

phosphoric acid ( )3 4H PO . Then triple phosphate of lime or triple superphosphate is formed.

( ) ( )3 4 3 4 2 42 2Ca PO 4H PO 3Ca H PO+ ®

The triple phosphate completely dissolves in water i.e., no waste product formed. SAQ 1. What is allotropy? Explain in N and its congeners? A: If the same element exists in two or more physical states having nearly similar chemical properties but

different physical properties. It is known as allotropy. Except bismuth all the VA group elements exhibit allotropy.

Solid 2N exists in anda b forms.

Phosphorous exists in many allotropic forms such as white P, red P, scarlet P, a - black, b - black and violet.

Arsenic exists as metallic or grey arsenic, non-metallic or yellow arsenic, black arsenic. Antimony exist in three allotropic forms * crystalline, metallic variety or common variety. * non-metallic or yellow or a -antimony * explosive antimony. 2. What is catenation? How does it vary in group 15? A: Combining capacity of the atoms of same element to form long chains is called catenation. Catenation

capacity depends on bond energy. In the group from top to bottom atomic size increases. So bond length increases and bond energy decreases, hence catenation capacity decreases. Since nitrogen is smaller atom it has more catenation power. So it can form 2 2H N NH- and 3N H having two and three atoms in a chain

respectively. Phosphorous can form 2 4P H having two P atoms in the chain. Other elements of VA group

don’t exhibit catenation power.

3. Draw the structures of 4 6P O and 4 10P O . In what respect do thy resemble each other, what happen when

they react with water? A:

P

P

P

P

O O

O O

O O

(dotted line does not represent bond)

P

P

P

P

O O

O O

O O

Phosphorous Oxygen

O

O

O

O

O In both 4 6P O and 4 10P O the four phosphorous atoms are arranged in tetrahedral shape. In both 4 6P O and

4 10P O six oxygen atoms are acting as bridges between phosphorous atoms forming P O P- - bonds.

In 4 10P O one oxygen atom is in dative bond with phosphorous atoms.


When 4 6P O react with water orthophosphorous acid is formed.

4 10 2 3 4P O 6H O 4H PO+ ®

4. Write balanced equations for the formation of 3NCl and 3PCl . Give their hydrolysis reactions.

A: 3NCl can be prepared by the action of excess chlorine on ammonia.

3 2 3NH 3Cl NCl 3HCl+ ® +

3PCl can be prepared by the direct reaction between phosphorous and chlorine.

4 2 3P 6Cl 4PClD+ ¾¾®

Hydrolysis: 3NCl hydrolyses in water forming ammonia and hypochlorous acid

3 2 3NCl 3H O NH 3HOCl+ ® +

3PCl hydrolyses in water forming phosphorous acid hydrochloric acid.

3 2 3 3PCl 3H O H PO 3HCl+ ® +

5. In a compound 5MX , M is any V group element except Nitrogen why (X) is halogen?

A: Except nitrogen all the VA group elements can form penta halides of the type 5MX

Ex: 5PCl , 5AsCl , 5SbCl , 5BiF etc. Nitrogen do not contain d-orbitals, in its valency shell but other VA

group elements contain vacant d-orbitals in their valency shell. So when excited one of the ns electron can go into the d-orbital.

Ground state electronic configuration ns np nd

Excited state electronic configuration ns np nd

Since there are five unpaired electrons in the excited state except nitrogen other VA group elements can form five covalent bonds with halogens. So they can form pentahalides of the type 5MX . Nitrogen cannot form

pentahalides because of the absence of d-orbitals in its valency shell. 6. Which of the acids of phosphorous does not show monomeric state but cyclic structure? Name any polymer

of the acid. A: Metaphosphoric acid do not exist as monomer. It exist as cyclic polymer. Ex: Cylcic metaphosphate.

O

P P

O P

O O

O O

O

O O

7. A sample of most ammonia is to be dried. What method do you suggest? A: Moisture from ammonia can be removed by passing over dry lime or CaO. But it cannot be dried over Conc.

2 4H SO , fused 2CaCl or 2 5P O as they react with ammonia.

8. Write the formula of superphosphate of lime. Why is it converted into triple phosphate?

A: The common formula of superphospate of lime is ( ) ( )2 4 4 22Ca H PO 2 CaSO .2H O+ . It acts as good

fertilizer. The 4CaSO in its insoluble waste product and its presence has no significance to the plants. To

avoid this waste product 4CaSO superphosphate is changes into triple phosphate which completely

dissolves in water. 9. How is superphosphate of lime formed? Give an equation. Explain why the product is a hard mass? A: Superphosphate of lime can be prepared by treating powdered phosphate rock with calculated quantity of

2 4H SO

( ) ( ) ( )3 4 2 4 2 2 4 4 22 2Super phosphate of lime

Ca PO 2H SO 4H O Ca H PO 2 CaSO .2H O+ + ® +

The reason for the hardness of the final product obtained in the dens is due to the presence of Gypsum.


VSAQ 1. Write the composition of “phosphate rock”.

A: ( )3 4 2Ca PO

2. Give any two examples to show negative oxidation state of nitrogen.

A: Ammonia 3NH 3- ; Hydrazine 2 4N H 2- ; Hydroxyl amine 2NH OH 1-

3. Why is NO paramagnetic in nature? When does it become diamagnetic? A: NO, is an odd electron molecule containing odd number of electrons. The total number of electrons in No

molecule is 15. When all electrons are paired one electron remains unpaired. So NO is paramagnetic. But when temperature decreases it dimerises and become diamagnetic.

4. How is dinitrogen tetroxide formed? Give equation.

A: When the temperature of 2NO is decreased it dimerises to convert into dinitrogen tetroxide.

2 2 42NO N O¾¾®

5. How many oxygens surround a phosphorous in phosphorous pentoxide? A: In phosphorous pentoxide each phosphorous atom is surrounded by four oxygen atoms.

6. Write the equations for the hydrolysis of 3NCl . How does it differ from hydrolysis of 3PCl ?

A: 3NCl hydrolysis in water giving 3NH and HOCl while 3PCl hydrolysis in water giving orthophosphorous

acid and HCl.

3 2 3NCl 3H O NH 3HOCl+ ® +

3 2 3 3PCl 3H O H PO 3HCl+ ® +

7. What are the orbitals of P that are involved in the formation of 5PCl ?

A: In the formation of 5PCl , 3sp d hybrid orbitals are involved.

8. What is nitrolim? How is it formed?

A: Nitrolium is a mixture of calcium cyanamide ( )2CaCN and graphite. When 2N is passed over calcium

carbide containing 2CaCl or 2CaF as catalyst at 1273 – 1372 K nitrolim is formed.

1273 1378

2 2 2Calcium cynamide Grpahite

CaC N CaCN C-+ ¾¾¾¾® +

9. What drying agent is suitable to dry 3NH ?

A: Ammonia can be dried by passing over dry lime (CaO) because it does not react with ammonia.

10. What is the function of 2CaCl in cyanamide process?

A: 2CaCl acts as catalyst in cynamide process during 2CaC react with 2N to form nitrolim.

11. Which oxides of 2N are neutral oxides?

A: Among the oxides of nitrogen, nitric oxide (NO) and nitrous oxide ( )2N O are neutral.

12. Which of the two oxides 2 5N O and 2 5P O is better dehydrating agent? Give an example for the same

reaction.

A: 4 10P O is a strong dehydrating agent than 2 5N O

Ex: Dehydrating of 3HNO with 4 10P O gives 2 5N O

3 4 10 2 5 34HNO P O 2N O 4HPO+ ® +

13. Give reasons for the chemical inactivity of nitrogen at ordinary conditions.

A: 2N molecule contains triple bond ( )N Nº . To break the triple bond large amount of energy

( )1945.4 kJ mol- is required. Due to this high bond dissociation energy nitrogen is apparently inactive

under normal conditions. 14. What is the stability order of VA group hydrides? Explain the gradation in the reducing property of these

hydrides?


A: The stability order of the hydrides of VA group hydrides is 3 3 3 3 3NH PH AsH SbH BiH> > > > . As the

stability decrease they dissociate easily and can act as strong reducing agents. So the order of reducing power of VA group hydrides is 3 3 3 3 3NH PH AsH SbH BiH< < < < .

15. Why does nitrogen does not form pentahalides? A: To form pentahalides the ns electron should be excited to nd orbital. In the case of nitrogen there is no d

orbital in its valency shell i.e., second orbit. So nitrogen cannot form pentahalides.

16. Can 5NCl be prepared by direct union of the elements? Why or why not?

A: No. 5NCl cannot be prepared by the direct union of elements because nitrogen do not react with chlorine

directly. Further due to the absence of d orbitals in its valency shell nitrogen cannot form 5NCl

17. Write the structure of 3HNO

A: The structure of nitric acid is the resonance hybrid of the following structures.

N

O O

OH

N

O O

OH

Nitric Acid 18. 2 5P O is strong dehydrating agent. Why is it not used to dry 3NH ?

A: 2 5P O is acidic and 3NH is basic. They bond react to form salt. So 3NH cannot be dried using 2 5P O

19. How do you convert 4 3NH NO into 3NH give reactions?

A: When ammonium nitrate is heated with a base ammonia gas will be liberated.

4 3 3 3 2NH NO NaOH NaNO NH H O+ ® + +

When 4 3NH NO is heated with alkaline solution of Zinc, it is converted into ammonia.

4 3 2 2 3 2NH NO 4Zn 8NaOH 4Na ZnO 2NH 3H O+ + ® + +

20. How do you prepare hypophosphorus acid in the laboratory? A: When white phosphorous is boiled with dilute solution of barium hydroxide barium hypophosphite will be

formed.

( ) ( )4 2 2 2 32 22P 3Ba OH 6H O 3Ba H PO 2PH+ + ® +

To the barium hypophosphite solution if dil. Sulphuric acid is added hypophosphorous acid will be formed.

( )2 2 2 4 4 3 22Ba H PO H SO BaSO 2H PO+ ® +

4BaSO being insoluble can be removed by filtration.

21. How many Valence shell electrons are utilized by each phosphorous atom in 4P molecule?

A: From the structural representation of 4P , we can infer that each phosphorous shares three of its valence

electrons with other phosphorous atoms. A lone pair of electrons is seen on each phosphorous. 22. Why is nitrogen is a diatomic gaseous molecule while phosphorous is a tetra atomic solid? A: Nitrogen atoms are small in size and can approach very close to one another. This facilities the lateral

overlap of the p-orbitals to form p -bonds. In phosphorous only single bonds are formed due to the larger sizes of phosphorus atoms. Multiple bond formation and through it acquiring octet of electrons is not possible. Hence 4P molecules are formed.

23. 3PH is quite stable in Air. But it catches fire when heated to 150 C° . Why?

A: 3PH frequently contains 2 6P H (disphosphine) in trace amounts as impurity. This catches fire on heating in

air.

24. Write the names of the compounds formed by the union of 3PH and 3AsH separately with HI.

A: The reaction must be similar to the union of 3NH with HI giving ammonium iodide. Therefore, 3PH gives

4PH I (phosphonium iodide) and 3AsH gives 4AsH I (Arsonium Iodide)


25. If pure P-orbitals in As or Sb overlap with the S-orbitals of hydrogens, the bond angle is expected to be 90° , why is it 91 .48 '° in their hydrides?

A: The HMH bond angle in 3AsH and 3SbH would be expected to be 90° But due to repulsions between M-H

bonds, the angle increases to 91 .49 '°

26. Many penta halides of VA group elements are known. NO hydrides of 5MH exist. Why?

A: To attain the pentavalent state, d-orbitals must be used. Hydrogen is not sufficiently electronegative to make the d-orbitals effective by contraction.

27. Identify the oxidant and reductant in the given reaction 2 3 23HNO HNO 2NO H O® + +

A: 2HNO acts as both the oxidant and reductant. 2HNO as a reductant changes to 3HNO . 2HNO as an

oxidant changes to NO.

28. What is the change in oxidation state of nitrogen in auto oxidation, auto reduction of 2HNO ?

A: 2 3 23HNO HNO 2NO H O® + +

In 2HNO to 3HNO , the change of oxidation state of nitrogen is from +III to +V. From 2HNO to NO the

change in oxidation state is from +III to +II 29. What is the difference between tautomeric and resonance structures?

A: In tautomers the skeleton of atoms in the structures differs ( )2Ex : HNO . In resonance, the skelton of atoms

does not change (Ex: Benzene) 30. What is nitration mixture?

A: A mixture (1:1) of 3HNO (Conc.) and Conc. 2 4H SO is known as nitration mixture. This is used in nitration

reaction.

( ) 2 4H SO6 6 3 6 5 2 260 C

NitrobenzeneC H HNO Conc. C H NO H O

< °+ ¾¾¾® +


VIA GROUP ELEMENTS 1. How is Ozone prepared in the laboratory? Give any three oxidation reactions of 3O ?

A: Principle: Ozone is prepared by passing silent electric discharge through pure cold and dry oxygen.

( ) ( )Silent Electric

2 g 3 gdisch arg e3O 2O , H 284.5 kJ¾¾¾¾¾® D = +

Siemen’s and Brodie’s ozonizers are used to prepare ozone in the Laboratory.

Siemen’s Ozonizer: * This ozonizer consists of two coaxial glass tubes sealed at one end. * The inner sides of the inner tube and the outer sides of the outer tube are coated with tin foils. * These tin foils are connected to the terminals of a powerful induction coil. * Cold and dry oxygen is passed through the annular space from one end. * Oxygen undergoes silent electric discharge partially and 10% Ozone is formed. * Ozone and oxygen mixture known as ozonized oxygen is collected from the other end. Oxidising properties of Ozone:

* Ozone oxidizes black lead sulphide (PbS) to white lead sulphate ( )4PbSO

3 4 2PbS 4O PbSO 4O+ ® +

* Hydrogen chloride is oxidized to chlorine by ozone.

3 2 2 22HCl O H O Cl O+ ® + +

* Ozone oxidizes moist potassium iodide ( )KI to iodine ( )2I

2 3 2 22KI H O O 2KOH O I+ + ® + +

2. How is ozone is prepared in Brodie’s method? Write any three reduction reactions of 3O with equations.

A: Preparation of Ozone, Brodie’s method: Principle: Ozone is prepared by subjecting cold dry oxygen gas to the silent electric discharge using dilute sulphuric acid as conducting medium and Cu wires as electrodes.

( ) ( )2 g 3 g3O 68K.Cal. 2O+ ®

* Brodie’s ozonizer consists of double wall U tube in which dilute 2 4H SO is placed.

* Copper wires are immersed in dilute 2 4H SO

* Copper wires are connected to the terminals of powerful induction coil. * Cold and dry oxygen is passed through annular space between the two walls of U tube. * Oxygen undergoes silent electric discharge and 15% of oxygen is converted into ozone.

Reducing properties of ozone: * Ozone reduces hydrogen peroxide to water

2 2 3 2 3H O O H O 2O+ ® +

* Ozone reduces barium peroxide to barium oxide


2 3 3BaO O BaO 2O+ ® +

* Ozone reduces silver oxide to metallic silver

2 3 2Ag O O 2Ag 2O+ ® +

3. How do you prepare hypo in the laboratory? Giving proper equations detail the reactions of hypo.

A: Crystalline hydrated sodium thiosulphate ( )2 2 3 2Na S O .5H O is known as “hypo”

Laboratory preparation: In the laboratory hypo is prepared by the following methods. * By boiling alkaline or neutral sodium sulphite solution with flowers of sulphur.

2 3 excess 2 2 3Na SO S Na S O+ ®

* By the oxidation of sodium sulphide or sodium polysulphide with air.

Heat in

2 5 2 2 2 3air2Na S 3O 2Na S O 6S+ ¾¾¾® +

* By treating sodium sulphide solution with sulphur dioxide.

2 2 2 2 32Na S 3SO 2Na S O S+ ® +

Reactions of Hypo:

* Action of heat: On heating hypo undergoes thermal decomposition to give 2 2H S,SO and S. Hypo

looses all the molecules of water (water by crystallization) when heated to about 488 K.

* Reaction with dilute acids: When hypo reacts with dilute acids like HCl or 2 4H SO to give 2SO and S.

2 2 3 2 2Na S O 2HCl 2NaCl H O SO S dilute+ ® + + +

* Reaction with 3AgNO solution: When hypo reacts with 3AgNO solution, two kinds of reactions may

take place.

a) When dilute hypo is added to 3AgNO solution, a white precipitate of 2 2 3Ag S O is formed which readily

changes to a black solid ( )2Ag S . The reactions are:

( )2 2 3 3 3 2 2 3Na S O 2AgNO 2NaNO Ag S O white ppt+ ® + ¯

( )2 2 2 3 2 4 2H O Ag S O H SO Ag S Black ppt+ ® + ¯

b) When concentrated hypo is added to 3AgNO solution a white precipitate ( )2 2 3Ag S O is obtained first.

The precipitate readily dissolves in excess of sodium thiosulphate due to the formation of complex compound.

( )2 2 3 3 2 2 3 3Na S O 2AgNO Ag S O 2NaNO White ppt.+ ® ¯ +

( ) ( ) ( )2 2 3 2 2 3 3 2 3 2Ag S O 3Na S O 2Na Ag S O Sodium argentothiosulphate complexcompoundé ù+ ® ë û

* Reaction with iodine:

Sodium thiosulphate reacts with iodine to give sodium tetrathionate ( )2 4 6Na S O

2 2 3 2 2 4 62Na S O I 2NaI Na S O+ ® +

This reaction is used in volumetric analysis to estimate iodine. * Reaction with exposed photographic film or AgBr: a) In photography the fixing is done by washing the film with hypo solution.

b) The silver bromide (or the silver halide) on the film reacts with sodium thiosulphate to give a complex compound.

( )2 2 3 3 2 3 2AgBr 2Na S O Na Ag S O NaBré ù+ ® +ë û

* Reaction with moist Cl2: Hypo reacts with moist 2Cl to give 2 4Na SO and HCl.

2 2 3 2 2 2 4Sodium sulphate

Na S O Cl H O Na SO S 2HCl+ + ® + +

In this reaction hypo is used as “antichlor” to remove excess of chlorine


* Reaction with salts: Sodium thiosulphate reacts with ferric chloride, cupric chloride or auric chloride etc. And converts them into complex thiosulphates.

4. Write the reactions of the following with 3O .

i) 2 2C H ii) 2 4C H iii) 6 6C H

Give the structures of the products in each case. What happens if these products are treated with 2Zn / H O

A: Reactions of O3:

* With Acetylene ( )2 2C H : Ozone when treated with acetylene, an addition compound called acetylene

ozonide is obtained. This on hydrolysis in presence of zinc, glyoxal is produced.

HC CH + O3 HC CH

O O

O Zn H2O

CHO

CHO

+ H2O2

(Acetylene ozonide) (Glyoxal)

* With Ethylene ( )2 4C H : Ozone when treated with ethylene, an addition compound called ethylene

ozonide is obtained. This on hydrolysis in presence of zinc, formaldehyde is produced.

H2C CH2 + O3 CH

CH2

O O

O Zn H2O

2HCHO + H2O2

(Ethylene ozonide)

(Formaldehyde)

* With Benzene ( )6 6C H : Ozone on treating with benzene an addition compound called benzene tri

ozonide is obtained, which on hydrolysis gives 3 moles of glyoxal.

+ 3O3 O O

O

O O

O

O

O O

Zn 3H2O

CHO

3CHO + 3H2O2

(Glyoxal)

SAQ

1. What are the bond angles in 2H O and 2H S? Why are they different?

A: * 1) In 2H O oxygen undergoes 3sp hybridization. Due to the presence of two lone pair of electrons

shape of 2H O molecule is angular or V-shape with H – O – H bond angle 104 28'°

2) Due to the greater repulsion between lone pair and lone pair of electrons than lone pair and bond pair, the tetrahedral angle is decreased to 104 28'°

* 1) In 2H S only pure “p” orbitals of sulphur are involved in bond formation. Hence the bond angle is less

than of 104 28'° and it is only 92 30 '°

2) 2H S is angular or V-shaped molecule.

2. What are the structures of 2 3SO and SO ? Explain them in terms of VBT.

A: * Structure of 2SO : 2SO is angular molecule. The O S O bond angle is 119 30 '° . In 2SO sulphur

atom undergoes 2sp hybridization in first excited state 3s, x3p and y3p orbitals undergo 2sp

hybridization. z3p and one d orbital are unhybridised and are used in the formation of p pp - p and

d pp - p bonds.


S

O O

S

O O( )x y

p p- ( )z z

p d- ( )x yp p-( )

z zp d-

o1.43A

o1.43A

* Structure of 3SO : In gaseous state 3SO has a planar triangular structure. The O S O bond angle is

120° . In 3SO sulphur atom undergoes 2sp hybridization in second excited state. 3s, x3p and y3p

orbitals underto 2sp hybridization. One z3p and two 3d orbitals are unhybridized state and are used in

the formation of one p pp - p and two d pp - p bonds.

S

O O

O

3. Give the structures of ( ) ( )4 6i SF ii SF . Explain them.

A: Structure of 4SF :

* The structure of 4SF is trigonal bipyramidal with one equatorial position occupied by a lone pair.

* Sulphur in 4SF undergoes 3sp d hybridization.

* S utilizes four hybrid orbitals for bonding while the fifth orbital accomidates a lone pair of electrons.

F

F

F

F

S

Structure of 6SF :

* The shape of 6SF is octahedral or square bipyramidal.

* Sulphur in 6SF undergoes 3 2sp d hybridization.

* All hybrid orbitals are used in bonding to form 6 sigma bonds between S and F atoms.

F

F

F

F F

F

S

VSAQ

1. Write the structure of gaseous sulphur molecule at low temperatures. A:

O = S 2. What is allotropy? Give the allotropes of oxygen?

A: Existence of an element in two or more physical forms is called as allotropy. Allotropes of oxygen and 2O

and 3O .

3. Write the names of the allotropic forms of S. A: Sulphur exists in many allotropic forms. The important forms are a or rhombic sulphur, b or monoclinic

sulphur, g or monoclinic sulphur, plastic or c sulphur.


4. At room temperature 2H O is a liquid while 2H S is a gas. Explain.

A: At room temperature water exists as liquid because of inter molecular hydrogen bonding. While 2H S exist as

a gas because of the absence of intermolecular hydrogen bonding.

5. What are the bond angles in 2H O and 2H S? Why they differ in their bond angles?

A: Bond angles in 2H O and 2H S are 104 36 '° and 92 30 '° respectively. In 2H O molecule oxygen undergoes

3sp hybridization where as in 2H S pure p-orbitals are involved in bond formation.

6. What is tailing of mercury? What chemical changes takes place in this process?

A: On passing 3O , mercury looses it metallic luster and meniscus and it sticks on the glass walls due to

formation of mercurous oxide ( )2Hg O . This is known as tailing of mercury. The chemical change takes

place in this process is 3 2 22Hg O Hg O O+ ® +

7. Give an example of a reaction which consumes all of the atoms of oxygen in 3O .

A: 2 3 33SO O 3SO+ ®

In this reaction 3O is completely consumed.

8. What happens when hypo reacts with 3AgNO ?

A: * When diluted hypo reacts with 3AgNO solution, it gives a white ppt. of silver thiosulphate which

on hydrolysis gives a black ppt. of silver sulphide.

2 2 3 3 2 2 3 3dil. hypo Silver thiosulphate

Na S O 2AgNO Ag S O 2NaNO+ ® +

( )

2 2 3 2 2 2 4Silver sulphide

Black.ppt

Ag S O H O Ag S H SO+ ® ¯ +

* When concentrated hypo reacts with 3AgNO solution, it gives a complex compound.

2 2 3 3 2 2 3 3Na S O 2AgNO Ag S O 2NaNO+ ® +

( )2 2 3 2 2 3 3 2 3 2Sodium argento thiosulphate

3Na S O Ag S O 2Na Ag S Oé ù+ ® ë û

9. How is hypo useful in photography? A: Hypo is used as fixing agent in photography.

( )2 2 3 2 2 3 2Sodium argento thiosulphate (Complex compound)

AgBr 2Na S O Na Ag S O NaBré ù+ ® +ë û

10. Mention two advantages of contact process over other processes.

A: * 2 4H SO obtained is extremely pure and concentrated.

* The impurities can be tested and the reactants can be recycled.

11. What is the reaction of 3O with PbS? Give the equation.

A: 3O oxidizes black PbS to white 4PbSO

3 4 2PbS 4O PbSO 4O+ ® +

12. What is antichlor? Give example. A: The reagent used to remove excess of chlorine in textile industry is called antichlor.

2 2 3 2 2 2 4Na S O Cl H O Na SO 2HCl S+ + ® + +

In this hypo is used as antichlor agent.

13. Write any two uses of ozone? A: * As an insecticide and bactericide * In sterilization of water.


VII GROUP ELEMENTS 1. Describe Whytlaw-Gray method for the preparation of fluorine?

A: * Fluorine is prepared by the electrolysis of fused potassium hydrogen fluoride ( )2KHF . The

electrode reactions are:

KHF2 Fusion K+ + H+ + 2F- Electrolysis

at cathode

at anode

2H+ + 2e- H2

2F- F2 + 2e-

1 2

3 4

5

8

7 6

9

1. Fused 2KHF 2. Heating coil 3. Grpahite (anode) 4. Copper diaphragm

5. Fluorspar stopper 6. Fluorine 7. Copper cell (cathode) 8. 2H

9. Inlet for HF Manufacture: * In this method, electrolysis is carried out in an electrically heated copper cell. * The copper vessel serves as cathode also. * Anode is made of graphite. The anode is surrounded by a copper diaphragm perforated at the bottom.

* This diaphragm prevents the mixing of 2H and 2F which reacts explosively if they come into contact.

* 2F , liberated at the anode, is passed through the U-tube containing sodium fluoride.

* Hydrogen fluoride vapours accompanying fluorine as impurity, are removed by NaF.

( )2NaF HF NaHF+ ®

* 2H is liberated at the cathode.

* The gaskets used in the cell are coated with Teflon to prevent corrosion of the parts.

2. Write the chemical properties of 2F with relevant equations?

A: * Reaction with water:

2F reacts with water and gives ozonized oxygen.

2 2 22F 2H O 4HF O ;+ ® + 2 2 33F 3H O 6HF O+ ® +

* Reaction with alkalies:

a) When 2F reacts with cold, dilute NaOH gives sodium fluoride and oxygen difluoride ( )2OF .

2 2 22NaOH 2F 2NaF OF H O+ ® + +

b) When 2F reacts with hot, concentrated NaOH gives sodium fluoride and oxygen.

2 2 24NaOH 2F 4NaF O 2H O+ ® + +

* Reaction with other halides: Fluorine oxidizes all other halide ions to the corresponding halogens.

2 2F 2KCl 2KF Cl+ ® +

* Reaction with inert gases: Heavier inert gases like Kr and Xe form compounds with fluorine.

2 2Xe 3F XeF+ ®

* Reaction with KHSO4 (Potassium hydrogen sulphate):


Fluorine oxidises potassium hydrogen sulphate to potassiumper sulphate ( )2 2 8K S O .

2 4 2 2 8F 2KHSO K S O 2HF+ ® +

* Reaction with H2S:

Fluorine oxidizes 2 6H S to SF

2 2 6H S 4F 2HF SF+ ® +

* Reaction with non metals: Except oxygen and nitrogen, other non-metals directly combine with fluorine and give binary compounds.

2 6S 3F SF+ ®

* Reaction with metals: All the metals (including noble metals like Au, Pt, etc.) form metal fluorides.

2 2Cu F CuF+ ®

3. What is the principle of preparing 2Cl in the laboratory? Describe Nelson’s method for its.

A: Principle: 2Cl is manufactured by electrolysis of brine solution.

Manufacture: * Nelson cell consists of U-shaped porous steel vessel lined inside with asbestos. * It serves as cathode. * This vessel is suspended in a rectangular iron tank. * Brine solution (10% NaCl) is taken in the vessel. * A carbon rod is dipped in the Brine solution which acts as Anode. * Steam is passed into the cell. On electrolysis, the following reactions take place.

( ) ( ) ( )aq aq aq2NaCl 2Na 2Cl+ -® +

At cathode: ( ) ( )2 aq 2 g2H O 2e 2OH H+ ® +

At anode: ( ) ( )aq 2 g2Cl Cl 2e® +

* 2Cl gas liberated at anode is collected and is compressed in steel cylinders.

* Sodium ions penetrate through the asbestos paper lining and reach the cathode.

* Here sodium ions combine with OH- to form NaOH. In this process, 2H and NaOH are important

byproducts.

2Na 2OH 2NaOH+ -+ ®

4. Write the structures of all the oxyacids of 2Cl .

A: NAME FORMULA OXIDATION STATE OF

CHLORINE BASICITY

Hypochlorous acid HClO +1 1

Chlorous acid 2HClO +3 1

Chlorine acid 3HClO +5 1

Perchloric acid 4HClO +7 1


Structures of oxyacids of chlorine: * Hypochlorous acid: HClO

Cl

OH

The chlorine atom in this acid undergoes 3sp hybridisation. This conjugate base of Hypochlorous acid is

Hypochlorite ion ( )ClO-

* Chlorous acid: 2HClO

Cl

OH O

The chlorine atom in this acid undergoes 3sp hybridization. The chlorine atom has 2s bonds one p

bond ( )p dp - p and two lone pairs.

The conjugate base of chlorous acid is chlorite ion ( )2ClO- which is angular in shape with a bond angle

of 111° .

* Chloric acid: 3HClO

Cl

OH O O

The chlorine atom in this acid undergoes 3sp hybridization. The chlorine atom has 3s bonds, 2p

bonds (both are p dp - p ) and one lone pair.

The conjugate base of chloric acid is chlorate ion ( )3ClO- which is pyramidal in shape with a bond angle

of 106°

* Perchloric acid: 4HClO

Cl

OH O O

O

The chlorine atom in this acid undergoes 3sp hybridization. The chlorine atom has 4s bonds, three p

bonds (all d pp - p ) and no lone pairs.

The conjugate base of perchlorine acid is perchlorate ion ( )4ClO- which is tetrahedral in shape with a

bond angle of 109.5° * The acid strength of different oxyacids of chlorine increases with an increase in the oxidation state of the

chlorine.

2 3 4HOCl HClO HClO HClO

acid strength increases

< < <uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuur

5. How is bleaching power prepared industrially? Give any of its four chemical properties with equations. A: Industrial preparation of bleaching powder (Baechmann’s plant): * Bachmann’s plant consists of a vertical iron tower provided with a hopper at the top and inlets for

chlorine and hot air slightly above the base. * The tower is fitted with a number of horizontal shelves at regular heights. Each shelf is fitted with a

rotating rake. * Dry slaked lime is introduced into the tower through the hopper at the top. * The slaked line moves downwards with the help of the rotating rakes and comes in contact with the

current of chlorine rising upwards. * This process of moving the reactants in opposite directions is called principle of counter currents.


* Slaked lime reacts with chlorine and gets converted into bleaching power which is collected in the container placed at the bottom.

* The hot air drives away unreacted chlorine ( ) 2 2 22Ca OH Cl CaOCl H O+ ® +

Chemical properties of bleaching powder: * Reaction with water: With cold water, bleaching powder gives chloride and hypochlorite ion.

22CaOCl Ca Cl ClO+ - -® + +

In hot water, it undergoes auto oxidation and gives chloride and chlorate ions. * (a) Reaction with insufficient amount of dilute acids: When small amounts of dilute acid is added, bleaching powder liberates oxygen.

2 2 4 2 4 22CaOCl H SO CaCl CaSO 2HCl O+ ® + + +

(b) Reaction with excess of dilute acids: On treating bleaching powder with excess of dilute acid, chlorine is liberated. This liberated chlorine is known as “available chlorine”.

2 2 4 4 2 2CaOCl H SO CaSO H O Cl+ ® + +

* Effect of a catalyst:

Bleaching powder decomposes to give 2O in the presence of a catalyst 2CoCl .

2CoCl2 2 22CaOCl 2CaCl O¾¾¾® +

* Oxidising property: Bleaching powder oxidizes lead salts to lead dioxide and ethanol to acetaldehyde.

22 2 2Pb 2CaOCl PbO 2CaCl+ + ® +

( ) 2CaOCl3 2 3 2

Ethyl alcohol AcetaldehydeCH CH OH O CH CHO H O+ ¾¾¾¾® +

6. Write all the chemical properties of bleaching powder. Give equations. How is it useful to man. A: * On long standing, bleaching powder undergoes auto-oxidation and changes into chloride and chlorate.

( )2 2 3 26CaOCl 5CaCl Ca ClO® +


* Reaction with water: With cold water, bleaching powder gives chloride and hypochlorite ion.

22CaOCl Ca Cl ClO+ - -® + +

In hot water, it undergoes auto oxidation and gives chloride and chlorate ions. * Reaction with insufficient amount of dilute acids: When small amounts of dilute acid is added, bleaching powder liberates oxygen.

2 2 4 2 4 22CaOCl H SO CaCl CaSO 2HCl O+ ® + + +

* Reaction with excess of dilute acids: On treating bleaching powder with excess of dilute acid, chlorine is liberated. This liberated chlorine is known as “available chlorine”.


* Effect of a catalyst:

Bleaching powder decomposes to give 2O in the presence of a catalyst 2CoCl .

2CoCl2 2 22CaOCl 2CaCl O¾¾¾® +

* Oxidising property: Bleaching powder oxidizes lead salts to lead dioxide and ethanol to acetaldehyde.

22 2 2Pb 2CaOCl PbO 2CaCl+ + ® +

( ) 2CaOCl3 2 3 2

Ethyl alcohol AcetaldehydeCH CH OH O CH CHO H O+ ¾¾¾¾® +

Uses * It is used in the sterilization of water, as bleaching agent, as an oxidizing agent and in the preparation of

chloroform. SAQ

1. How does 2F react with (i) 2H O , (ii) NaOH. Give equations for them.

A: * Reaction with 2H O :

2F reacts with water and gives ozonized oxygen.

2 2 22F 2H O 4HF O+ ® + ; 2 2 33F 3H O 6HF O+ ® +

* Reaction with NaOH:

a) When 2F reacts with cold, dilute NaOH gives sodium fluoride and oxygen difluoride ( )2OF .

2 2 22NaOH 2F 2NaF OF H O+ ® + +

b) When 2F reacts with hot, concentrated NaOH gives sodium fluoride and oxygen.

2 2 24NaOH 2F 4NaF O 2H O+ ® + +

2. Give the reaction of 2Cl with the following:

(i) 2SO (ii) NaOH (iii) Iron metal

A: * Reaction with 2SO :

2Cl when reacts with 2SO under the influence of sunlight and gives 2 2SO Cl (sulphuryl chloride).

sunlight

2 2 2 2SO Cl SO Cl+ ¾¾¾¾®

* Reaction with NaOH: a) When chlorine reacts with cold, dilute NaOH gives sodium chloride and sodium hypochlorite

(NaOCl). 2 2Cl 2NaOH NaCl NaOCl H O+ ® + +

b) When chlorine reacts with hot, concentrated NaOH gives sodium chloride and sodium chlorate.

( )3NaClO . 2 3 23Cl 6NaOH 5NaCl NaClO 3H O+ ® + +

* Reaction with Iron metal:

Chlorine reacts with iron metal and gives ferric chloride ( )3FeCl

2 33Cl 2Fe 2FeCl+ ®


3. What is “Available Chlorine”? Give chemical equation(s) which determine the same.

A: Available chlorine is the amount of 2Cl set free when bleaching powder is treated with excess of dilute

2 4H SO or 2CO

A good sample of bleaching powder contain about 35-38% available chlorine.


2 2 3 2CaOCl CO CaCO Cl+ ® +

VSAQ

1. Why is the E.A. of 2Cl greater than that of 2F ?

A: Fluorine has unexpectedly low electron affinity than chlorine. This is due to very small size of the fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p subshell of fluorine and thus, the incoming electron does not feel much attraction. Therefore, its electron affinity is less.

2. Write an equation for 2F reaction with 4KHSO and tell the nature of the chemical change.

A: 2F oxidizes potassium hydrogen sulphate to potassium persulphate.

2 4 2 2 8F 2KHSO K S O 2HF+ ® +

3. Write the balanced equation(s) for the reaction of 2Cl with 3NH

A: a) With excess of 2Cl , nitrogen trichloride ( )3NCl and HCl are formed.


b) With excess of 3NH , ammonium chloride and nitrogen are formed.

3 2 4 28NH 3Cl 6NH Cl N+ ® +

4. 1 mole of 3NH is mixed with 8 moles of 2Cl in a reaction vessel. Write the equation for the reaction.

A: Since chlorine is taken in excess amount, 1 mole of Ammonia reacts with 3 moles of chlorine.


5. What is the reaction between bleaching powder and excess of dil. 2 4H SO ?

A: On treating bleaching powder with excess of dilute 2 4H SO , chlorine is liberated. This liberated chlorine is

known as “available chlorine”


6. Give any two uses of bleaching powder. A: a) It is used in the sterilization of water b) It is used as a bleaching agent for cotton and paper pulp.

7. Give the uses of florine. A: * It is used in rocket fuels. * HF is used in etching of glass.

* NaF, 3 6Na AlF are useful insecticides.

* As a refrigerant (Freon 2 2CCl F ) and as a plastic ( )2 4 nTeflon C Fé ùë û


TRANSITION ELEMENTS 1. Giving examples explain Werners theory of complex compounds.

OR Why was Werners theory nessesary? Discuss the theory.

A: Werner explained the mechanism of formation of complexes. The important postulates of this theory are * Every complex compound contains a central metal atom (or) ion. * The central metal shows two types of valences. a) Primary valency:

* Primary valency is numerically equal to the oxidation state of the metal. These valencies are non directional and are represented by dotted lines (……………………..)

* Species (or) groups bound by primary valencies undergo complete ionization. These valencies are

identical with the number of ionic bonds. Ex:- In 3CoCl ( 3Co+ and 3Cl- are present). These are three

primary valencies for Co i.e. three ionic bonds are present.

Ex: Similarly ( )3 36Co NH Clé ùë û . In this complex the primary valencies of Co are three.

b) Secondary Valency: The secondary valencies of metal are directed m space around it in a symmetric order each metal in a

given oxidation state has a characterstic number of secondary valencies. * The number of secondary valencies is numerically equal to the coordination number of the metal ion in

the complex * Since the secondary valencies are directional in space a complex has a specific shape. * Some negative ligands may satisfy both primary and secondary valencies. Such complexes do not

ionize. * The primary valency of a metal is known as its outer sphere of attraction (or) ionizable valency. The

secondary valencies are known as the inner sphere of attraction (or) coordination sphere or non-ionizable valency.

i) 3 3CoCl 6NH :- Since the coordination number (secondary valency) of 3Co+ is 6, 6 3NH groups must

be linked to the central metal ion which do not ionize and 3Cl- ions are held by primary valencies.

CO

NH3 NH3 NH3

NH3

NH3 NH3

Cl-

Cl-

Cl-

ii) 3 3CoCl 5NH :- Since the coordination no. (secondary valency) of 3Co+ is 6 to satisfy its coordination

number.

* Ammonia molecules and one Cl- ion are linked to central metal ion which are non ionizable. The rest

of the chloride ions i.e. two Cl- ions are held by primary valencies.

CO

NH3 NH3 NH3

NH3

NH3

Cl-

Cl- Cl-

iii) 3 3CoCl 4NH :- Since the coordination number (secondary valency) of 3Co+ is 6 to satisfy its

coordination number, 4 ammonia molecules and two Cl- ion are linked to central metal ion which are

non ionizable. The rest of the chloride ions i.e. one Cl- ions are held by primary valencies.


CO+3 NH3

NH3

NH3

H3N

Cl- Cl-

Cl-

iv) 3 3CoCl 3NH :- Since the coordination number (secondary valency) of 3Co+ is 6 to satisfy its

coordination number, 3 Ammonia molecules and three Cl- ion are linked to central metal ion which are non ionizable.

CO+3 NH3

NH3

H3N Cl-

Cl-

Cl-

SAQ

1. Define EAN. Calculate the EAN of the following metals in their respective complexes.

i) ( ) ( )3 24Cu NH OHé ùë û ii) ( ) ( )2 36 3

Co H O NOé ùë û iii) ( )4 6K Fe CNé ùë û

A: The total number of electrons the central metal in a complex possess after formation of coordination complex is effective atomic number (EAN) of the metal in that complex.

EAN = [Z(atomic number of metal) – No. of electrons lost (oxidation states) + No. of electrons gained from ligands]

* ( ) ( )3 24Cu NH OHé ùë û EAN 29 2 8 35Þ = - + =

* ( ) ( )2 36 3Co H O NOé ùë û EAN 27 3 12 36Þ = - + =

* ( )4 6K Fe CNé ùë û EAN 26 2 12 36Þ = - + =

2. What do you understand by the term ligand? A: Atom, ion or molecule which can donate a pair of electrons to central metal in complex is called ligand. * Ligands are classified into 3 types based on their charge.

1) Negative ligands: 2 24 2 4X,SO ,CN and C O . The ligands satisfy both primary and secondary valencies

of central metal atom. (where X = halogens)

2) Neutral ligands: 2 3H O, NH . These ligands satisfy only secondary valency

3) Positive ligand: NO+ * Based on the number of coordination positions 1) Monodentate ligands: Donate only 1 pair of electrons to central metal.

Ex: 2 24 2 4 2 3X,SO ,CN,S O ,H O, NH

2) Bidentate ligands: Donate 2 pair of electrons to central metal.

Ex: Oxalate ion ( )22 4C O- , Glycinate ion ( )2 2NH CH COO-&&

3) Polydentate ligands: Donate more than 2 pair of electrons (tri, tetra, penta, hexa etc.) Ex: Diethylenetriamine tridendate, Triethylenetetramine tetradentate Another ways of classification of ligands is based on Donor and Acceptor properties of ligands.

1) Ligands with one or more lone pair of electrons. Ex: 2 3H O, NH

2) Ligands without a lone pair of electrons but with p bonding electrons. Ex: 2 2C H ,CO, NO


3. Give the coordination number of the metal in the following compounds.

i) ( )3 44Cu NH SOé ùë û ii) ( )3 36

Co NH Clé ùë û iii) ( )( )4 6K Fe CN

A: i) ( )3 44Cu NH SOé ùë û in this coordination number of copper is 4, because it is surrounded by 34NH

molecules.

ii) ( )3 36Co NH Clé ùë û in this coordination number of cobalt is 6, because each cobalt is surrounded by ‘6’

3NH molecules.

iii) ( )( )4 6K Fe CN in this coordination number of Iron is 6, because central atom Fe is surrounded by 6

cynide groups VSAQ

1. 4 2CuSO .5H O has a pale blue colour. While 4 2ZnSO ,7H O is white. Explain the difference.

A: In 4 2CuSO .5H O , 2Cu+ ion has one unpaired electron in d-orbitals hence it is coloured. It absorbs red

colour radiation and transmits blue colour radiation.

In 4 2ZnSO ,7H O , 2Zn+ ion has completely filled d10 configuration, hence it is colourless.

2. Ferrous salts are more unstable compared to Ferric salts. Explain interms of their configuration.

A: Ferrous salts are unstable, because in 2Fe+ , the d-orbitals are partially filled ( )6 03d 4s . Where as 3Fe+ ion

has 5d configuration (Half filled) hence it is stable.

3. 4CuSO is paramagnetic while 4ZnSO is diamagnetic explain interms of electronic configuration.

A: In 4CuSO , ( )2 9Cu 3d+ ion has unpaired electron so it is paramagnetic in nature. Where as in 4ZnSO ,

2Zn+ ion has no unpaired electron ( )10d hence it is diamagnetic.

4. What kind of magnetic property do Fe, Co and Ni show? A: Fe, Co and Ni exhibits ferromagnetism.

5. Give composition of Nichrome. A: Nichrome contains 60% Ni, 25% Fe and 15% Cr.

6. What elements are present in Brass? Write its composition. A: Brass contains Cu and Zn metals. Its composition is 60-80% Cu and 20-40% Zn.

7. What is an alloy? Give the name and composition of an alloy? A: An intimate mixture having physical properties similar to that of the metal, formed by a metal with other metals

(or) metalloids (or) some times a non-metal, is called as an alloy. Ex: Brass is alloy which contains 60-80% Cu and 20-40% Zn.

8. Explain EAN with suitable examples. A: Sum of number of electrons donated by all ligands and those present on central atom or ion in a complex is

called EAN (Effective Atomic Number). General EAN is equal to nearest noble gas configuration for extra stability of metal complexes.

EAN of ( )( )3 3 45Co NO NH SOé ùë û Pentamminenitro Cobaltate (II) Sulphate.

EAN = 27 – 3 = 24 + (6 ´ 2 electrons from ligands) = 24 + 12 = 36

EAN of 3Co+ = 36 which is the atomic number of Krypton (Noble gas). Thus the complex is stable.


ORGANIC CHEMISTRY

ETHYL CHLORIDE Q.1) How is Ethyl Chloride obtained from ethyl alcohol by (a) Grove’s process (b) the action of 3PCl

and 5PCl (c) the action of Thionyl chloride

Ans: (a) Grove’s Process:- Ethyl alcohol reacts with HCl in presence of anhydrous 2ZnCl to form ethyl chloride.

2ZnCl2 5 2 2 2C H OH HCl C H Cl H O+ ¾¾¾® +

(b) Action of 3PCl or 5PCl on ethyl alcohol: -

3PCl and 5PCl react with 2 5C H OH and form ethyl chloride.

2 5 3 2 5 3 33C H OH PCl 3C H Cl H PO+ ® +

2 5 5 2 5 3C H OH PCl C H Cl HCl POCl+ ® + +

(c) Action of Thionyl Chloride on ethyl Alcohol:- Thionly chloride reacts with ethyl alcohol to form ethyl chloride. 2 5 2 2 5 2C H OH SOCl C H Cl SO HCl+ ® + +

Q.2) How is ethyl alcohol prepared from (a) Ethylene (b) Ethane? Ans: (a) From Ethylene:- Ethylene reacts with HCl in the presence of anhydrous 3AlCl to give ethyl

chloride.

3AlCl2 2 3 2Anhydrous

H C CH HCl H C CH Cl= = ¾¾¾¾¾® - -

(b) From ethane:- Ethane reacts with chlorine at ordinary temperature in the presence of light to give ethyl chloride.

h2 6 2 2 5400 C

Class ConcentrationC H Cl C H Cl HCln

°+ ¾¾¾® +

Q.3) What happens when ethyl chloride is treated with (a) Alcoholic KOH (b) Moist silver oxide (c) Potassium cyanide (d) Silver cyanide

Ans: (a) Alcoholic KOH reacts with ethyl chloride to give ethyl alcohol. 3 2 3 2CH CH Cl KOH H CCH OH KCl+ ® +

(b) Moist silver oxide reacts with ethyl chloride to give ethyl alcohol. 2 2Ag O H O 2AgOH+ ®

2 5 2 5C H Cl AgOH C H OH AgCl+ ® +

(c) Potassium cyanide reacts with ethyl chloride to give ethyl cyanide as the major product.

2 5 2 5C H Cl KCN C H CN KClD+ ¾¾® +

(d) Silver cyanide reacts with ethyl chloride to give ethyl isocyanide as major product.

2 5 3 2C H Cl AgCN H C CH NC AgClD+ ¾¾® - - +

Q.4) How does ethyl chloride reacts with (a) potassium or sodium nitrite (b) silver nitrite? Ans: (a) Ethyl chloride reacts with potassium nitrite to give nitro ethane as major product.

Dimethyl2 5 2 3 2formaldehyde

Ethyl nitriteC H Cl KNO H C H C O N O KCl+ ¾¾¾¾¾® - - - = +

(b) Ethyl chloride reacts with silver nitrite ( )2AgNO solution to give ethyl nitrite as the major product.

2 5 2 5 2 2Nitro ethane

C H Cl AgNO H C NO AgCl+ ® - +


Q.5) How does ethyl chloride reacts with ammonia? Write the chemical equations. Ans: Ethyl chloride reacts with alcoholic ammonia under pressure to form a mixture of primary,

secondary, tertiary amines and quaternary ammonium salt. 2 5 3 2 5 2C H Cl NH C H NH HCl+ ® +

( )2 5 2 2 5 2 5 2C H NH C H Cl C H NH HCl+ ® +

( ) ( )2 5 2 5 2 53 3C H NH C H Cl C H N HCl+ ® +

( ) ( )2 5 2 5 2 53 4Quaternary ammonium salt

C H N C H Cl C H N Cl+ -+ ®

Q.6) Explain Williamson’s synthesis with chemical equations. Ans: Ethyl chloride reacts with sodium ethoxide to form diethyl ether. This is called, Williamson’s

synthesis of ether. 2 5 2 5 2 5 2 5

Sodiume ethoxide DiethyletherC H Cl NaOC H C H O C H NaCl+ ® - - +

Q.7) How does ethyl chloride reacts with the following. Explain with chemical equations. (a) Benzene / 3AlCl (b) Silver Acetate (c) Mg / Dryether (d) NaBr or KI

+ C2H5Cl AlCl3

C2H5

Ans: (a) Ethyl chloride reacts benzene in presence of 3AlCl catalyst to give ethyl benzene.

(b) Ethyl chloride reacts with silver acetate in alcoholic solution to form ethyl acetate.

3 2 5 3 2 5Ethyl acetate

CH COOAg C H Cl H CCOOC H AgCl+ ® +

(c) Ethyl chloride reacts with magnesium metal in presence of dry ether to give ethyl magnesium chloride. This is called as Grignard’s reagent.

Dry ether2 5 2 5

Ethyl magnesium chlorideC H Cl Mg C H MgCl+ ¾¾¾¾®

(d) Ethyl chloride reacts with NaBr and KI to give ethyl bromide and ethyl iodide respectively. 2 5 2 5C H Cl NaBr C H Br NaCl+ ® +

2 5 2 5C H Cl KI C H I KCl+ ® +

Q.8) Explain the reduction of ethyl chloride and what are the reagents used in this reaction. Ans: Ethyl chloride is reduced to ethane with (i) Zn / HCl (ii) 4LiAlH (iii) 2H /Ni or Pd

4 2

Zn/HCl2 5 2 6LiAlH Or H /Ni

C H Cl C H¾¾¾¾¾¾¾®

Q.9) Explain wurtz reaction. Ans: Ethyl chloride reacts with sodium metal in presence of dryether to form butane. This is called

Wurtz reaction

Dry5 2 2 5 5 2 2 5Ether

n bu tan eH C Cl 2Na Cl C H H C C H 2NaCl

-- + + - ¾¾¾® - +

Q.10) What are the used of ethylchloride? Ans: 1) 2 5C H Cl is a refrigerant

2) Ethylating agent 3) Preparation of Grignard’s reagent. 4) It is a local anaesthesia


CHLORO FORM

Q.1) How is chloro form prepared from (i) ethyl alcohol (ii) Acetone

Ans: Heating ethyl alcohol with bleaching powder and water gives chloro form.

( )+ ® +2 2 22CaOCl H O Ca OH Cl

+ ® +3 2 2 3Acetaldehyde

CH CH OH Cl H CCHO 2HCl

+ ® +3 2 3Chloral

H CCHO 3Cl Cl CCHO 3HCl

( ) ( )+ ® +3 32 2Chloroform Calcium formate

2Cl CCHO Ca OH 2CHCl HCOO Ca

With Acetone:

( )+ ® +2 2 22CaOCl H O Ca OH Cl

+ ® +3 3 2 3 3Trichloro acetone

H CCOCH 3Cl Cl CCOCH 3HCl

( ) ( )+ ® +3 3 3 32 2Calcium Acetate

2Cl CCOCH Ca OH 2CHCl H CCOO Ca

Q.2) How is chloroform prepared from (a) carbon tetra chloride and (b) from chloral hydrate.

Ans: (a) Chloroform is prepared on large scale by the reaction of 4CCl with iron filings and water.

D+ ¾¾¾¾® +2Fe,H O4 3 3CCl 2CH HCCl HCl

(b) Chloral hydrate treated with sodium hydroxide to give pure chloroform

( ) ( )D+ ¾¾® + +3 3 2aq2 Sodium formateChloral hydrate

Cl CCH OH NaOH HCCl HCOONa H O

Q.3) Explain oxidation of chloroform in presence of air and light.

Ans: In the presence of air and light, chloroform is slowly oxidised to phosgene a highly poisonous

gas.

( )

+ ¾¾® +hν3 2 2

Carbonyl chloridePhosgene

1HCCl O COCl HCl

2

Anaeshetic chloroform has to be pure. Therefore, chloroform is kept in well stoppered dark

brown or blue bottles by filling them to the brim.

About 1% of ethyl alcohol is also added which is expected to retard the oxidation of chloroform

and also to convert phosgene to harm less ethyl carbonate.

Q.4) How does chloroform reacts with (a) 3HNO (b) DAg / (c) Aq.KOH (d) Acetone

Ans: (a) Chloroform reacts with vapours of 3HNO to give chloropicrin a poisonous liquid.

+ ® +3 2 3 2 2HCCl HONO CCl NO H O

(b) Chloroform is heated with silver to form acetylene


D+ + ¾¾® º +3 3Acetylene

HCCl 6Ag Cl CH HC CH 6AgCl

(c) Aqueous KOH reacts with chloroform to give formicacid.

( ) -+ ® ¾¾¾® ¾¾¾® +2H O KOH3 23 Potassium formateUnstable

HCCl 3KOH CH OH HCOOH HCOOK H O

(d) Acetone condense with chloroform in presence of KOH to give chloroetone.

C OH3C

H3C + HCCl3 C

H3C

H3C

OH

CCl3

Chloretone

Chloretone is a hypnotic drug.

Q.5) Explain (a) Reimer-Tiemann reaction (b) Carbylamine or isocyanide test

Ans: (a) Chloroform reacts with phenol in presence of sodium hydroxide at °65 C to give

orthohydroxybenzaldehyde (Salicylaldehyde)

°+ + ¾¾¾® + +65 C6 5 3 2C H OH CHCl 3NaOH 3NaCl 2H O

OH

C

O

H

(b) Primary amines reacts with chloroform in presence of alcoholic KOH to give phenyl iso

cyanide with offensive odour.

- + + ¾¾¾® + +Warm6 5 2 3 6 5 2

Phenyl isocyanideC H NH HCCl 3KOH C H NC 3KCl 3H O

Q.6) What are the uses of chloroform?

Ans: (1) Solvent for fats, oils and waxes

(2) Detection of primary amines

(3) Preparation of chloropicrin, chloretone

(4) In the preservation o of anatomical species.


CHLORO BENZENE 1. Write any two preparations of Chloro Benzene? Ans: 1) Electrophilic substitution reaction:-

Chlorobenzene is formed by electrophilic substitution when benzene reacts with 2Cl in presence

of Lewis catalyst Fe or Iron (III) chloride ( )3FeCl

+ Cl2

Fe/Dark

Cl

+ HCl

2) Sandmeyer’s rection:- Aniline is dissolved in cold aqueous HCl and treated with sodium nitrite to give diazonium salt. NH2

NaNO2 + HCl

N2Cl

+ HCl 0 to 50C

Benzene dia zonium chloride Cl

Cu2Cl2

N2Cl

+ N2

This solution is treated with cuprous chloride to give chlorobenzene. 2. Explain electrophilic substitutions reactions of chlorobenzene with a) Anhydrous 3 2AlCl /Cl b) 2 4Conc.H SO c) 1 : 1 reaction 3Conc.HNO and 2 4H SO

d) Anhydrous 3 3AlCl /CH Cl e) Anhydrous - -3 3

||O

AlCl /H C C Cl Ans: a) Chorobenzene reacts with chlorine to give p-dichlorobenzene as major product and O-dichloro

benzene as minor product

Cl

Cl2/Anhydrous AlCl3

Cl

D Cl

+

Cl Cl

b) Chlorobenzene reacts with Conc. 2 4H SO to give chlorobenzene sulphonic acid as major product

and 2-chlorobenzene sulphonic acid as minor product.

Cl

Conc. H2SO4

Cl

D SO3H

+

Cl SO3H

4-chloro benzene sulphonic acid

2-chloro benzene sulphonic acid

c) Chlorobenzene reacts with 1 : 1 ratio 3HNO and 2 4H SO to give 1-chloro-4-nitro benzene as

major product and 1-chloro -2- nitro benzene as minor product. d) Chlorobenzene reacts with 3CH Cl in presence of Anhydrous 3AlCl to give 1-chloro-4-methyl

benzene as major product and 1-chloro-2-methylbenzene as minor product.

Cl

Anhydrous AlCl3

Cl

CH3

+

Cl

1-chloro- 4-methyl benzene

+ CH3Cl

major 1-chloro- 2-methyl benzene

minor

CH3

e) Chlorobenzene reacts with Acetyl chloride æ öç ÷- -ç ÷è ø

3

||O

CH C Cl in presence of Anhydrous 3AlCl to

give 4-chloro acetophenone as major product and 2-chloro acetophenone as minor product.

Cl

Anhydrous AlCl3

Cl

+

Cl

+ H3CCOCl

major minor O = C – CH3

C – CH3

O


3. Explain Wurtz-Fitting reaction and Fitting reaction with chlorobenzene. Ans: a) Chloro benzene reacts with Alkylhalide in presence dry ether and sodium metal to give alkyl benzene.

This is called as Wurtz-Fitting reaction.

R

Dry ether

Cl

+ + 2Na + RX NaX + NaCl

b) Chloro benzene reacts with sodium metal in presence of dry ether to give diphenyl. This is called as fitting reaction.

Dry ether

Cl

+ + 2Na 2NaCl

Diphenyl

2

4. Write the structure of DDT.

Cl

CCl3 Cl C

H

ALCOHOLS

1. How is ethyl alcohol prepared from (a) Ethyl halide (b) Ethyl acetate (c) Acetaldehyde (d) Ethylene (e) Grignard’s reagent and formaldehyde. Ans:

a) Ethyl halide on Hydrolysis with aqueous solution of NaOH or AgOH or 2Ag O in boiling

water forms 2 5C H OH

+ ® +3 2 3 2H CCH Cl NaOH CH CH OH NaCl

D+ ¾¾® +3 2 3 2H CCH Cl AgOH CH CH OH AgCl

b) From Ethyl Acetate: Ethyl alcohol is formed by the hydrolysis of ethyl acetate with aqueous alkali. ( )+ ¾¾® +3 2 5 3 2 5aqH CCOOC H KOH CH COOK C H OH

c) From Acetaldehyde: Acetaldehyde reduced by 4LiAlH or 4NaBH in ether gives Ethyl alcohol

¾¾¾¾¾®4

4

LiAlH /Ether3 3 2or NaBHH CCHO CH CH OH

d) From ethylene: Ethylene treated with 2 4H SO to give ethyl hydrogen sulphate. Then Ethyl hydrogen sulphate treated with water to give ethyl alcohol.

°- °= + ¾¾¾¾® -4

|75 80 C

2 2 2 4 3 2

SO H

H C CH H SO H C CH

+ ® +3 2 2 3 2 2 4|

4SO H

H C CH H O H CCH OH H SO

e) Ethyl alcohol is obtained by the action of 3H CMgX on formaldehyde followed by the hydrolysis.

C = O + H3CMgBr H

H C H

H

OMgBr

CH3

H3C – CH2OH + Mg OHBr

2. How is ethyl alcohol obtained by fermentation process of molasses. Ans: Molasses is the mother liquor left behind after the crystallization of sugar from sugarcane juice.

This dark syrupy liquid is diluted with water to have the percentage of sugar in the solution to about 10%. To the diluted molasses solution 2 4H SO is added to maintain the pH of the solution at 4. Ammonium sulphate and ammonium phosphate are added as food for the yeast. Maltose solution or diluted molasses solutions is cooled to °30 C and fermented with yeast for 24 – 72 hours. If the maltose solution is taken:

+ ¾¾¾¾® +Maltose12 22 2 6 12 6 6 12 6Enzyme

Maltose Glu cose FructoseC H OH H O C H O C H O

If sucrose solution is taken

+ ¾¾¾¾® +Invertage12 22 2 6 12 6 6 12 6Enzyme

Glu cose FructoseC H OH H O C H O C H O


¾¾¾¾® +Zymase6 12 6 2 5 2Cenzyme

Glu cose/FructoseC H O 2C H OH 2CO

Generally 95% aqueous alcohol is called rectified spirit. 3. How is ethyl alcohol reacts with the following: a) Hydrogen halide b) Phosphorous halide c) Thionyl chloride d) °Cu /300 C e) 2 2 7 2 4K Cr O /H SO f) 2Cl

g) 2CaOCl h) 2I /KOH

Ans: a) Ethyl alcohol reacts with hydrogen halide to form ethyl halide. The reactivity order is HI > HBr > HCl > HF

HCl reacts with Alcohol in the presence of anhydrous 2ZnCl catalyst to give ethyl chloride. Conc. HCl +

Anhydrous ZnCl2 is called as Lucas reagent. + ¾¾¾® +2ZnCl

2 5 2 5 2C H OH HCl C H Cl H O

b) Ethyl Alcohol reacts with PCl3 or PCl5 to give ethyl chloride. + ¾¾® +2 5 3 2 5 3 33C H OH PCl 3C H Cl H PO

( )+ ¾¾® +2 5 5 2 5 3C H OH PCl C H Cl POCl Phosphorous Chloride

c) Thionyl chloride ( )2SOCl reacts with ethyl alcohol to give ethyl chloride.

+ ¾¾® + +2 5 2 2 5 2C H OH SOCl C H Cl SO 2HCl

d) Dehydrogenation: When ethyl alcohol vapours are passed over freshly reduced copper at °300 C acetaldehyde is formed.

Cu2 5 3 2300 C

||O

C H OH CH C H H°¾¾¾® - - +

e) Oxidation of ethyl alcohol with acidified 4KMnO or 2 2 7K Cr O forms Acetic Acid.

+¾¾¾¾® ¾¾¾¾®2 2 7 2 2 7

2 4 2 4

K Cr O K Cr O3 2 3 3H SO H SOCH CH OH CH CHO H CCOOH

f) Action of halogen: Halogens oxidize ethyl alcohol first to acetaldehyde. Then it undergoes halogenation to form chloral.

( )¾¾® ¾¾¾®2 2Cl 3Cl3 2 3 3H CCH OH H CCHO Cl CCHO chloral

g) Reaction with Bleaching powder: 2 5C H OH reacts with 2CaOCl in water to give chloroform.

( )+ ¾¾® +2 2 22CaOCl H O Ca OH Cl

[ ]+ ¾¾® +2 2Cl H O 2HCl O

[ ]+ ¾¾® +3 2 3 2H CCH OH O H CCHO 2H O

+ ¾¾® +3 2 3H CCHO 3Cl CCl CHO 3HCl

( ) ( ) ( )+ ¾¾® +3 32 22CCl CHO Ca OH 2CHCl HCOO Ca Calcium formate

h) Iodoform reaction:- When ethyl alcohol is treated with 2I solution and KOH solution if forms Yellow crystals of 3CHI (Iodoform).

3 2 2 3

||O

H CCH OH I H C C H 2HI+ ® - - +

3 2 3Tri iodo acetaldehyde

|| ||O O

H C C H 3I I C H 3HI- - + ® - - +

3 3

|| ||O O

I C C H KOH CHI H C OK- - + ® + - -

Overall reaction is 2 5 2 3 2C H OH 4I 6KOH CHI HCOOK 5KI 5H O+ + ® + + +

4. How the maltose is prepared from starch. Ans: Common sources of starch are wheat, barely, potato etc. The grain is mashed with hot water. It is then

heated with freshly germinated barley (malt) at 50 C° for 1 hour. Malt contains the enzyme diastase. Diastase converts starch into the sugar maltose by hydrolysis.


( ) Diastase2 6 10 5 12 22 11n

MaltoseStarch

nH O C H O C H O

2+ ¾¾¾¾®

5. How is methyl alcohol prepared. Ans: Water gas is heated at 573-673 K at 200 to 300 atm in presence of 2 3ZnO Cr O- to give methyl alcohol.

2 3ZnO Cr O2 3200 300 atm

573 673K

CO 2H CH OH---

+ ¾¾¾¾¾®

6. What is the composition of Azeotropic mixture of 2 5C H OH with 2H O .

Ans: A mixture of 95.6% ethyl alcohol and 4.4% water forms constant boiling point mixture, known as Azeotropic mixture.

7. Explain the reaction of ethyl alcohol with (a) Sodium metal (b) Esterification (c) Grignard ragent

Ans: (a) Soldium metal reacts with 2 5C H OH to give sodium ethoxide.

2 5 2 5 22C H OH 2Na 2C H ONa H+ ® +

(b) Esterification:- Ethyl alcohol undergoes acid catalysed condensation with a carboxylic acid to form an ester. The reaction is called Fisher esterification.

2H O3 2 5 3 2 5 2

Ethyl acetateH CCOOH C H OH H C COOC H H O

Å

+ - +ˆˆˆ †̂‡ˆˆˆ̂

Ethyl alcohol reacts with acid chlorides such as acetyl chloride or acid an hydride such as acetic anhydride to form the ester such as ethyl acetate.

2 5 3 5 2 3Ethyl acetate

|| ||O O

C H OH Cl C CH H C O C CH HCl+ - - ® - - - +

(c) Ethyl alcohol reacts with alkyl magnesium halide to form alkane

C2H5OH + H3CMgI CH4 + Mg OC2HI

8. How to identify primary, secondary and tertiary alcohols by using Lucas reagent. Ans: Lucas reagent is Conc. HCl+ Anhydrous 2ZnCl

Lucas reagent reacts with 3° alcohol to give turbidity within 30 seconds.

2

| |

| |

R R

R R

R C OH HCl R C Cl H O- - + ® - - +

Lucas reagent reacts with 2° alcohol to give turbidity after 5 minutes.

2

| |

| |

R R

R R

R C OH HCl R C Cl H O- - + ® - - +

Lucas reagent does not react with 1° alcohols. 9. How is 1 ,2 and 3° ° ° alcohols are identified by using Victarmeyer’s test.

Ans: The alcohol compound is treated with red phosphorous and 2I , and the product is treated with Silver nitrite

and then with nitrous acid ( )2 2 4NaNO H SO+ and finally made alkaline.

Primary alcohols give red coloration, Secondary alcohols give blue colouration. Tertiary alcohols give no reaction. 10. Explain the action of conc. 2 4H SO on ethyl alcohol at different temperatures.

Ans: (i) 100 110 C5 2 2 4 2 5 4 2

Ethyl hydrogen sulphateH C OH H SO C H HSO H O- °+ ¾¾¾¾® +

(ii) 2 4Conc. H SO2 5 2 5 5 2 2 5 2140 C

EthERC H OH C H OH H C O C H H O°+ ¾¾¾¾¾® - - +

(iii) 2 3Al O2 5 2 5 5 2 2 5 2260 CC H OH C H OH H C O C H H O°+ ¾¾¾® - - +

(iv) 2 4170 C; Conc.H SO2 5 2 4 2C H OH C H H O°¾¾¾¾¾¾¾® +

2 3Al O ,350 C2 5 2 4 2C H OH C H H O°¾¾¾¾¾® +


PHENOLS 1. How are Phenols prepared form (a) Haloarenes (b) diazonium salts(c) Benzene sulphonic acid (d)

Cumene Ans: (a) From haloarenes:-

Halobenzene is fused with NaOH at 320 atm and 350 C° . It gives sodium Phenoxide and then it is treated with HCl to give phenol. ONa

350oC, 3220 atm

X OH

+ NaOH HCl

+ NaCl

(b) From diazonium salts: An aromatic primary amine reacts with nitrous acid ( )2NaNO HCl+ at 0 5 C- ° to from

diazoinium chloride. Diazonium salts on warming with water or dilute acids hydrolyses to phenol. N2Cl

NaNO2 + HCl

OH H2O

+ N2 + HCl

NH2

0 – 5oC warm (c) From benzene sulphonic acid:-

Benzene is sulphonated with oleum to get benzene sulphonic acid which on heating with molten NaOH forms sodium phenoxide. Acidification of sodium phenoxide with acid like HCl forms phenol

H2SO4, SO3

O Na

NaO + NaCl

NH2 SO3H HCl

OH

(d) From Cumene:

Phenol is manufactured from the hydrocarbon known as cumene i.e., isopropyl benzene. Cumene is oxidized in presence of air to cumene hydroperoxide which on treating with dilute acid give phenol and acetone.

O2

OH H+

+ H3C – C – CH3

CH3

H2O Cumene

CH H3C H3C – C – O – O – H

CH3

O

2. Write the resonance structures of phenol and explain acidic character of phenol.

O – H O – H O – H O – H O – H

Ans: When phenol gives HÅ it gives phenoxide ion. Phenoxide ion is resonance statilised.

O – O O O O

+ H+

O

3. Explain the esterification reaction of phenols. Ans: Phenols react with carboxylic acids and their derivatives like acid chlorides and acid anhydrides to from

esters.

O – C – R

+ H2O

OH

+ RCOOH ˆ †̂‡ˆ̂H+

O

O – C – R

+ HCl

OH

+ R – C - Cl

O

O

4. How is aspirin prepared? Write the equation.

Ans: Introduction of 3

||O

C CH- - on phenolic oxygen is called acetylation. Salicylic acid on acetylation gives acetyl salicylic acid or aspirin.


O – C – CH3

OH + (CH3CO)2O

O

COOH

Salicylic acid

H+

Acetyl salicylic acid (Aspirin)

OH

5. Explain the nitration and Halogenation of phenol. Ans: -OH graph on benzene in phenol is a ring activating and ortho para directing group. a) Nitration: Phenol reacts with dilute 3HNO to give ortho nitrophenol and p-nitrophenol.

OHdil.HNO3

OH NO2

P-nitro phenol

+

OH

NO2 O-nitro phenol

With Conc. 3HNO it gives 2, 4, 6 trinitro phenol known as picric acid.

OH

NO2

NO2 O2N

OH

Conc. H2SO4

OH

SO3H

SO3H

b) Halogenation:

Phenol reacts with 2Br in presence of 2CS to give ortho-Bromophenol and para bromo phenol.

OHBr2/CS2

OH Br

+

OH

Br

273K

With 2Br it give 2, 4, 6- tri bromo phenol.

OH

+ 3Br2

OH BrBr

Br

6. Explain Reimer-tiemann reaction.

Ans: Phenol on treating with chloro form in the presence of NaOH gives ||O

C H- - group ortho position.

ONa

CHCl3 Aq. NaOH

OH CHCl2 NaOH

ONa CHO H+

OH CHO

Salicyladehyde 7. Explain Kolbe’s reaction: Ans: Phenol reacts with NaOH to give sodium phenoxide then it is treated with 2CO and an acid to give 2-

hydroxy benzoic acid.

Phenol

OH ONa

Sodium phenoxide

(i) CO2 OH

COOHNaOH (ii) H+

major

8. Explain Fries rearrangement?

OH

Acetic anhydride

H3C – C

H3C – C O +

Anhydrous AlCl3

O

O

O – C – CH3

O

AlCl3

OH C – CH3

O

+

O – C – CH3

OH

O-hydroxy acetophenone P-hydroxy

acetophenone


9. Explain the reaction of phenol with (a) Zn dust and (b) Chromic acid Ans: (a) Phenol reacts with Zn to give Benzene

Phenol

OH

+ Zn + ZnO

Benzene (b) Phenol reacts with chromic acid to give benzoquinone.

Chromic Acid

OH

Benzoquinon

[Na2Cr2O7 + H2SO4]

O

O

ETHERS

1. How is diethyl ether prepared from (i) Ethyl alcohol (ii) ethyl bromide (iii) By Williamson’s synthesis Ans: (i)

(a)From Ethyl alcohol:- Ethyl alcohol is taken in excess is treated with Conc. 2 4H SO at °140 C to give diethyl ether.

°- + - - ¾¾¾¾¾® - - +2 4Conc. H SO5 2 2 5 5 2 2 5 2140 CH C OH H O C H H C O C H H O

(b)Dehydration of ethyl alcohol to diethylether is also carried out by passing the alcohol vapours at °260 C

over aluminum ( )2 3Al O catalyst °¾¾¾® - - +2 3Al O2 5 5 2 2 5 2260 C2C H OH H C O C H H O

(ii) From ethyl bromide:- Ethyl bromide reacts with dry silveroxide to from diethyl ether. + + - ® - - +5 2 2 2 5 5 2 2 5H C Br Ag O Br C H H C O C H 2AgBr

(iii) Williamson’s synthesis: Ethyl halide reacts with sodium or potassium ethoxide to form diethyl ether. + ® - - +2 5 2 5 5 2 2 5C H ONa IC H H C O C H NaI 2. How does ethyl alcohol reacts with (a) Halogen (b) Oxygen (c) Mineral acids (d)

2 4H SO (e) HI (f) Boiling water (g) 5PCl (h) - -3

||O

H C C Cl (i)

( )3 2H CCO O (j) 2 2 7K Cr O (k) °2 3Al O / 360 C (l) + 3CO BF

Ans: (a) Halogenation: Diethyl ether reacts with chlorine or bromine to form halogen substituted ethers.

- - - - ¾¾¾® - - - -| |

2

Cl ClCl

3 2 2 3 3 3DarkH C CH O CH CH H C CH O CH CH

In the presence of sunlight all hydrogens are displaced. - - ¾¾¾¾® - -2Cl

5 2 2 5 5 2 2 5SunlightPerchloro diethyl ether

H C O C H Cl C O C Cl

(b) Oxygen: Diethyl ether reacts with atmospheric oxygen or ozonised oxygen to form peroxide. The peroxide is highly

explosive.

[ ]

- - + ¾¾® - -

&&

5 2 2 5 5 2 2 5Peroxide

:O:

H C O C H O H C O C H

(c) Formation of oxonium salts:- Diethyl ether reacts with strong mineral acids such as HCl, HBr and 2 4H SO stable oxonium slats at low

temperature.

Å- - + ® - - e

&&5 2 2 5 5 2 2 5

H

H C O C H HBr H C O C H .Br

(d) Action of Sulphuric Acid:- Diethyl ether forms oxonium slats with cold sulphuric acid, it ruptures C-O bond with hot sulphuric acid ( )

D- - + ¾¾® +5 2 2 5 2 2 5 2 5 44 Conc.H C O C H H SO C H OH C H HSO

(e) Action of hydrobromic acid or hydro iodic acid - - + ® +2 5 2 5 2 5 2 5C H O C H HI C H I C H OH


O – C2H5 OH

+ HI + C2H5I

(f) Diethyl reacts with Boiling water to give ethyl alcohol. - - + ®2 5 2 5 2 2 5C H O C H H O 2C H OH

(g) Action of 5PCl :

Diethyl ether is treated with hot 5PCl to form ethyl chloride.

- - + ® +5 2 2 5 5 2 5 3H C O C H PCl 2C H Cl POCl (h) Action of acetyl chloride Acetyl chloride reacts with diethyl ether in presence of 3AlCl to form ethyl chloride and ethyl acetate.

- - + ¾¾¾® +3AlCl5 2 2 5 3 2 5 3 2 5

Ethyl AcetateH C O C H H CCOCl C H Cl H CCOOC H

Diethyl ether reacts with acetic anhydride to give ethyl acetate.

( )- - + ¾¾¾® - -2ZnCl5 2 2 5 3 3 2 52

||O

H C O C H H CCO O 2H C C OC H

(j) Action of a strong oxidizing agent like 2 2 7K Cr O :- Diethyl ether oxidized acetaldehyde and finally acetic acid with strong oxidizing agent. - - ¾¾¾¾® ¾¾¾¾®2 2 7 2 2 7K Cr O K Cr O

5 2 2 5 3 3Oxidation OxidationH C O C H 2H CCHO 2H CCOOH

(k) Dehydration:- When diethyl ether vapours are passed over alumina at °360 C ethylene is formed

2 3Al O5 2 2 5 2 2 2360 CH C O C H 2H C CH H O°- - ¾¾¾® = +

(l) Action of carbon monoxide:- In the presence of 3BF at °150 C and 500 atm directly ether reacts with CO to form ethyl propionate.

°- - + ¾¾¾¾® - - -3BF /150 C5 2 2 5 5 2 2 5500 atm

Ethyl propionate

||O

H C O C H CO H C C O C H

3. Explain electrophilic substitutions of alkyl aryl ether. Ans: Alkoxy group is O, P-directing and ring activating. They give ortho and Para substituted products.

Br2 / CH3COOH

OCH3

Br Anisole

OCH3

Anhydrous AlCl3

OCH3

CH3

+ H3CCl

OCH3

+

OCH3

CH3

P-methoxy tolene

O-methoxy tolene

H3CCOCl

OCH3

O = C – CH3

OCH3

+

OCH3

C – CH3

O

H2SO4/HNO3

OCH3NO2

OCH3

+

OCH3

NO2 4. Write the uses of diethyl ether. Ans: (i) It is a refrigerant. A mixture of diethyl ether and dry ice produces - °110 C , very low temperature (ii) As substituent for petrol after mixing with 2 5C H OH under the trade name natalite.


ALDEHYDES & KETONES Choose the correct answer: 1. How are Aldehydes prepared from the following compounds?

(a) Alcohols (b) Calcium salts of carboxylic acids

(c) Carboxylic acids with mono catalyst (d) From alkenes (Wacker process)

(e) From Alkynes (f) Hydrolysis of Alkylidene chlorides

(g) Acid chlorides

Sol: (a) (i)oxidation of ethyl alcohols:

Oxidation of ethyl alcohol with pyridium dichromate (PDC) or pyridium chlorochromate (PCC) in anhydrous

media like dichloro methane gives acetaldehyde

( )2 2 2

PDC or PCC3 2 3CH Cl H O

Ethyl alcohol AcetaldehydeH C CH OH H CCHO-- - ¾¾¾¾¾®

(ii) By Catalytic dehydrogenation of ethyl alcohol:

When ethyl alcohol vapours are passed over Ag or Cu catalyst at 300 C° to give acetaldehyde

Cu or Ag3 2 3Air ||

O

CH CH OH H C C H¾¾¾¾® - -

(b) From calcium salts of carboxylic acids:

Acetaldehyde is prepared by heating a mixture of calcium salts of formic acid and acetic acid.

( ) ( )3 3 32 2AcetaldehydeCalcium formate CalciumAcetate

HCOO Ca CH COO Ca 2CH CHO 2CaCOD+ ¾¾® +

(c) From carboxylic acids:

Acetaldehyde is obtained by passing a mixture of vapours of formic acid and acetic acid over manganous

oxide (MnO) catalyst at 300 C°

MnO, 300 C3 3 2 2HCOOH H CCOOH H CCHO CO H O°+ ¾¾¾¾¾® + +

(d) From Alkenes (Wacker Process)

Acetaldehyde is obtained by passing ethylene through an acidified aqueous solution of palladium chloride and

cupric chloride.

2CuCl2 2 2 2 3H

H C CH PdCl H O H CCHO Pd 2HCl+= + + ¾¾¾® + +

(e) From Alkynes:- Acetaldehyde is obtained by passing acetylene through an aqueous solution of 40%

Sulphuric acid and 1% mercuric sulphate at 60 C°

2 4

4

40% H SO , 60 C2 21% HgSO |

OH

H C CH H O H C CH tautomerism°- º + ¾¾¾¾¾¾® = ƒ 3H CCHO

(f) From the hydrolysis of alkylidene chlorides:

Heating ethylidene chloride with KOH or NaOH give acetaldehyde

( )2

2KOH3 2 3 3H O2

AcetaldehydeH C CHCl H CCH OH H C CHO-- ¾¾¾® ¾¾¾® -

(g) Reduction of acid chlorides:

Reduction of acetyl chloride with hydrogen in presence of palladium catalyst which is supported on barium

sulphate. This reaction is called Rosenmund’s reduction

4

2

Pd BaSO3 3HH CCOCl H CCHO HCl-¾¾¾¾® +


2. How are ketones are prepared by the following compounds.

(a) Oxidation alcohols (b) Catalytic dehydrogenation of 2 5C H OH

(c) Heating calcium acetate (d) From carboxylic acid

(e) From aalkenes (wacker process) (f) From alkynes

(g) From hydrolysis of alkylidere chlorides (h) From nitriles

Ans:

(a) Oxidation of isopropyl alcohol:-

Isopropyl alcohol on oxidation with pyridum dichromate (PDC) or pyridium chloro chromate (PCC) in

anhydrous media like dichloromethane gives acetone.

PCC3 3 3 3Oxidation

AcetoneH CCHOHCH H C CO CH¾¾¾¾® - -

(b) By catalytic dehydrogenation of ethyl alcohol:-

When isopropyl alcohol at 300 C° heated in presence of Cu or Ag catalyst to give acetone.

Cu or Ag3 3 3 3 2AirH CCHOHCH H CCOCH H O¾¾¾¾® +

(c) From salts of carboxylic acid salt:

Acetone is prepared by heating calcium acetate.

( )3 3 3 32H CCOO Ca H CCOCH CaCOD¾¾® +

(d) From carboxylic acids:-

Acetone is obtained by passing the vapours of acetic acid over manganous oxide (MnO) catalyst at 300 C° .

MnO, 300 C3 3 3 2 22H CCOOH H CCOCH CO H O°¾¾¾¾¾® + +

(e) From alkenes (wacker process)

Propylene is passed through an acidified aqueous solution of palladium chloride ( )2PdCl and cupric chloride

( )2CuCl

2CuCl3 2 2 2 3 3H

||O

H C CH CH PdCl H O H C C CH Pd 2HCl+- = + + ¾¾¾® - - + +

(f) From Alkynes:

Acetone is obtained by passing propyne into a solution of 40% 2 4H SO and 1% 4HgSO at 60 C°

2 4

4

tautomerise40% H SO3 2 3 2 3 360 c

1% HgSO| ||OH O

H CC CH H O H C C CH H C C CH°º + ¾¾¾¾® - = - -ˆˆˆˆˆ †̂‡ˆˆˆˆˆ̂

(g) Hydrolysis of isopropylidene chloride:-

( ) 2H O2KOH3 2 3 3 3 3 32

Acetone

||O

H C CCl CH H C C OH CH H C C CH-- - ¾¾¾® - - ¾¾¾® - -

(h) From nitriles:-

1-phenol propanone is prepared from propane nitrite and phenyl magnesium halide.

H3C – CH2 – C N + C6H5MgX

Ether H3C – CH2 – C

NMgBr

C6H5

H3O+ C2H5 – C – C6H5

O

1-Phenol propanone


3. How is Benzaldehyde prepared form (a) Methyl benzene (b) Benzene?

(a) (i) Oxidation of methyl Benzene:- (Etard reaction)

Methyl benzene reacts with chromyl chloride followed by hydrolysis.

CH3

+ CrO2Cl2

toluene

CS2

CH(OCrOHCl2) 2

H3O CHO

(ii) Methyl benzene reacts with 3CrO in acetic anhydride to give Benzaldehyde.

CH3

+ CrO3 + (H3CCO)2O 273 – 283 K

CH(OCOCH3) 2

H3O CHO

(iii) Side chain chlorination followed by hydrolysis

CH3

Cl2/ hn HCCl2

H2O CHO

373 K

(iv) Gatterman-Koch reaction:-

Benzene is treated with CO and HCl in the presence of Anhydrous 3AlCl and 2 2Cu Cl

CO, HCl

CHO

Anhydrous AlCl3, Cu2Cl2

4. How does acetaldehyde reacts with

(a) HCN (b) 3NaHSO (c) 3NH (d) 2H N OH-

(e) 2 2H N NH- (f) 2 6 5H N NHC H- (g) 2 2

||O

H NNH C NH- - (h) 2, 4-DNP

(i) 5PCl (j) RMgX (k) ROH (l) 4 4NaBH or LiAlH

(m) Zn Hg /HCl- (n) 2 2H N NH /KOH- (o) 2 2 7 2 4K Cr O / H SO (p) 2I /NaOH

(q) ( )2 40 C /H SO Polymerisation°

Ans:

(a) Action of HCN: Addition of HCN to acetaldehyde gives acetaldehyde cyanohydrin.

C = O + HCN C

CH3 OH

H CN

H3C

H

Acetaldehyde cyanohydrin

(b) Action of 3NaHSO : Addition of sodium bi sulphate to acetaldehyde gives acetaldehyde sodium bisulphite

salt.

C = O + NaHSO3 C

OH

H SO3

H3C

H

Acetaldehyde sodium bi sulphate

H3C Na+

(c) Action of 3NH :-

3 3 3 2 3Acetaldehyde ammonia Ethanalim ine

|OH

H C CHO NH H C CH NH H C CH NH- + ® - - ® - =

(d) Action of hydroxylamine:- It gives a acetaldeoxime


C = O + H2NOH

OH

H H3C

H Acetaldoxime

H3C – C – N

OH

H

H3C – C = N – OH - H2O

H

(e) Action of hydrazine: ( )2 2H N NH-

C = O + HN2 – NH2 C

H3C NH – NH2

H OH

H3C

H hydrazine C = N – NH2 + H2O

H3C

H hydrazone

(f) Action of phenylhydrazine:-

C = O + H2N – NH – C6H5

H3C

H Phyl hydrazine C = N – NH – C6H5

H3C

H Phenyl hydrazone

- H2O

(g) Action of semi carbazide:-

C = O + H2N – NH – C – NH2 H3C

H Phyl hydrazine

C = N – NHC6H5 H3C

H - H2O

O

C H3C NH – NH – C – NH2

H OH

O

Semi carbozone

(h) With 2,4 DNP

C = O + H2N – NH H3C

H 2,4-dinitro phenyl hydrazine

C H3C NH – NH

H OH NO2

NO2 NO2

NO2

- H2O C = N - NH NO2

NO2 H3C

H 2,4-dinitro phenyl hydrazone

(i) Action of 5PCl : Acetaldehyde reacts with acetaldehyde to give ethylidene chloride.

H3C – C = O + PCl5

H

H3C – C – Cl + POCl3

H

Cl

Ethylidene chloride

(j) Addition of Grignard’s reagent: Acetaldehyde reacts with RMgX to give an adduct and after Acid hydrolysis

to give secondary Alcohol

H3C

H

H+ C = O + RMgX C

R

OMgX H3C

H C

R

OH

20 Alcohol

H3C

H

(k)

R

H C = O C

OH

OR

Acetal

H3C

H

ROH

dry HCl

ROH

H3O+

R

H C

OR

OR

+ H2O

Hemi Acetal

(l) Acetaldehyded reduction:

Ni3 2 3 2|

H

H C C O H H C CH OH- = + ¾¾® - -

4

4

NaBH3 3 2or LiAlH|

H

H C C O H CCH OH- = ¾¾¾¾®

(m) Clemenson reduction:


H3C

H

CH2 + H2O C = O

Ethane

H3C

H

Zn – Hg HCl

(n) Wolff-Kishner reduction:

H3C

H C = O

Ethane

H2N – NH2

-H2O

H3C

H C = N – NH2

KOH

Ethylene glycol H3C – CH2 – H + N2

(o) Oxidation:

2 2 7

2 4

K Cr O3 3H SOAcetaldehyde Acetic acid

H C CHO H CCOOH- ¾¾¾¾®

(p) Iodoform reaction:

3 2 3 2H CCHO 3I 4NaOH HCOONa HCI 3H O 3NaI+ + ® + + +

(q) Polymerisation of acetaldehyde

3H3CCHO

Paradehyde Hypriotic drug

At room temp.

H2SO4 few drops

O O

O H3C – CH CH - CH3

CH

CH3

4H3CCHO

Metaldehyde

At 00 C

H2SO4 few drops

H3C – CH – O – CHCH3 O O

H3C – CH – O – CHCH3

(Solid fuel)

5. How does acetone reacts with

(a) HCN (b) 3NaHSO (c) 3NH (d) 2H N OH-

(e) 2 2H N NH- (f) 2 6 5H N NHC H- (g) 2, 4-DNP (h) 5PCl

(i) RMgX (j) Alcohol addition (k) Chlorination, Bromination

(l) 4 4 2 5NaBH or LiAlH or Na /C H OH (m) Clemensen reduction

(n) Wolff-Kishner reduction (o) 2 2 7 2 4K Cr O /H SO

(p) 2I /NaOH (q) Few drops of 2 4H SO

(a) Addition of HCN:-

OH C = O + HCN

H3C

H C

H3C

H3C CN

Acetaldehyde cyano hydrin

(b) Addition of 3NaHSO :-

OH C = O + NaHSO3

H3C C

H3C

H3C SO3

Acetone sodium bisulphite

H3C Na+

(c) Addition of 3NH


OH C = O + NH3

H3C C

H3C

H3C NH2

Acetone Ammonia

H3C -H2O C = NH

H3C

H3C

Propanonimine

(d) Addition of Hydroxyl amine:-

C = O + H2N – OH

H3C C = N – OH + H2O

H3C

H3C

Acetoxime

H3C

(e) Addition of Hydrazine

C = O + H2N – NH2

H3C C = N – NH2

H3C

H3C

Hydrazone

H3C

(f) Addition of Phenyl hydrazine:-

NH – NH

C = O + H2N – NH – C6H5 H3C

C H3C

H3C OH H3C

-H2O C = N – NHC6H5

H3C

H3C

Phenyl hydrazone

C6H5

(g) Addition of semicarbazide:-

C = O + H2NNHCONH2

H3C C = NHCONH2

H3C

H3C H3C

-H2O

Semi carbazone

(h) Addition of 2, 4 – DNP

C = O + H2N – NH

H3C C = N – NH

H3C

H3C H3C

-H2O

2,4 – dinitrophenyl hydrazone

NO2

NO2

NO2

NO2

(i) Addition of 5PCl

Cl

C = O + PCl5 H3C

C H3C

H3C Cl H

Ethylidene chloride

+ POCl3

(j) Addition of Grignard reagent

OMgX

C = O + RMgX H3C

C H3C

H3C R H3C

H+ / H2O OH

C H3C

R

30 Alcohol

H3C + Mg

X

OH

(k) Addition of Alcohol


OR

C = O H3C

C

H3C

H3C OH H3C

OR C

R

Acetal

R

ROH

HCl

ROH

H+ OR

Hemi acetal

(l) Reduction of Acetone

C = O + 2 [H]

H3C

H3C

Na/C2H5OH H3C – CH – CH3

OH

20 Alcohol

(m) Clemmensen reduction

C = O

H3C

H3C

Zn - Hg H3C – CH2 – CH3 +H2O

Propane HCl

(n) Wolff-Kishner reduction

C = O

H3C

H3C

H2N – NH2 H3C – CH2 – CH3 +N2

Propane -H2O

C = N – NH2 H3C

H3C

KOH / CH2 – OH

CH2 – OH

(o) Oxidation:

2 2 7

2 4

K Cr O3 3 3H SO

Acetic acid

||O

H C C CH CH COOH- - ¾¾¾¾®

(p) Iodo form reduction:

3 3 2 3 3 2

||O

H C C CH 3I 4NaOH H CCOONa CHI 3H O 3NaI- - + + ® + + +

(q) Polymerisation of acetone:

3H3CCOCH3

H2SO4 CH3 CH3

CH3

2, 4, 6 – trimethyl benzene (mesitlylene)

6. Explain the tests for acetaldehyde:-

Ans:

Acetaldehyde responds to Tollen’s test, Fehling’s test and Schiff’s base test. But acetone does not respond to

these tests.

(a) Tollen’s test: - Tollen’s reagent is Ammonical silver nitrate solution. i.e. ( )3 2Ag NH OHé ùë û . It gives silver

mirror with acetaldehyde .

( )3 3 3 4 3 22H CCHO 2 Ag NH OH H CCOONH 2Ag 3NH H Oé ù+ ® + + +ë û

Acetaldehyde oxidised to acetic acid

(b) Fehling’s test:-

Fehling’s solution is the mixture of equal volumes of solution –I ( 4CuSO solution) and solution –II (

Rochellesalt solution with NaOH (sodium potassium tartreate))

23 3 2H CCHO 2Cu 5OH H CCOO Cu O+ - -+ + ® +

A yellow red ppt 2Cu O is formed ( )2 1Cu Cu+ +® + 3 2H O

(c) Benedict’s solution also gives above test. Benedicts solution is alkaline copper (II) sulphate solution, with

citrate solution. 2Cu + converted to yellow or red 12Cu O+


(d) Schiff’s base test:-

Schiff’s base is p-rosaniline hydrochloride pink aqueous solution decolourised with 2SO . Acetaldehyde

restores the pink colour of Schiffs base.

7. Explain the tests for Acetone:

Ans:

1. Legal test:- If aqueous solution of acetone is treated with freshly prepared sodium nitroprusside in NaOH

solution wine red solution is formed and on standing to yellow.

2. Indigotest:- When acetone is treated with ortho nitro Benzaldehyde and KOH solution gives blue colour

indigotin.

8. Write the uses of Acetone:-

1) To store acetylene 2) Manufacture of cordite (A smokeless explosive powder)

3) Preparation of plexiglass (unbreakable glass)

4) Manufacture of synthetic rubber

9. What are the uses of acetaldehyde?

A: 1) Antiseptic inhalent in nose troubles

2) In the preparation of paraldehyde (Hypnotic drug)

3) In the preparation of metaldehyde (a solid fuel)

4) Preparations of chloral, rubber, dyes.

10. Explain Aldol condensation.

Ans: In presence of dilute NaOH, 2 3K CO or HCl acetaldehyde undergoes condensation to form aldol.

OH or H3 3 2

Aldol

|OH

2H CCHO H C CH CH CHO- +

¾¾¾¾® - - -

Aldol on heating gives crotanaldehyde.

23 2 3H O

Cro tan aldehyde

|OH

H C CH CH CHO H C CH CH CHOD-- - - ¾¾¾® - = -

Acetone gives Ketol

( )

23 3 3 2 3 3 3H O

3| |

2

|

3

CHOHBa OH

CH

2H CCOCH H C C CH COCH H C C CHCOCHD-- - - ¾¾¾® - =ˆˆˆˆˆ †̂‡ˆˆˆˆˆ̂

11. Explain cannizzaro reaction.

A: Aldehydes that have no a - hydrogen atom under go this reaction involving disproportionation (self oxidation

and reduction) on treating with strong concentrated alkali.

Eg:

1)

C = O

H Conc. NaOH

formaldehyde

H – C – OH + H – C - OK H

C = O + H

H

H

H

O

methanol Potassium formate

2)


CHO

2 Conc. KOH

CH2OH

+

COOK

Benzaldehyde Benzyl alcohol Sodium benzoate

12. What is the product of the following reaction?

CHO H2SO4 + HNO3

CHO

273 – 283K M – Nitro Benzaldehyde

13. Explain cross aldol condensation?

A: If the aldol condensation is between two different Aldehydes or ketones is called crossed aldol condensation.

Eg:

( )

i ) NaOH3 3 2 3 3 2ii )

( i ) 2 butenal

ii 2 methyl 2 pentenal

|

3| |

CHH H

H C C O H C CH C O H C CH CH CHO H C H C CH C CHOD-

- - -

- = + - - = ¾¾ ¾¾® - = - + - - = -

( )

3 3 2iv pent 2 enal

( iii ) 2 methyl 2 butenal

|

3CH

H C CH C CHO H C CH CH CHCHO- -

- - -

+ - = - + - - =

Four products

iii and iv are crossed Aldol products. i and ii are simple or self aldols


CARBOXYLIC ACIDS

1. How is acetic acid prepared form (a) 1° alcohol (ethyl alcohol) (b) RMgX (Grignard’s reagent) (c) From alkyl cyanide (Alkylnitrites) (d) From 3CH OH and carbon monoxide

(e) From acetaldehyde (f) Bio chemical oxidation of ethyl alcohol (g) From alkyl benzene

A: (a) Oxidation of ethyl alcohol: In presence of 4KMnO or 2 2 7 2 4K Cr O /H SO ethyl alcohol gives acetic acid.

4 4

2 2 7 2 4 2 2 7 2 4

KMnO KMnO3 2 3 3or K Cr O /H SO or K Cr O /H SOCH CH OH H C CHO H CCOOH¾¾¾¾¾¾¾® - ¾¾¾¾¾¾¾®

(b) From Grignard’s reagent: Methyl Magnesium bromide reacts with 2CO followed by acid hydrolysis to give acetic acid

H3CMgBr + C

H+ / H2O H3C – C = O H3C – C – OH + Mg

OH

Br

O OMgBr O

O (c) Methyl chloride with KCN gives methyl cyanide, then on hydrolysis it gives acetic acid.

3 3H CCl KCN CH CN KCl+ ® +

CH3CN + H2O H3C – C = NH

O OH

OH

H3C – C – NH2 H2O O

H3C – C – OH + NH3

(d) From methyl alcohol and carbon monoxide:

Cobalt or3 3Rhodium , , pressureH COH CO H CCOOHD+ ¾¾¾¾¾¾¾®

(e) From Acetaldehyde:- Acetaldehyde is oxidised with air inpresence of manganese acetate commercial quantities of acetic acid are formed.

2air Mn2 3 3

1O H CCHO H CCOOH

2+

+ ¾¾¾¾®

(f) From biochemical oxidation of ethyl alchol: Dilute ethyl alcohol solution is oxidised in air by the bacteria micoderma aceti to give 6 to 10% of acetic acid (Vinegar)

3 2 2 3 2H CCH OH O H CCOOH H O+ ® +

(g) Preparation of Benzoic acid from alkyl benzenes:

CH3 KMnO4, KOH

Heat

COOK H3O+

COOH

Toluene

CH2 – CH2 – CH3 KMnO4, KOH

COOK H3O+

COOH

2. How does acetic acid reacts with

(a) active metals (b) Carbonates and bicarbonates (c) NaOH (d) ethyl alcohol (e) 3 5 2PCl ,PCl SOCl+ (f) 3NH

(g) 2Red P Cl+ (h) 4 10P O (i) Sodalime

(j) Kolbe’s electrolysis (k) Calcium acetate (l) Lithium Aluminium hydride (m) HI / Red P (n) Ni / D

A. (a) Active metals like sodium, potassium reacts with acetic acid to give salt. 3 3 2

Sodium acetate2H CCOOH 2Na 2CH COONa H+ ® +

(b) Action of carbonates and bicarbonates: 3 2 3 3 2 22H CCOOH Na CO 2H CCOONa CO H O+ ® + +

2 3 3 3 3Na CO CH COOH H CCOONa NaHCO+ ® +

3 3 3 2 2NaHCO H CCOOH CH COONa CO H O+ ® + + (c) Reaction with alkali: 3 3 2H CCOOH NaOH H CCOONa H O+ ® +


(d) Action of ethyl alcohol:

3H O3 2 5 3 2 5 2H CCOOH C H OH H CCOOC H H O

Å

+ ¾¾¾® +

(e) Action of 3 5 2PCl ,PCl ,SOCl

3 3 3 3 33H CCOOH PCl 3H CCOCl H PO+ ® +

3 5 3 3H CCOOH PCl H CCOCl HCl POCl+ ® + +

3 5 3 2H CCOOH SOCl H CCOCl SO HCl+ ® + +

(f) Reaction of 3NH

3 3 3 4 3 2 2Ammonium Acetate Acetamide

||O

H CCOOH NH H CCOONH H C C NH H OD+ ® ¾¾® - - +

(g) Action of Red P/ 2Cl :

Red P3 2 2HCl |

Cl

H CCOOH Cl CH COOH-+ ¾¾¾® -

Red P2 2 HCl

|

||

H

ClCl

H C COOH Cl Cl C COOH-- + ¾¾¾® - -

Red P2 HCl

| |

| |

ClH

Cl Cl

Cl C COOH Cl Cl C COOH-- - + ¾¾¾® - -

(h) Action of phosphorous pentoxide:

2H3CCOOH P4O10

-H2O

H3C – C

H3C – C O + H2O

O

O Acetic anhydride

(i) Action of Sodalime: CaO

3 4 2 3H CCOONa NaOH CH Na CO+ ¾¾¾® +

(j) Kolbe’s electrolysis: 3 2 3 3 2 2

At CathodeAnode2H CCOOK 2H O H C CH 2CO 2KOH H+ ® - + + +

(k) (a) Distillation of Calcium Acetate:

( )3 3 3 32Acetone

||O

H CCOO Ca H C C CH CaCOD¾¾® - - +

(b) Mixture of calcium acetate and calcium formate on distillation gives acetaldehyde ( ) ( )3 3 32 2

H CCOO Ca HCOO Ca 2H CCHO 2CaCO+ ® +

(l) Reduction of acetic acid by 4LiAlH

4LiAlH3 3 2H CCOOH H C CH OH¾¾¾® - -

(m) Reduction of Acetate Acid by HI / Red P

HI/Red P3 3 3H CCOOH H C CH¾¾¾¾® -

Ni/3 2 3 3 2H CCOOH 3H H C CH 2H OD+ ¾¾¾® - +


ANILINE

1. How are (i) 6 5C H Cl (ii) 6 5C H OH converted to aniline?

A. (i) Chloro benzene on heating with ammonia at 200 C° under pressure in the presence of 2Cu O forms

aniline.

2Cu O6 5 3 6 5 2 4C H Cl 2NH C H NH NH Cl+ ¾¾¾® +

(ii) Conversion of 2 5C H OH to aniline:

Phenol on treating with 3NH at 300 C° under pressure in presence of 2ZnCl forms aniline.

2ZnCl6 5 3 6 5 2 2300 CC H OH NH C H NH H O°+ ¾¾¾® - +

2. How is nitrobenzene converted to aniline?

A. Zinc and HCl or Sn and HCl, reduces nitrobenzene to aniline.

NO2

+ 6[H] + 2H2O Zn + HCl

or Sn + HCl

NH2

Aniline

3. How does aniline react with the following:

(a) 3 3H COOCH

(b) H3C – C

H3C – C O

O

O

(c) 3CHCl /KOH /Alcohol

(d) 2NaNO /HCl (e) 2Br /0 5 C- ° (f) 6 5

||O

Cl C C H- -

(g) Benzene suphonyl chloride (h) Benzaldehyde / 2 4H SO

(i) 3

||O

H C C Cl- - (j) 2 4H SO

A. (a) Aniline reacts with methyl acetate to give N-acetyl aniline (Acetanilide)

NH2

+ H3C – C – OCH3 + H3COH

O H – N – C – CH3

O

(b) NH2

+ + HCl

H – N – C – CH3

O

H3C – C

H3C – C O

O

O

Acetic an hydride

(c) Carbylamine reaction: -

Aniline or aprimary amine reacts with chloroform and alcoholic potash to form isocyanide which

possess fouling odour. This is a test to detect a primary amine.


Alcohol6 5 2 3 6 5 2

Phenyl isocyanideC H NH HCCl 3KOH C H NC 3KCl 3H O+ + ¾¾¾¾® + +

(d) Diazotisation:

In this reaction aniline reacts aniline reacts with nitrous acid at 0 - 5 C° to form diazonium chloride.

NH2 0 – 50C + NaNO2 + 2HCl

N

+ 2H2O + NaCl

NCl

Benzene dia-zonium chloride

(e) Aniline reacts with bromine at 0 5 C- ° to give 2,4,6-tribromo aniline.

NH2 0 – 50C + 3Br2

NH2

+ 3 HBr

Br

Br Br

(f) Nitration: Aniline is converted to acetanilide then nitration of acetanilide gives p-nitroacetanilide,

after hydrolysis of the compound gives p-nitro aniline.

NH2

CH3COCl

NO2

NHCOCH3 NH2

NO2

HNC – CH3

or

H3C – C

H3C – C O

O

O

O

HNO3 hydrolysis

(g) Aniline reacts with benzene sulphonyl chloride to give N-phenyl benzene sulphanamide.

NH2

+ Cl – SO2C6H5 - HCl

C6H5 – N – SO2C6H5

H

Benzene sulphonyl chloride

N-phenyl Benzene sulphonamide

(h) Aniline reacts with benzaldehide to give benzalidene aniline.

2 4Conc. H SO6 5 2 6 5 6 5 6 5

Benzalidene aniline

|H

C H NH O C C H C H N C C H+ = - ¾¾¾¾¾® - = -

(i) Aniline reacts with acetyl chloride to give acetanilide.

Pyridine6 5 2 3 6 5 3HCl

Acetan ilide

||O

C H NH Cl C CH C H NH COCH-+ - - ¾¾¾¾® - -

(j) Aniline when treated with sulphuric acid firstly it gives anilinium hydrogen sulphate. Then it gives

zwitter ion

NH2

H2SO4 NH3HSO4 NH2

SO3H

NH2

SO3 Anilinium hydrogen sulphate Zwitterion


4. What are the uses of aniline?

A. (a) In the manufacture of benzene dia zonium chloride (This is for the preparation of several organic

compounds)

(b) In making schifits bases

(c) In the preparation of acetanilide, sulphanilic acid, sulpha drugs and azodyes.

5. Discuss the mechanism of nucleophilic substitution reactions 1NS and 2

NS .

A. Nucleophilic substitution reactions take place in two ways depending on the nature of alkyl halide. These are

(i) unimolecular ( )1NS and (ii) bimolecular ( )2

NS reaction mechanism.

(i) Unimolecular reaction ( )1NS : The rate of hydrolysis of alkyl halide depends only on the concentration of

alkyl halide. It does not depend on the concentration of nucleophile ( )OH- . Hence this reaction is called

unimolecular nucelophilic substitution reaction 1NS .

Tert. Alkyl halides follow this mechanism.

Rate µ [tert. Butyl bromide]

Rate = k [tert. Butyl bromide]

Hence the reaction follows first order kinetics.

Mechanism: This mechanism involves two steps.

Step-1: In the first step, ‘C – X’ bond breaks resulting in the formation of carbonium ion.

Slow

3 3

tert. butyl bromide tert. carbonium ion

33 ||

| |

3 3

CHCH

CH CH

H C C Br H C C BrÅ- - - + eˆˆˆ†̂‡ˆˆˆ̂

This step is slow step and hence rate determining.

Step – 2: In the secondstep, the nucleophile attacks the carbonium ion to give the product, alcohol.

Fast

3 2 3 3

3 3 3| | |

| | | |

3 3 3

CH CH CH..

HCH CH CH

H C C H O : H C C O H H C C OH HÅ

Å Å- + ¾¾¾® - - - - - +ˆ †̂‡ˆ̂

ii) bimolecular reaction ( )2NS : The rate of hydrolysis of alkyl halide depends on the concentration of alkyl

halide and also on concentration of nucleophile. Hence this is a second order reaction and biomolecular.

Primary alkyl halides undergoes hydrolysis preferably in this way.

Rate µ [Primary alkyl halide] [Nucleophile]

Mechanism: 2NS reaction mechanism involves only one step. The breaking of ‘C – X’ bond and the making

‘C – OH’ take place simultaneously.

HO + C – Br HO C Br HO – C + Br

alcohol transition state alkyl halide

d d


NITROBENZENE

1. How is Nitro benzene prepared?

A. Nitrobenzene is prepared in the laboratory by the nitration of Benzene with the acid mixture of concentrated

nitric acid sulphuric acid at a temperature below 60 C° .

+ HNO3

Conc. H2SO4 NO2

> 600C + H2O

2. How does Nitrobenzene reacts with

(a) Metal + acid (b) Metal + base (c) Metal + Neutral medium

(d) 4LiAlH (e) Ni /Pt (f) 2 2Cl /Br /Fe

(g) 3 2 4Conc.HNO , Conc. H SO 60 C> °

A. (a) Zinc and hydrochloric acid, tin and hydrochloric acid or stannous chloride and HCl reduces nitrobenzene to

aniline.

+ 6 [H]

Zn + HCl NH2

or SnCl2 / HCl + 2H2O

NO2

(b) Alkaline medium:

Nitrobenzene gets reduced with zinc and alcoholic potash to hydrazobenzene.

+ 10 [H]

Zn / KOH N – N

NO2

2 + 4H2O

H H

Hydrozo benzene

(c) Neutral medium:

Nitrobenzene gets reduced by Zinc and 4NH Cl solution to phenyl hydroxylamine

+ 4 [H]

Zn /NH4Cl

H – N – OH

+ 2H2O

NO2

Phenyl hydroxyl amine

(d) With lithium aluminium hydride:

Nitrobenzene is reduced by 4LiAlH to Azobenzene.

+ 8 [H]

LiAlH4 N – N

NO2

2

Azobenzene

(e) Nitrobenzene is reduced catalytically with hydrogen and finely divided Ni or Pt at room temperature to give

aniline

+ 3H2

Ni/Pt NO2

Aniline

+ 2H2O

NH2

(f) Halogenation :

+ Cl2

Fe NO2

m-chloro nitro benzene

NO2

Cl


+ Br2

Fe NO2

m-bromo nitro benzene

NO2

Br

(g) Nitration above 60 C° :

Conc. HNO3 + Conc. H2SO4

NO2

m-dinitro benzene

NO2

> 600 C NO2

3. What are the uses of Nitrobenzene?

A. (a) Preparation of floor polishes

(b) Cheap perfumes preparation coil of mirbane

(c) Oxidising agent

(d) As a solvent, in making dyes


BIOMOLECULES

Very Short answer Questions

1. What are lipids? Give examples.

A. Lipids are carbon compound (oily and greezy) insoluble in water but soluble in organic solvents like

chloroform, ether and benzene. Ex: Fats, Oils, Waxes etc.

2. How are lipids classified?

A. Lipids are classified into simple lipids (homo lipids). Compound lipids (hetero lipids), derived lipids (compound

obtained from simple and compound lipids)

3. What is the alcohol present in neutral lipids? Give its formula.

A. The alcohol present in neutral lipids is glycerol.

-

-

2

2

CH OH|CH OH Glycerol|CH OH

4. What are waxes? Give one example.

A. Waxes are insect secretions or protective coatings on animal furs and plant leaves.

Ex: Bee’s wax

5. What are compound lipids? Give one example.

A. Lipids which contain glycerol and other groupings other than fatty acids.

Ex: Lecithin, Cephalin.

6. What are derived lipids? Give one example.

A. Derived lipids are hydrolysis products of simple and compound lipids. Ex: Ergosterol, bile acids.

7. Give two biological function of lipids.

A. 1) Lipids are the energy sources 2) Lipids are the food veserves

8. What is cholesterol?

A. It is a white crystalline substance of molecular formula 27 45C H OH found in the tissues of animals,

synthesized by liver.

9. What are hormones and how are they classified?

A. Hormones are molecules of carbon compounds that transfer biological information from one group of cells to

distant tissues or organs. Hormones are classified into 3 groups on the basis of their chemical structures.

They are steroid hormones, protein hormones, amino acid derivatives.


10. Give the structure of any three hormones?

CH3 OH

H

H

H

O , OH

H H

CH3 OH

H

, O

H H

CH3

H

COCH3

CH3

Testosterone Estradiol Progesterone 11. What are plant hormones, give examples.

A. Plant hormones are carbon compounds produced by higher plants. They regulate growth and physiological

functions at a site remote from the place of secretion.

Ex: Auxins, Gibberlines, Cytokinins

12. What are the main functions of sex hormones?

A. Sex hormones are responsible for the development of secondary sexual characteristics such as deep voice,

facial hair, sturdy physical structure in men.

13. What is the structure of insulin and what is its function?

A. Insulin is peptide hormone and has great influence on carbohydrate metabolism.

Structure of Insulin

Leu

Glu

Gln

Gly

Ile

Leu

ValGlu

CysCys

Thr Ser Ile Cys Ser

Asn

TyrGln

LeuGlu

Tyr

Asn

Cys

Gln

1

2

34

56

78 9 10 11 12 13

1415

1617

18

19

20

21

1

2

3

4

5

67

89

1011

12 13 14 15 16 17 18 1920 21

2223

24

2526

27

28

29

30Phe

Val

Asn

HIS

CysGly

SerHIS

Leu ValAln Tyr LeuCys

ValCyn Gly

GluArg

GlyPhe

PheTyr

Thr

Cys

Pro

Thr

SS

S S

S

S

A-chain

B-chain

14. Give two examples of steroid hormones.

A. Ex: Androgen, estrogen.

15. Give two examples of protein hormones.

A. Insulin, Thyroxin


16. What are vitamins? Give one example.

A. Vitamins are low molecular weight compound, their absence in the human body causes deficiency diseases or

disorders.

Ex: Vitamin B-complex

17. How are vitamins classified?

A. Vitamins are classified into two broad groups.

1) Fat soluble vitamins. Ex: Vitamin A, D, E and K

2) Water soluble vitamins. Ex: Vitamin C and B-complex

18. What are water soluble vitamins?

A. Vitamins which are soluble in water are called water soluble. Ex: Vitamin C and B-complex

19. What are fat soluble vitamins?

A. Vitamins which are insoluble in water are called fat soluble vitamins. Ex: Vitamin A, D, E and K

20. *What are the sources for vitamins A, B, C, K?

A. Source of vitamin A – Fish oils, liver, rice

Source of vitamin B – Cereals, milk, egg

Sources of vitamin C – Green leafy vegetables, citrous fruit

Sources of vitamin K – Green leafy vegetables, intestinal bacteria

21. *Give the deficiency diseases caused by A, C, D, E, K

Vitamin A – Night blind ness, redness in eyes (xeropbthalmia)

Vitamin C – Scurvy, delay in wound

Vitamin D – Rickets in children

Vitamin E – Sterility, neurosis of heart muscles

Vitamin K – Blood clotting does not occur


CHEMISTRY IN EVERYDAY LIFE

Very Short answer Questions

1. Define drug. Give two examples.

A. The chemicals of low molecular masses ranging from 100 to 500 U that react with macromolecular targets to

produce a biological response are called drugs.

Ex: Antihistamine, Cardio vascular drug.

2. Define medicine. Give two examples.

A. The drugs that produce biological response therapeutically and that are useful in diagnosis, prevention and

treatment of disease are known as medicines.

Ex: Aspirin, Ibuprofen

3. How do you differentiate between drug and medicine?

A. Drug which produce biological responses and prevent a disease is called a medicine. Drug produces a

biological response by reacting with target. It may or may not prevent a disease.

4. What are Narcotic drugs? Give an example.

A. Narcotic drugs cause depression of central nervous and are strong analgesics.

Ex: Morphine, Codeine

5. What are non-Narcotic drugs? Give an example.

A. The analgesics which have limited use in mild aches, pains like backache and headache.

Ex: - Aspirin, Ibuprofen

6. What are Analgesics? Give two examples.

A. Analgesics reduce or totally abolish pain without causing any damage to consciousness, mental confusion ,

disturbances of nervous system.

Ex: Aspirin, Ibuprofen

7. Define Antipyretics. Give two examples.

A. Drugs used to control fever are called antipyretics.

Ex: Paracetmol, Analgin etc.

8. Define transquilizers. Give two examples.

A. Drugs that cause quieting effect accompanied by relaxation and rest but need not necessarily induce sleep.

Ex: Luminal, Seconal.

9. Define antiseptics. Give example.

A. The chemical compounds that kill or prevent the growth of micro organisms. Ex: Dettol.

10. What are disinfectants?

A. Disinfectants are used for killing or preventing the growth of micro organisms. They are applied to objects like

floors, drainage system etc.

Ex: 1% phenol.

11. Explain antimicrobials. Give examples.

A. Antimicrobials kill or inhibit the growth of organism that cause disease. They increase immunity and

resistance to infection of the body.

Ex:- Lysozyme, Lactic acid.

12. What are antifertility drugs? Give examples.

A. Antifertility drugs are oral contraceptives which are generally steroids.

Ex: Norethindrone, Ethynylestrdiol


13. Define antibiotics. Give example.

A. It is a substance that is produced wholly or partly by chemical synthesis and in low concentrations inhibits the

growth or destroys micro organisms by interfering in their metabolic process.

Ex:- Penicillin, Chloramphenicol.

14. What are antacids? Give examples.

A. Chemical that remove excess acid in the stomach and maintain the pH to normal level are antacids.

Ex: Omeprazole, Lansoprozole.

15. Define antihistamines. Give examples.

A. Antihistamines prevent the interaction of histamine with receptors of stomach wall theses producing less

amount of acid.

Ex: Dimetane, Sardane.

16. What difference do you find between antacid and antihistamine?

A. Antacid remove acidity in stomach while antihistamine lowers acid content in the stomach indirectly by

preventing the interaction of histamine with receptors of stomach wall.

17. What are food preservatives? Give examples.

A. Chemicals which are used to enhance the appeal and preservation of the food.

Ex: BHT, BHA, 2SO etc.

18. Why are synthetic food colours not advisable?

A. Generally dyes are used as food colours. They do not have nutritive value, may be harmful for children with

asthma.

Ex: Tetrazine

19. What are artificial sweetening agents? Give examples.

A. Artificial sweetening agents are chemicals used in place of sugar or sucrose. They decrease the calorific

intake and at same time several time sweeter than sucrose.

Ex: Aspartame – 100 times sweeter to sugar.

20. What are the advantages with artificial sweetening agents?

A. (1) It decrease the calorific intake, and useful for diabetic patient.

(2) Excreted through urine easily.


POLYMERS Q.1) Which of the following polymer is Copolymer? (a) Polythene, (b) Buna – S, (c) Nylon – 6, 6 and

(d) Polyvinyl Chloride

Ans: (b) Buna – S is co-polymer because constituent monomers of it are different (i.e.,) 1, 3-butadiene

and Styrene.

(c) Nylon -6, 6 is copolymer because constituent monomers of it are different (i.e.) hexamethylene

diamine and adipic acid.

Q.2) Give any two examples for semi-synthetic polymers.

Ans: Examples for semi-synthetic polymers are cellulose rayon and cellulose nitrate.

Q.3) Define an elastomer.

Ans: Elastomers: These are the polymers which have rubber like elastic properties. OR

These are the polymers having very weak intermolecular forces of attraction between the polymer

chains. Vulcanished rubber is a very important example of an elastomer.

Q.4) What is cross linking agent used in Vulcanization?

Ans: Sulphur(or) sulphur containing compounds react with rubber molecules to effect cross linking in

vulcanization.

Q.5) Give the complete name for PHBV.

Ans: Poly b -hydroxy butyrate – Co - b - hydroxyl valerate is PHBV.

Q.6) Write one difference between the polymers Buns-S and Buna -N.

Ans: Constituent monomers of Buna-S are 1,3-butadiene and Styrene (Vinyl Benzene).

Constituent monomers of Buna-N are 1,3-butadiene and Acrylonitrile(Vinyl Cyanide).

Q.7) What are the most common type of molecular weights used in the case of polymers?

Ans: Following two are most common type of molecular weights used in polymers.

(i) Number – average molecular weight ( )nM (ii) Weight – average molecular weight ( )wM

Q.8) What do you meant by polydispersity index?

Ans: The ratio of the weight average ( )wM and the number-average ( )nM molecular weights of a

polymer is called “Polydispersity indes (PDI”.

w

n

MPDI

M=

Q.9) What is Vulcanization of rubber?

Ans: The process of heating the raw rubber with sulphur (or) sulphur containing compounds at 373-

415K temperature is called Vulcanization of rubber.


PHYSICAL CHEMISTRY DILUTE SOLUTIONS

STRAIGHT OBJECTIVE TYPE (SINGLE OPTION CORRECT) 1. Explain with suitable examples the terms Molarity, Molality, Normality and Mole fractions. Which method of

expressing concentration is better than the others? Why? A: Molarity: i) The number of moles of solutes present in one litre of solution is known as molarity. ii) It is denoted by M.

iii) Molarity ( ) ( )( )

No. of moles of solute nM

Volume of solution in litres v=

iv) Molarity ( ) Weight of solute 1M

Molecular wt. of solute Volume of solutiion in litres= ´

v) Molarity is expressed in moles/litre Ex: If 40g of NaOH is present in one litre of solution, then it is called 1 M solution or molar solution. Molality: i) The number of moles of solute present in 1000 g of solvent is known as molality. ii) It is denoted by ‘m’

iii) Molality ( ) Weight of solute 1m

Molecular wt. of solute wt. of solvent in kg= ´

v) Molality is expressed in moles / kg of solvent. Ex: If 40 gm of NaOH is present in one kg of solvent, then it is called 1 m solution or molal solution. Normality: i) The number of gram equivalents of solute present in one litre of solution is known as normality. ii) It is denoted by N.

iii) Normality ( ) No. of gram equivalents of soluteN

Volume of solution in lit=

iv) Normality ( ) Weight of solute 1N

Equivalent weight of solute Volume of solution in litres= ´

v) Normality is expressed in gram equivalents/lit. Ex: If 53g of anhydrous sodium carbonate is present in one litre of solution, then it is called 1N or normal solution. Molefraction: i) The ratio of number of moles of one component to the total number of moles of all the components of a solution is known as mole fraction of that component. ii) It is denoted by x.

iii) ( )soluteNo. of moles of solute

xNo. of moles of solute + No. of moles of solvent

=

iv) ( )solventNo. of moles of solvent

xNo. of moles of solute + No. of moles of solvent

=

v) Mole fraction has no units vi) In a binary solution, the sum of mole fractions of solute and solvent is equal to unity. vii) Molality is the convenient method for measuring concentration of solutions, because it is independent of temperature.

2. Calculate the equivalent weights of 2 4 2 3 4H SO , Na CO , KMnO in acid medium and ( )2Ba OH .

A: * Equivalent weight of an acid Molecular weight of the acid

Basicity of the acid=

Basicity:- Number of replaceable Hydrogen atoms.

Eq. Wt. of 2 498

H SO 492

= =

* Equivalent weight of a salt Molecular weight of the salt

Total charge of cation or anion=

22 3 3Na CO 2Na CO+ -+ƒ


Eq. Wt. of ( )2 3 2 3106

Na CO 53 molecular weight of Na CO 1062

= = =

* Equivalent weight of an oxidizing agent Molecular weight of the substance

Number of electrons involved during oxidation=

In acid medium, 7Mn+ ion in 4KMnO is reduced to 2Mn+ ion ( )7 2Mn 5e Mn+ - ++ ®

i.e. Eq. Wt. of 4158

KMnO 31.65

= =

* Equivalent weight of base Molecular weight of base

Acidity of the base=

Acidity:- Number of replaceable OH- ions

( )2Ba OH Mol.Wt 171.4= ; Acidity = 2

Eq. Wt. of ( )2

171.4Ba OH 85.7

2= =

3. Define and explain Raoult’s law. How is it useful in determining the molecular weight of a non-volatile solute? Or

Define Raoult’s law for lowering of vapour pressure. How is the law useful in determining the molecular weight of solvent when a known non-volatile solute is dissolved?

A: Raoult’s law: * The relative lowering of vapour pressure of a dilute solution of a non-volatile solute is equal to the mole fraction of the solute. * RLVP = mole fraction of solute (x)

* Mathematically, the law is represented as 0 s s

0 s 0

p p np n n-

=+

Where 0p is the vapour pressure of pure solvent

sp is the vapour pressure of solution of a non-volatile solute

sn is the number of moles of solute

0n is the number of moles of solvent

And s 0a b

n , nA B

= =

Where a = weight of solute; A = molecular weight of solute; b = weight of solvent; B = molecular weight of solvent

Therefore 0 s s

0 s 0

ap p n A

a bp n nA B

-= =

+ +

* For dilute solutions, the number of moles of solute ( )sn can be neglected in comparision to the number of

moles of solvent ( )0n

0 s

0

p p a Bp A b-

= ´

Molecular weight of solute ( ) ( )0

0 s

p a BA

p p b´ ´

=-

4. What are colligative properties? Explain each of them with necessary examples. * The properties of dilute solutions which depend upon the number of solute particles [molecules (or) ions] not upon their nature are called colligative properties * Colligative properties are used to determine the molecular weight of solutes. * There are four colligative properties. i) Lowering of vapour pressure ii) Elevation of boiling point iii) Depression in freezing point iv) Osmotic pressure i) Lowering of vapour pressure ( PD )


* The decrease in vapour pressure of pure solvent when a non volatile solute dissolved in it is called lowering of vapour pressure. * Ex: Vapour pressure of salt water (solution) is less than vapour pressure of pure water (solvent)

* ( )

0

0 s

p a BA

p p b´ ´

=-

ii) Elevation of boiling point ( bTD )

* The increase in the boiling point of a liquid when a non volatile solute dissolved in it is called elevation of boiling point. * Ex:- The boiling point of salt water (solution) is greater than the boiling point of pure water (solvent)

* ( ) bb b

k a 1000T k m

A b´ ´

D = ´ =´

where bk = molal elevation constant or ebulliscopic constant.

iii) Depression in freezing point ( fTD )

* The decrease in freezing point of a solvent when a nonvolatile solute dissolved in it is called depression in freezing point. * Ex:- The freezing point of slat water (solution) is less than pure water (solvent)

* ( )f f fa 1000

T k m kA b

D = ´ = ´ ´

Where b = wt. of solvent; a = wt. of solute; B = M.wt. of solvent; A = M.wt. of solute; fk = molal depression

constant. iv) Osmotic pressure( p ) * The process of solvent flowing into the solution when the solvent and the solution are separated by a semi permeable membrane is called osmosis * The pressure required to be applied on the solution to prevent osmosis is called osmotic pressure * CSTp = Where C = concentration of the solution, b = wt. of solvent; S = solution constant; a = wt. of solute T = temperature in “K” ; A = M. wt. of solute

5. What is Osmosis? Define Osmotic Pressure? Describe Berkely-Hartley method of determining osmotic pressure.

A: Osmosis: The process of solvent flowing into the solution when the solvent and the solution are separated by a semi permeable membrane is called osmosis.

Osmotic pressure: The pressure required to be applied on the solution to prevent osmosis is called osmotic pressure. CSTp = Where C = concentration of the solution, S = solution constant; T = temperature in “K” Berkely – Hartely Method * In this method, the pressure required is applied on solution to just prevent osmosis. * A porous tube whose both ends are open is taken and copper ferrocyanide is precipitated in the pores of membrane by electrolytic method. * The two ends are now closed with two rubber corks. * Cork thistle funnel is fixed at one cork and a capillary tube is fixed at another cork. This tube is fixed with outer cylindrical tube made from gun metal. Manometer is fixed on the piston to measure the pressure. * Outer cylindrical tube is used to measure the pressure. * Water (or solvent) is added into the porous tube through the funnel * The space between the inner and outer – tubes is filled with the solution * As osmosis starts, the level of the liquid in the capillary decreases due to flow of water from inner tube into outer tube. * Now pressure is applied externally on the piston and the level of liquid in the capillary is brought to initial position. This applied external pressure is called osmatic pressure.


1

7

2 3

4

5 6

Berkely – Hartley method 1. Solvent reservoir 2. Semipermeable membrane 3. Solution

4. Capillary tube 5. Piston 6. Pressure guage 7. Solvent

SAQ 1. Define molarity, molality. A: Molarity (M): * The number of moles of solute present in one litre of solution is called molarity. It is denoted by M. * Molarity is expressed in moles / lit.

* ( ) ( )No. of moles of solute

Molarity MVolume of solution lit

=

Molality (m): * The number of moles of solute in 1000g (1 kg) of solvent is called molality. It is denoted by m. * Molality is expressed in moles / kg of solvent.

* ( ) ( )No. of moles of solute

Molality mweight of solvent kg

=

2. What is meant by mole fraction? Explain. A: Mole fraction: The ratio of number of moles of one component (solute or solvent) to the total number of

moles of all the components present in the solution is called mole fraction of that component (solute or solvent)

In a binary solution, mole fraction of solute ( ) 22

1 2

nn n

c =+

Mole fraction of solvent ( ) 11

1 2

nn n

c =+

Where 1n = no. of moles solvent and 2n = no. of moles of solute

1 2 1c + c =

3. Derive the equation CRTp =

A: * At constant temperature, the osmotic pressure ( )p of a dilute solution is directly proportional to the

concentration. Cp µ ; where C = concentration : moles /lit

The volume (V) of a solution is inversely proportional to concentration: 1

VC

µ

[V is volume of solution in litres containing one mole of the solute]

1V

\p µ …………………….(1)

Or V Kp = K = proportionality constant.

This is similar to Boyle’s law ( )PV K=

* The osmotic pressure ( )p of a solution of constant concentration (C) is directly proportional to the

temperature in the Kelvin scale (T) Tp µ ……………………….(2)


K T¢p =

K¢ = proportional constant This is similar to Charles law. On combining the two equations we get,

1, T

Vp µ p µ

TS.

Vp =

S = proportionality constant V STp =

S is called solution constant

But 1

VC

=

CST or CRT\p = p =

4. What is meant by an ideal solution? A: * A solut5ion which obeys Raoult’s law at all conditions concentration and temperature is called an ideal

solution. * In all ideal solutions the solute and the solvent molecules have similar structure and polarity, the solute-solute interactions and the solvent-solvent interactions are the same.

* mixHD (Enthalpy change of mixing) is zero and

* mixVD (Volume change of mixing) is zero

A few examples of the ideal solutions are: * Benzene and Toluene * n-Hexane and n-Heptane * Carbon tetra chloride and Silicon tetra chloride * Ethyl bromide and ethyl iodide * Chloro benzene and bromo benzene * n-Butyl chloride and n-Butyl bromide etc. 5. How is molar mass related to the elevation in boiling point of a solution? Give the equation.

A: Elevation of boiling point ( )bTD

* The increase in the boiling point of a liquid when a non volatile solute dissolved in it is called elevation of boiling point. * Ex:- The boiling point of salt water (solution) is greater than the boiling point of pure water (solvent)

* ( )b bT K mD = ´

bK a 1000M b´ ´

=´

Where bK is molal elevation constant, a = wt. of solute, W = M.wt. of solvent, M = M.wt. of solute, fK =

Molal depression constant 6. Write the equation relating depression in freezing point of a solution and the molar mass of the solute.

Explain the terms.

A: Depression in freezing point ( )fTD

* The decrease in freezing point of a solvent when a nonvolatile solute dissolved in it is called depression in freezing point. * Ex:- The freezing point of salt water (solution) is less than pure water (solvent)

* ( )f fT K mD = ´

fa 1000

KM b

= ´ ´

Where b = wt. of solvent; a = wt. of solute; W = M.wt. of solvent, M = M.wt. of solute; fK = Molal depression

constant. 7. What is Van’t Hoff’s factor I and how is it related to ' 'a in the case of binary electrolysis. A: Van’t Hoff’s Factor: The ratio of experimental values of colligative property and the calculated value of

colligative property is called Van’t Hoff’s Factor.

Experimental values of colligative property

iCalculated value of colligative property

\ =


Calculated molar mass of solute

iExperimental molar mass of solute

\ =

For solute dissociation (or) Ionization If a solute on dissociation gives ‘n’ ions and ' 'a is degree of ionization 1 n- a aƒ

Total particles 1 n= - a + a ( )1 n 1= + a -

( )1 n 1

i1

+ - a=

i 1n 1

-a =

-

For solute associations

If ‘n’ particles of ‘A’ combines to give n"A "

nnA Aƒ

If ' 'a is degree of dissociation

1na

- a =

Total particles = 1na

- a +

1

1 1n

æ ö= + - aç ÷è ø

11 1

ni1

æ ö+ - aç ÷è ø=

i 11

1n

-a =

-

VSAQ 1. Define molality. A: The number of moles of solute present in 1000g (1kg) of solvent is known as molality. It is denoted by m.

( ) ( )No. of moles of solute

Molality mWt. of solvent kg

=

2. In a binary solution, what is mole fraction? A: Mole fraction is the ratio of number of moles of one component to the total number of moles of all components

in binary solution.

Ex: Mole fraction of solute number of moles of solute

total no. of moles in solution=

3. What is the equivalent weight of a salt?

A: Equivalent weight of a salt formula weight of the salt

total charge on the cation (or) on the anion of the salt=

4. How do you calculate the equivalent weight of an oxidant? A: Equivalent weight of an oxidant calculated by dividing the molecular weight of oxidant by number of electrons

involved in the reaction.

Equivalent weight of oxidant = formula weight of oxidant

electrons gained by the oxidant=

5. Write the relation between the molecular weight and equivalent weight of 3 4H PO ?

A: Equivalent weight of an acid Molecular weight of acid

Basicity=

Phosphoric acid ( )3 4H PO is a tribasic acid


Equivalent weight of 3 4H PO = 983

= 32.66

6. Define vapour pressure. A: Vapour Pressure: The pressure exerted by the vapour molecules on the surface of the liquid, when both the

liquid and the vapour are in equilibrium with each other at a given temperature is called vapour pressure of the liquid.

7. What prediction can you make for the relation between vapour pressure and temperature? A: Vapour pressure increases with increase in temperature decreases with decrease in temperature. The

increase in vapour pressure with temperature is exponential but not linear. 8. Define boiling point of a liquid in terms of its vapour pressure. A: The temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure is

known as boiling point of the liquid. 9. Define osmotic pressure. A: The minimum pressure which should be applied on the solution, so as to prevent the migration of solvent

molecules into the solution through a semi-permeable membrane is called osmotic pressure. 10. What is semi-permeable membrane? Give two examples. A: Semi-permeable membranes are the porous membranes which allow only the solvent molecules to pass

through them. Ex: Cellophane, parachment paper and animal protein membrane. 11. What is meant by abnormal behaviour of electrolytes in colligative properties? A: The solutions which are non-ideal show deviation in colligative properties. Some salts under ionization

(dissociation) in aqueous solutions [ ]2Ex : NaCl in H O where as some substances undergo association

[ ]3 6 6Ex : CH COOH in C H . The behavior is known as abnormal behaviour of electrolytes.

12. Will the molecular weight determined for an ionizing solute be greater or lesser than that calculated by elevation of boiling point method?

A: The molecular weight is less, due to ionization of solute the no. of particles increases and the molecular weight decreases.

Solved Problems 1. 4.0 of NaOH are dissolved in 4 lit. of the solution. Find the molarity of the solution. A: Volume of solution = 4 litres = (v) Weight of NaOH dissolved = 4.0 g = (G) Gram molecular weight (GMW) of NaOH = 40 g

( ) ( )G 1

Molarity M .GMW V in lit

\ =

4 1 1

0.025 M40 4 40

´ = =

\Molarity of the solution = 0.025 M 2. Calculate the molarity of 10.6% (W/V) 2 3Na CO solution.

A: 10.6% (W/V) solution means 10.6 % of 2 3Na CO are dissolved in 100 ml of solution.

\ Weight of 2 3Na CO dissolved = 10.6 g = (G)

Volume of the solution = 100 ml = V (in ml) Gram molecular weight of 2 3Na CO = 106 g = (GMW)

Molarity M = ? = (M) (say) then substitution the data in the formula

( ) ( )

G 1000M .

GMW V in ml= we get,

10.6 1000

M 1106 100

é ù= ´ =ê úë û

\ Molarity of 2 3Na CO solution = 1 M

3. Find the volume of water to be added to 250 ml of 0.05 N 2 3Na CO solution to make it 0.01 N solution.

A: Normality of the given solution = 0.05 N = ( 1N )

Initial volume of the solution = 250 ml = ( 1V ml)


Required normality after dilution = 0.01 N = ( 2N )

Final volume of the solution = ? = ( 2V ml)

According to the formula 1 1 2 2V N V N=

( ) ( )2250 0.05 V 0.01´ = ´

2250 0.5

V 1250 ml0.01

´= =

\ Volume of water to be added to get 0.01 2 3Na CO solution

( ) ( )2 1V V 1250 250 ml 1000ml= - = - =

4. Calculate the vapour pressure of a solution containing 10g of a non volatile solute in 80 g of ethanol at K. Given the molecular weight of the solute as 120; the vapour pressure of alchol is 22.45 mm of Hg at 298 K.

A: Vapour pressure of the pure solvent ( )0P = 22.45 mm

Weight of solute ‘a’ = 10 g

\ Number of moles of solute = s10

n 0.083120

= =

(since molecular weight of solute = M = 120) Weight of the solvent = b = 80 g

Molecular weight of the solvent (alcohol) ( )2 5C H OH = 46.0

Number of moles of solvent = 080

n 1.73946

= =

From Raoult’s law 0 s s

0 0 s

P P nP n n-

=+

( ) ( )s s0 s

0 0

P n1 for a dilute solution n n

P n- = >>Q

sP 0.0831

22.45 1.739é ù\ - =ê úë û

sP 0.083 1.6561

22.45 1.739 1.739é ù- =ê úë û

s1.656

P 22.45 21.378 mm1.739

\ = ´ =

5. The vapour pressure of a solution containing 108.24 g of a non-volatile solute, in 1000g of 2H O at 20 C° is

17.354 mm Hg. The vapour pressure of 2H O at 20 C° =17.54 mm Hg. Find the mol. Wt. of the substance.

A: Vapour pressure of 2H O = 0P =17.54 mm

Vapour pressure of solution sP = 17.354 mm

Lowering of vapour pressure ( )0 sP P= -

( )P 17.54 17.354 0.186 mm= D = - =

Weight of the solute = ‘a’ g = 108.24 g Weight of 2H O 'b ' g 1000g= =

Mol. Wt. of solute = M = ? Mol. Wt. of 2H O = W = 18

From Raoult’s law, 0 s s

0 0 s

P P nP n n-

=+

aa WM .

a b M bM W

é ùæ öç ÷ê ú é ùè øê ú = ê úæ ö æ ö ë ûê ú+ç ÷ ç ÷ê úè ø è øë û

(for dilute solution)


0.186 108.24 1817.54 M 1000

= ´

108.24 18 17.54

M 183.731000 0.186

´ ´= =

´

6. 0.46 g of ethanol dissolved in 1000g of 2H O . What is the molality of the ethanol solution?

A: Given data: Weight of solute = 0.46 gm; Weight of solvent = 1000 gm; Molecular weight of solute = 46

Molality ( ) Weight of solute 1000M

molecular weight of solute weight of solvent in grams= ´

=0.46 1000

0.01 M46 1000

´ =

7. Calculate the molarity and normality of a dibasic acid solution (mol. Wt. 132) having 33.0 g of the acid in 400 ml.

A: Given data: Weight of solute = 33 gm, Volume of solution = 400 ml; Moleuclar weight of solute = 132;

Equivalent weight of solute Moleuclar weight

662

= =

Molarity ( ) Weight of the solute 1000M

moleuclar weight of the solute volume of the solution in ml= ´

33 1000

0.625 M132 400

= ´ =

Normality ( ) weight of the solute 1000N

equivalent weight of the solute volume of solution in ml= ´

33 1000

1.25 N66 400

= ´ =

8. Calculate the normalities of the following solutions: a) 6.3 g of 3HNO in a litre solution b) 11.2 g of KOH in a litre solution

A: a) Given data: Weight of solute = 6.3 gm

Equivalent weight of solute = Moleuclar weight

63Basicity

=

Normality (N) Weight of the solute 1

Equivalnet weight of the solute volume of the solution in lit= ´

6.3 1

0.1 N63 1

= ´ =

b) Given data: Weight of solute = 11.2 gm

Equivalent weight of solute Moleuclar weight

56Acidity

= =

Normality (N) = weight of the solute 1

equivalent weight of the solute volume of the solution in lit´

11.2 1

0.2N56 1

= ´ =

9. To convert 1 litre of 1.123 N solution of an acid into 1 N solution, how much volume water should be added? A: Given data:

1 2V 1000 ml, V ?= =

1 2N 1.123 N, N 1N= =

For dilution process, the equation is 1 1 2 2 2V N V N 1000 1.123 V 1= Þ ´ = ´


21000 1.123

V 1123 ml1´

= =

Water to be added ( )2 1v v 1123 1000 123 ml- = - =

10. Find the molarity of 3% (W/V) 2 2H O solution.

A: Molarity ( )

Weight of the solute 1000GMW of the solute vol. of solution ml

= ´

3 % 2 2H O solution means that 3 g of 2 2H O present in 100 ml of solution.

Wt. of solute is = 3 g GMW of 2 2H O = 34 g

Volume of solution = 100 ml

Molarity 3 1000 30

0.88 M34 100 34

= ´ = =


CHEMICAL KINETICS, EQUILIBRIUM AND IONIC EQUILIBRIUM LONG ANSWER TYPE

1. Define rate of chemical reaction and explain the factors that effect the rate of a reaction.

A. The rate of a reaction at any instant is directly proportional to the product of the concentration of the reactants

taking part in the reaction at that instant.

Factors that affect reaction rate are

a) The chemical nature of the reactants b) The concentration of the reactants

c) The temperature of the reaction d) The catalyst in the reaction.

(a) The chemical nature of the reactants:

The reaction taking place between the ionic compounds is faster than the reaction taking place between the

covalent compounds. Reactions taking place between covalent compounds involve breaking and formation of

bonds, hence the reactions are slow.

Reaction between ionic compound: ( ) ( ) ( )aq 3 aq 3 aqNaCl AgNO AgCl NaNO+ ® ¯ +

Reaction between covalent compounds : -

( ) ( ) ( )H

2 5 3 3 2 25C H OH CH COOH CH COOC H H O+

+ ¾¾¾® +l l l

(b) Effect of concentration of reactants:

The rate of a reaction depends on the concentration of the reactants at any given temperature. Greater the

concentration of the reactants, greater is the rate

(c) Effect of temperature:-

The rate of reaction generally increases with increase in the temperature of the reaction. The specific rate of

a reaction is approximately doubled for every 10 C° rise of temperature for many reactions.

(d) Effect of catalyst:-

Catalyst increases the rate of reaction by changing the path of the reaction. The reaction takes place through

a different path of lower activation energy.

2. What is meant by rate law? Write the rate laws of elementary reactions of I and II order reactions. Deduce

the units for the rate constant (K) of zero and first order reactions.

A. The equation that describes mathematically the dependence of the rate of a reaction on the concentration

terms of the reactants is known as the rate equation or rate law.

I order reaction

A Products®

[ ]V A ¢µ

Or [ ]V K A ¢= (where K is specific reaction rate or rate constant)

II order reaction

2A Products®

[ ]2V Aµ

[ ]2V K A=

Units of rate constant K

Zero order reaction

[ ]V K A= °


V K=

C moles /litre

V moles /litre sect sec

= = =

Q V = K

Units of K = moles/litre sec

I order reaction

[ ]1V K A=

[ ]V moles/litre sec

KA moles /litre

= =

1K sec-=

Units for rate constant of I order reaction = 1sec-

3. Derive the rate equation for a first order reaction? Derive the expression for half life of first order reaction.

A. Let the first order reaction be represented as R P®

Let the initial concentration be ‘a’ 3moles /dm . After time ‘t’ the concentration of reactant left be

( ) 3a x moles /dm- where x is the amount of A decomposed in time ‘t’ seconds.

The rate dcdt

- or dxdt

at time t is proportional to the concentration of R after time ‘t’.

( )dxa x

dtµ -

( )dxK a x

dt= - Þ

( )dx

Kdta x

=-

On integration

( )

dxK.dt

a x=

-ò ò

( ) ( )ln a x Kt C............... 1- - = + when t = 0, x = 0

( )ln a 0 K(0) C- - = + Þ ln a C- =

Substitution this in equation (1)

( )ln a x kt ln a- - = -

Or ( )t

aK ln

a x=

-

1 a

K lnt a x

=-

or

2.303 a

K logt a x

=-

The above equation is for I order reaction.

Half life of I order reaction:

When 1/2a

x ; t t2

= =


2.303 a

K logt a x

=-

( )1/2

2.303 at log

K a a /2=

-2.303 a

logK a / 2

=

1/22.303

t log 2K

= Þ log 2 0.3010=

1/22.303 0.3010 0.693

tK K´

= = Þ 1/20.693

tK

=

4. What is Le chatlier’s principle? Explain the same with suitable example?

A. Le chatlier’s principle :

If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration (of the

reactants/products) the equilibrium shifts in the direction in which this change is reduced or nullified.

Ex:- ( ) ( ) ( )2 g 2 g 3 gN 3H 2NH ; H 92.0 kJ+ D = -ˆˆ†‡ˆˆ

The forward reaction (formation of 3NH ) is a exothermic reaction. According to the Le chatlier’s principle low

temperatures favour the formation of 3NH . The reaction is slow at low temperatures so an optimum

temperature of 725-775 K is used.

Catalyst : Fe Mo+ promoter.

The number of molecules of ammonia (2) is less (hence less volume) compared to the reactants (4).

According to Le chatlier’s principle high pressure favours the formation of 3NH . Hence 200 atmospheres

pressure is used

The backward reaction is endothermic and has more number of molecules (or volume). So according to Le

chatliers principle high temperature and low pressure favours the backward reaction.

5. Explain Le chaltliers principle and apply the same to the equilibrium: ( ) ( ) ( )2 g 2 g 3 g2SO O 2SO+ ˆˆ†‡ˆˆ

A. Le chatlier’s principle :

If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration (of the

reactants/products) the equilibrium shifts in the direction in which this change is reduced or nullified.

( ) ( ) ( )2 g 2 g 3 g2SO O 2SO H 189.0 kJ+ D = -ˆˆ†‡ˆˆ

Effective of temperature on forward reaction :

The forward reaction leading to the formation of sulphur trioxide is exothermic reaction. So according to

Lechatlier’s principle low temperature favours the forward reaction. But at low temperature the reaction is

very slow. Therefore, the optimum temperature of 673 K is used and a catalyst 2 5V O or platinised asbestos

is used to speed up the reaction.

Effect of temperature on backward reaction :

As forward reaction is exothermic, backward reaction formation of 2SO is endothermic. So according to

Lechatlier’s principle high temperature favours the formation of sulphur dioxide.

Effect of pressure on the reaction :


The forward reaction leading to the formation of 3SO involves decrease in number of molecules (2 molecules

of 3SO ) or decrease in volume. As per Le Chaltier’s principle high pressure favours the formation of 3SO .

But at high pressure, the towers used get corroded. Therefore optimum conditions are used.

The backward reaction which has more number of molecules is favoured at low pressure.

6. Explain the term “hydrolysis of salts” with suitable examples. Give the relation between hydrolysis constant

( )hK ionization constant ( )aK and ionic product ( )wK of water for sodium acetate solution.

A. Hydrolysis of Salt:

It is the phenomenon in which the anion or cation or both of salt react with water producing excess or both of

salt react with water producing excess of OH- ions or H+ ions or both in aqueous solution.

Ex : -

(1) Salt of strong acid and weak base.

In water : 4 4NH Cl NH Cl+ -® +

4NH+ React with water as it is strong conjugate acid of weak base 3NH .

4 2 4NH H O NH OH H+ ++ ® +

As the solution has more H+ concentration the solution is acidic having pH < 7

(2) Salt of Weak acid and Strong base

3CH COONa

In water 3 3CH COONa CH COO Na- +® +

Acetate ions are strong conjugate base of weak acid acetic acid, hence undergo hydrolysis to give a basic

solution.

3 2 3CH COO H O CH COOH OH- -+ +ˆˆ†‡ˆˆ

As OH H- +é ù é ù>ë û ë û the pH > 7

3 2 3CH COO H O CH COOH OH- -+ +ˆˆ†‡ˆˆ

[ ][ ]

3

3 2

CH COOH OHK

CH COO H O

-

-

é ùë û=

é ùë û

[ ][ ]

2

3hH O

3

CH COOH OHK K

CH COO

-

-

é ùë û= =

é ùë û

( ) [ ]3

a3

CH COO HK of acetic acid

CH COOH

- +é ù é ùë û ë û= …………..(1)

wK H OH+ -é ù é ù= ë û ë û ………………..(2)

[ ]w3

a 3

H OHKCH COOH

K CH COO H

+ -

- +

é ù é ùë û ë û=

é ù é ùë û ë û


[ ]3h

3

CH COOH OHK

CH COO

-

-

é ùë û=

é ùë û

wh

a

KK

K\ =

NUMERICAL PROBLEMS 1. The rate constant of a first order reaction is 0.693 1sec- , what is 1/2t ?

A. 1/20.698

tK

= for I order reaction

1K 0.693 sec-=

1/20.693

t 1 sec0.693

= =

2. The initial concentration of A is 0.3 moles / lit. The concentration after 2.3 minutes is 0.15 moles / litres. What

is the value of rate constant.

A. [ ]d A 0.3 0.15 0.15

r 0.065 moles /litre mindt 2.3 2.3

-= = = = -

3. The value of pK if the value of cK is 32 10-´ for ( ) ( ) ( )2 g 2 g gH I 2HI+ ˆˆ†‡ˆˆ at 27 C° .

A. ( ) np cK K RT D=

p cK K= as nD for the above reaction = 0

3pK 2 10-= ´

4. How many grams of NaOH are present per litre if the pH of NaOH solution is 10 ?

A. pH = 10; OHP =4; OHpH P 14+ =

4n OH 10- -é ù= =ë û

Given weight 1000

NormalityEquivalent weight Volume in mL

= ´

4 4 3x 100010 40 10 4 10 g

40 1000- - -= ´ = ´ = ´

5. What is the pH of 0.1 M NaOH?

A. 1OH 10- -é ù =ë û

1OHP log 10-é ù= - ë û ; OHP 1=

pH = 14 – 1 = 13


6. Find the pH of 0.05 ( )2Ba OH aqueous solution.

A. Each ( )2Ba OH molecules gives 2OH- ions.

OH 0.05 2 0.1-é ù = ´ =ë û

1OHP log 10 1-é ù= - =ë û

pH 14 1 13= - =

7. Find the pH of 810 M- HCl solution.

A. As 810 M- HCl is much diluted solution: the concentration of H+é ùë û in water should be added in the

equilibrium.

8 7H 10 10+ - -é ù = +ë û

( )7 7H 10 0.1 1 1.1 10+ - -é ù = + = ´ë û

7pH log 1.1 10 7 log 1.1 7 0.0414 6.95-= - ´ = - = - = [ ]log 1.1 0.0414=Q

8. The pH of HCl solution is 5.4. What is the hydrogen ion concentration?

A. pH 5.4=

log H 5.4+é ù = -ë û or 6.6

6H Anti log 6.6 3.981 10 M+ -é ù é ù= = ´ë ûë û

9. 50 mL of 0.2 N 2 4H SO were added to 100mL of 0.2 N 3HNO . Then the solution is diluted to 300

mL. What is the pH of solution?

A. 1 1 2 2

1 2

V N V NN

V V+

=+

500 0.2 100 0.2600

´ + ´=

100 20

N 0.2600

+Þ = = ; 0.2 600 900Þ ´ = ´x

0.2 600

0.13900´

= =

1pH log 0.13 1.3 10 1 log 1.3-= - = ´ = - 1 0.113 0.887= - =

10. At a certain temperature the ionic product of 2H O is 14 29.55 10 /lit-´ . Then what is the pH of the solution.

A. 14 7wH OH K 9.55 10 3.21 10+ - - -é ù é ù= = = ´ = ´ë û ë û

( )7pH log 3.21 10-= - ´

7 log 3.21 7 0.5051 6.5049= - = - =

11. The aK for 0.2 M 3CH COOH is 52 10-´ . What is the pH? (Assume 3CH COOH as a strong acid)

A. As we assume it to be strong acid,

1H 0.2M 2 10+ -é ù = = ´ë û


pH = 1 – log2 = 1 – 0.3010 = 0.6990

12. Find the pH of 0.5 M 4NH OH solution. bK for 4NH OH is 52 10-´ . (Assume 4NH OH as strong base)

A. OH 0.5M-é ù =ë û

POH log 0.5= - 15 10 1 log 5 1 0.6990 0.3010-= ´ = - = - =

pH 14 0.3010 13.6990= - =

13. What is the pH of HCl solution containing 3.65 g in 250 mL?

A. Given wt. 1000 3.65 1000

N HEq.wt. V in mL 36.5 250

+é ù= = ´ = ´ë û = 0.4

1pH log H log 0.4 4 10+ -é ù= - = - = ´ë û 1 log 4 1 0.6021 0.497= - = - =

14. A litre of buffer solution contains 0.1 moles of acetic acid and 1 mole of sodium acetate. Find its pH of pKa of

3CH COOH 4.8=

A. [ ][ ]

saltpH pKa log

acid= +

[ ]salt 1 mole /lit= ; [ ]acetic 0.1 mole /lit=

10.1pH 4.8 log 4.8 log 10

1-= + = + Þ pH 4.8 1 5.8= + =

15. 20 mL of 0.2 M 3CH COOH and 20 mL of 0.4 M sodium acetate were mixed together to form a buffer. What

is the pH if pKa of 3CH COOH is 4.8

A. [ ][ ]

saltpH pKa log

acid= +

20 0.440pH 4.8 log

20 0.240

´é ùê úë û= +

´é ùê úë û

4.8 log 2 4.8 0.3010 5.1= + = + =

16. Find the hydrolysis constant of 0.1 M sodium acetate Ka of 53CH COOH 2 10-= ´

A. wh

a

KK

K=

14 14 5w a5

1K 1 10 10 ; K 2 10

2 10- -

-= ´ = ´ = ´

´

9 100.5 10 or 5 10- -= ´ ´

17. What is the pH of a buffer formed by mixing 20 mL of 0.1 M 4NH Cl and 2 mL of 0.1 M 3NH ? pKb for

4NH OH 4.8=

A. [ ][ ]b

saltPOH pK log

base= +


b

20 0.140pK log

20 0.140

´é ùê úë û= +

´é ùê úë û

4.8 log 1 4.8= + =

pH 14 4.8 9.2= - =

18. The hydrolysis constant for 4NH Cl solution is 90.5 10-´ . Then what is the dissociation constant of the

base?

A. wh

b

KK

K= Þ

149

b

1 100.5 10

K

-- ´

´ =

14

14 4b 9 10

1 10 1K 10 0.2 10

0.5 10 5 10

-- -

- -´

= = ´ = ´´ ´

19. The dissociation constant of 4NH OH and 3CH COOH are 52 10-´ and 51.8 10-´ respectively. Find the

hydrolysis constant of ammonium acetate.

A. wh

a b

KK

K K=

´

14 144 5

5 5 101 10 1 10

0.2 10 2 102 10 1.8 10 3.6 10

- -- -

- - -´ ´

= = = ´ = ´´ ´ ´ ´

20. 50 mL of 1 M 3CH COOH solution when added to 50 mL of 0.5 M NaOH gives a solution with pH value of X.

Find the value of X (pKa of 3CH COOH = 4.8)

A. pH = [ ][ ]

50 0.5salt 100pKa log 4.8 log 4.8 log 1 4.8

50 0.5acid100

´é ùê úë û+ = ´ = ´ =

´é ùê úë û

21. What is the pH of a solution formed by mixing 50 mL of 1 M HCl and 50 mL of 0.1 M NaOH?

A. 150 1 50 0.1 45H 50 5 0.45 4.5 10

100 100+ -´ - ´é ù = = - = = = ´ë û

pH 1 log 4.5 1 0.658 0.34= - = - =

22. The 0.005 M monobasic acid has pH of 5. What is the degree of dissociate?

A. pH = 5; 5H 10+ -é ù =ë û

H c.+é ù = aë û

510 0.005- = ´ a

5

23

10: 0.2 10 or 0.2%

5 10

--

-= ´

´

23. The solubility product of salt 10 2AB 10 moles /lit.-= What is the solubility?

A. AB A B+ -+ˆˆ†‡ˆˆ

2spK S S S= ´ =

10 5spS K 10 10 moles /lit- -= = =


24. The solubility of 32A B 2 10 moles /lit.-= ´ What is solubility product?

A. ( )2 2spK 2S S 4S= ´ =

( )33 94 2 10 8 10- -= ´ ´ = ´

SHORT ANSWER TYPE 1. Define and explain the terms with suitable examples “order ” and “molecularity” of a reaction.

A. Order: The order of the reaction is sum of the powers of the concentration terms in the rate equation of the

reaction.

Ex: 1. ( ) ( ) ( )2 25 g 4 g 2 g1

N O N O O2

® +

[ ]2 5rate K N O= Hence I order.

2. ( ) ( ) ( )2 g 2 g 2 g2N O 2N O® +

[ ]22rate K N O= Hence II order reaction.

Molecularity : The number of atoms or ions taking part in the rate determining step is called molecularity of

the reaction.

Ex: Pseudo Unimolecular reaction.

H3 2 5 2 3 2 5CH COOC H H O CH COOH C H OH

++ ¾¾¾® +

As water is taken in excess it will be an example of pseudo unimolecular reaction.

2. Describe any two methods for the determination of order of reaction.

A. The methods used for the determination of the order of reaction include:

(1) integral equation method or trial and error method (2) half-time method

(3) Van’t Hoff differential method (4) Ostwald’s isolation method.

(1) Trial and Error Method or Integral form of rate equation method.

The concentration of the reactant or reactants (a) initially when time is 0 and the concentration of the reactant

after time t (a – x) are measured by suitable analytical methods and substituted in different rate equations of

different orders. The order of the reaction will be the one which gives constant k values.

Zero order ( )Z R P- ° ; [ ] [ ]0 tR R

kt

-=

I order [ ][ ]

0

t

R2.303 a 2.303k log or log

t a x t R=

-

(2) Half-time ( 1/2t ) method:

The time required for the initial concentration (a) of the reactant to become half of its value (a/2) during the

process of the reaction is called half time of the reaction.

The half – time ( 1/2t ) is inversely proportional to ( )n 1a - where a is the initial concentration of the reactant

and n is the order of the reaction.

1/2 n 11

ta -

µ


Therefore for a given reaction two half-time values with initial concentration a¢ and a¢¢ are determined

experimentally and the order is established in the equation.

n 11/2

1/2

t at a

-¢ ¢¢æ ö= ç ÷¢¢ ¢è ø

3. What are the main postulates of simple collision theory of reaction state?

A. (1) Molecules have to collide with each other for the reaction to occur.

(2) All collisions do not lead to the formation of the products.

(3) The colliding molecules have to possess minimum energy to form products. This minimum energy is

called Threshold energy.

(4) The energy of the molecules at STP is less than this threshold energy.

(5) The difference between the threshold energy and the energy of molecules in the normal state is called

activation energy.

Activation energy = Threshold energy – Energy of normal molecules.

(6) The molecules possessing the threshold energy are called activated molecules.

(7) Collisions taking places between activated molecules are activated collisions and these collisions lead to

the formation of products.

(8) The fraction of activated collisions among the total collisions is very much less.

4. What are equilibrium processes? Explain with examples equilibrium physical processes and equilibrium

chemical processes.

A. Any process whether physical or chemical that takes place in the forward as well as in the reverse (opposite)

direction under the same conditions is generally referred as reversible process.

Equilibrium physical process:

Equilibrium between any two phases of the same substance provided both the phases occurs at the given set

of conditions and no phase disappears in the process is called physical equilibrium.

Solid ˆˆ†‡ˆˆ liquid (melting or fusion)

Solid ˆˆ†‡ˆˆ vapour (sublimation)

Equilibrium between 2 allotropic forms of the same substance will be examples of physical equilibrium.

sulphur sulphura - b -ˆˆ†‡ˆˆ

Equilibrium chemical process :

A reaction is considered reversible, if the reaction mixture contains the reactants as well as the products and if

both forward and backward reactions are taking place under experimental conditions.

Ex: ( ) ( ) ( )3 s s 2 gCaCO CaO COD

+ˆˆ†̂‡ˆˆ̂

5. What is meant by dynamic equilibrium? Explain with suitable example.

A. The forward and the reverse reaction of a reversible reactions have to take place with equal rates

simultaneously at the equilibrium state also. Then the equilibrium is called dynamic equilibrium.

At equilibrium state the reaction does not stop. Both the forward and backward reactions continuously take

place. The equilibrium concentrations of the reactants and the products remain unchanged with time provided

the reaction conditions are not altered. The physical properties such as colour, density, pressure etc do not

change.


Ex: ( ) ( ) ( )3 s s 2 gCaCO CaO COD

+ˆˆ†̂‡ˆˆ̂

6. Give the characteristics of chemical equilibrium.

A. (1) Both forward and reverse reactions continue to take place.

(2) The rate of forward reaction is equal to the rate of backward reaction

(3) The properties such as pressure, concentration, density, colour etc remain unchanged with time.

(4) Addition of a catalyst to the reaction does not alter the position of the equilibrium. It only speeds up the

attainment of the equilibrium

(5) The same chemical equilibrium can be reached by carrying out the reaction starting with reactants or by

carrying out the reaction staring with products.

(6) Change in pressure or concentration of the reactants or the products may change the position of

equilibrium.

7. Explain cK and pK and derive the relation between them?

A. cK : Equilibrium constant taking the molar concentration of different substance in a reaction.

pK : The equilibrium constant taking the partial pressures of the reacting substances which are in gaseous

state.

Ex: aA bB cC dD+ +ˆˆ†‡ˆˆ

[ ] [ ][ ] [ ]

c d

c a b

C . DK

A B= ;

c dC D

p a bA B

P PK

P P=

Derivation of relation between pK and cK

From ideal gas equation: PV nRT= ; n

P RTV

=

Q n concentration

V=

P CRT= …………….(1)

c dC D

p a bA B

P PK

P P= ………….(2)

Substituting the value of P in equation 2 we get,

[ ] [ ][ ] [ ]

c d

p a b

CRT DRTK

ART BRT= ;

[ ] [ ][ ] [ ]

( ) ( )c d

c d a bp a b

C DK .RT

A B

+ - +=

( ) np cK K RT D= ;

[ ] [ ][ ] [ ]

c d

c a b

C DK

A B=Q

8. Explain the concepts of Bronsted acids and Bronsted bases. Illustrate the same with example.

A. Bronsted Acid: A substance which loses or donate a proton is called an acid.

Ex: HCl, 3 4H PO etc.

Bronsted base: A substance which has a tendency to gain a proton or accepts is a proton called a base.

In an acid base reaction proton transfer takes place. This process of transfer of proton from an acid to a base

is referred as neutralization.


2 3HCl H O H O Cl+ -+ +ˆˆ†‡ˆˆ

Water accepts a proton from the HCl hence acts as a base.

3 2 4NH H O NH Cl+ -+ +ˆˆ†‡ˆˆ

When ammonia is dissolved in water, 3NH accepts a proton from water. Water acts as acid. In the above

two examples 3H O+ and 4NH+ are Bronsted acids. Cl- and OH- are Bronsted bases.

9. What are conjugate acids an conjugate bases? Give 4 examples.

A. In all acid-base reactions at equilibrium (according to Bronsted Lowry theory) two acids and two bases are

involved. An acid is converted to a corresponding base and vice versa. Such a related pair of an acid and a

base which differ by a single proton is said to be conjugate pair.

2 3HCl H O H O Cl+ -+ +ˆˆ†‡ˆˆ

HCl and Cl- are the conjugate pair, 2H O and 3H O+ are the conjugate pairs in the above example.

Cl- is the conjugate base of HCl and 3H O+ is the conjugate acid of 2H O . Every protonic acid (HCl in the

above example) has its conjugate base ( Cl- ) and every bronsted base ( )2H O has its conjugate acid

( )3H O+

10. Explain with suitable examples Lewis acid-base theory?

A. Lewis-acid: A substance than can accept an electron pair to form a coordinate covalent bond with donor.

Ex: H+ , 3BF , 4SnCl

Lewis base:- A substance that can donate a lone pair of electrons to form a coordinate covalent bond with the

acceptor.

Ex: 2H O , 3NH , lignads in complex compounds.

Acid-Base reaction:

An acid base reaction should involve formation of coordinate covalent bond

| | | |

| || |

F H F H

F FH H

F B : N H F B N Hé ùê ú- + - ® - ¬ -ê úê úë û

11. What is degree of ionization in respect of weak acids and weak bases? Derive relation between the degree of

ionization ( )a and the ionization constant ( )aK

A. All acids and all bases do not ionize to the same extent. Extent of ionization depends on the polarity of HX

(Acid) or BOH(base). If the extent of the ionization of substance is large it is called strong acid or strong base.

If the extent of ionization of the substance is less, the compound is called weak acid or base.

Extent of ionization large – Strong

Extent of ionization less – Weak

The extent of ionization is represented by a

( ) ( )

HX H XInitial Conc C

After time t C 1 C C

+ -+- -

- a a a

ˆˆ†‡ˆˆ

If a is the degree of dissociation


[ ] ( )a

H X C .CK

HX C 1

+ -é ù é ù a aë û ë û= =- a

( )

2

aC

K1

aÞ =

- a

Where aK is dissociation constant for weak acid HX

For bases; ( )

2

bC

K1

a=

- a

12. Define pH. What is a buffer solution? How are buffer solutions prepared?

A. The negative value of the logarithm to the base 10 of the hydrogen ion concentration.

10 101

pH log H logH

++

é ù= - =ë û é ùë û

A buffer solution is that solution which resists any change in its pH value on dilution or addition of small

amount of strong acid or strong alkali.

Buffer solutions are prepared generally, by mixing equal or different volumes of equimolar solution of a weak

base and its salt.

Ex: 3 3CH COOH CH COONa+ ; 4 4NH OH NH Cl+

VERY SHORT ANSWER TYPE

1. Define rate of reaction.

A. Rate of reaction is defined as change in the molar concentration of reactants or products per unit time.

2. What is rate law?



3. Define order of a reaction.

A. The order of a reaction is the sum of the powers of the concentration terms in the rate equation of the

reaction.

4. Name two first order reactions you know?

A. 1. ( ) ( ) ( ) [ ]12 2 2 22 aq 2 g

1H O H O O ; rate K H O

2® + =l 2. ( ) ( ) ( ) [ ]2 2 2 55 g 4 g 2 g

1N O N O O ; rate K N O

2® + =

5. What is half life?

A. The time required for the initial concentration (a) of the reactant to become half of its value (a/2) during the

progress of the reaction.

6. What is zero order reaction?

A. Reactions in which the rate of reaction is independent of the concentration of the reacting substances is called

zero order reactions.

7. Give the integrated form of the rate equilibrium for first order reaction.

A. ( )

[ ][ ]

0

t

R2.303 a 2.303K log log

t a x t R= =

-

8. What is activation energy?

A. The difference between the threshold energy and the energy of the molecules in the normal state is called

activation energy.


9. State law of mass action.

A. The rate of a reaction at any instant is directly proportional to the product of the concentrations (active

masses) of the reactants taking part in the reaction at that instant.

10. Write the relation between pK and cK .


11. What is the effect of pressure on ( ) ( ) ( )2 g 2 g 3 gN 3H 2NH+ ˆˆ†‡ˆˆ ?

A. When pressure is increased on the reaction at equilibrium, favours the reaction in the direction in which the

volume or number of molecules decreases. When pressure is decreased on the reaction, it favours the

reaction towards reactants (in the direction in which the volume or number of molecules increases) When

pressure is increased it favours forward reaction.

12. What is Bronsted base? Give one example.

A. A chemical substance that has a tendency to gain a proton from donor, i.e., proton acceptor.

Ex. 3NH

13. What is Lewis acid? Give one example.

A. A substance that can accept an electron pair to form a coordinate covalent bond with the donor. Ex:

3H ,BF+

14. Define ionic product of water.

A. The ionic product of water, wK at a given temperature, is defined as the product of the concentration of H+

and OH- in water or in aqueous solution.

15. What is a buffer solution?

A. A solution which resists any change in its pH value on dilution or on addition of a small amount of strong acid

or strong alkali.

16. Give one example each for acid and base buffer solution.

A. ( ) 3 31 CH COOH CH COONa acid buffer+ - ( ) 4 42 NH OH NH Cl basic buffer+ -

17. Explain hydrolysis of a salt?

A. It is the phenomenon in which the anion or cation or both of a salt react with water producing excess of OH-

ions or H+ ion or both in aqueous solution.

18. Which ions are released in the hydrolysis of Cl- ions with water?

A. 2Cl H O HCl OH- -+ ® +

HCl being strong electrolyte undergo complete ionization. Hence Cl- do not undergo hydrolysis.

19. Is the pH of aqueous solution of 4NH Cl greater or smaller than 7? Why?

A. 4 4NH Cl NH Cl+ -® +

4NH+ being strong undergo hydrolysis and the solution is acidic.

4 2 4NH H O NH OH H+ ++ ® +

Hence pH is less than 7.


20. (a) For the hydrolysis of ammonium acetate, write the relation between h a b wK ,K ,K and K .

(b) What is solubility product of salt?

A. (a) Ammonium acetate ( )3 4CH COONH is 9 salt of weak acid and weak base. wh

a b

KK

K .K=

(b) 3 4 3 4CH COONH CH COO NH- ++ˆˆ†‡ˆˆ

3 4CH COO NH- +é ù é ù=ë û ë û

2spK S S S= ´ =

Very Short answer Questions 1. Define rate of reaction.

A. Rate of reaction is defined as change in the molar concentration of reactants or products per unit time.

2. What is rate law?



3. Define order of a reaction.

A. The order of a reaction is the sum of the powers of the concentration terms in the rate equation of the

reaction.

4. Name two first order reactions you know?

A. 1. ( ) ( ) ( ) [ ]12 2 2 22 aq 2 g

1H O H O O ; rate K H O

2® + =l 2. ( ) ( ) ( ) [ ]2 2 2 55 g 4 g 2 g

1N O N O O ; rate K N O

2® + =

5. What is half life?

A. The time required for the initial concentration (a) of the reactant to become half of its value (a/2) during the

progress of the reaction.

6. What is zero order reaction?

A. Reactions in which the rate of reaction is independent of the concentration of the reacting substances is called

zero order reactions.

7. Give the integrated form of the rate equilibrium for first order reaction.

A. ( )

[ ][ ]

0

t

R2.303 a 2.303K log log

t a x t R= =

-

8. What is activation energy?

A. The difference between the threshold energy and the energy of the molecules in the normal state is called

activation energy.

9. State law of mass action.

A. The rate of a reaction at any instant is directly proportional to the product of the concentrations (active

masses) of the reactants taking part in the reaction at that instant.

10. Write the relation between pK and cK .



11. What is the effect of pressure on ( ) ( ) ( )2 g 2 g 3 gN 3H 2NH+ ˆˆ†‡ˆˆ ?

A. When pressure is increased on the reaction at equilibrium, favours the reaction in the direction in which the volume or number of molecules decreases. When pressure is decreased on the reaction, it favours the reaction towards reactants (in the direction in which the volume or number of molecules increases) When pressure is increased it favours forward reaction.

12. What is Bronsted base? Give one example.

A. A chemical substance that has a tendency to gain a proton from donor, i.e., proton acceptor. Ex. 3NH

13. What is Lewis acid? Give one example.

A. A substance that can accept an electron pair to form a coordinate covalent bond with the donor. Ex:

3H ,BF+

14. Define ionic product of water.

A. The ionic product of water, wK at a given temperature, is defined as the product of the concentration of H+

and OH- in water or in aqueous solution.

15. What is a buffer solution?

A. A solution which resists any change in its pH value on dilution or on addition of a small amount of strong acid or strong alkali.

16. Give one example each for acid and base buffer solution.

A. ( ) 3 31 CH COOH CH COONa acid buffer+ - ( ) 4 42 NH OH NH Cl basic buffer+ -

17. Explain hydrolysis of a salt?

A. It is the phenomenon in which the anion or cation or both of a salt react with water producing excess of OH-

ions or H+ ion or both in aqueous solution.

18. Which ions are released in the hydrolysis of Cl- ions with water?

2Cl H O HCl OH- -+ ® +

HCl being strong electrolyte undergo complete ionization. Hence Cl- do not undergo hydrolysis.

19. Is the pH of aqueous solution of 4NH Cl greater or smaller than 7? Why?

A. 4 4NH Cl NH Cl+ -® +

4NH+ being strong undergo hydrolysis and the solution is acidic.

4 2 4NH H O NH OH H+ ++ ® +

Hence pH is less than 7.

20. (a) For the hydrolysis of ammonium acetate, write the relation between h a b wK ,K ,K and K .

(b) What is solubility product of salt?

A. (a) Ammonium acetate ( )3 4CH COONH is 9 salt of weak acid and weak base. wh

a b

KK

K .K=

(b) 3 4 3 4CH COONH CH COO NH- ++ˆˆ†‡ˆˆ

3 4CH COO NH- +é ù é ù=ë û ë û 2spK S S S= ´ =


SOLID STATE

1. Write short notes on the following giving suitable examples. a) Schottky defect b) Frenkel defect

A: a) Schottky defect : * Schottky defect is a point defect. * This defect arises when an atom or an ion is missing from its normal site in the lattice.

* Ionic crystals maintain electrical neutrality. Hence no. of vacancies at cation sites are equal to no. of vacancies at the anion sites.

Ex: AB is ionic compound

A+ B- A+ B-

B- B- A+

A+ B- A+ B-

B- A+A+

Cation Vacancy

Anion Vacancy

* This defect occurs mainly in (a) Highly ionic compound where the cationic and anionic sizes are similar (b) Compounds with high coordination numbers. * The density of the crystal may be decreased or may become unpredictable. Ex: NaCl, CsCl b) Frenkel defect: * Frenkel defect is point defect.

* This defect arises when an atom or ion is missing from its normal site and occupies interstitial space in lattice.

* In ionic crystals, usually cation escapes from its site and occupies the space among anions because of small size.

Ex: AB is ionic compound.

A+ B- A+ B-

B- B- A+

A+ B- A+ B-

Interstitial site A+

* This defect occurs mainly in a) Ionic compound where the large difference exists in sizes between the cation and anion b) Compounds with low coordination numbers. * The density of the crystal is not changes. Ex: AgCl, AgI, ZnS

2. What do you know about ‘amorphous’ substances? Discuss. A: * Amorphous solids are those substances that do not have sharp melting points but they melt over a range of temperature. Ex: Glass * In amorphous solids, the arrangement of component particles are in random manner. * Amorphous solids do not have long range orders or crystal, they have short range orders like in liquids. * Any given material can be converted into amorphous solids or glassy by sudden cooling (quenching) a

melt or freezing vapour of that substance.

Molten crystalline solid quenching¾¾¾¾® amorphous solids

Ex: Quartz is melted and quenched then it turns into amorphous solid. * The melting of amorphous solids when slowly cooled (annealed) changes as crystalline solids at a

definite temperature.

Molten amorphous annealing¾¾¾¾® crystalline solid

* Amorphous solids can be moulded or blown into articles of different shapes. * Amorphous solids are used in domestic constructions and appliances like photovoltaic cells.

3. How many types of semi conductors are known? Explain the influence of doping on the conductivity of crystalline solids.

A: A) Semiconductors:

The solids whose conductivity is in order of 610- to 410 1 1ohm cm- - are known as semiconductors. These are two types:


(i) Intrinsic semiconductors: Pure substances whose conductivity increases with rise in temperature are called as intrinsic semiconductors. Conductivity of these substances is due to thermal electrons and holes in the lattice of semiconductors. These are insulators at room temperature.

Ex: Pure Si or Ge (ii) Extrinsic or impurity semiconductors

Extrinsic semiconductors are those in which conductivity depends on the concentration of electron donor or acceptor i.e. impurity. Addition of impurity to the semiconductors is called as ‘doping’ and the added substance is called as ‘dopant’. Based on impurity these are of two types.

(a) n-type semiconductors: When Si or Ge is doped with VA or 15 th group elements (P, As, Sb or Bi) is known as n-type semiconductors. ‘n’ standing for negative means that negatively charged electrons are responsible for electrical conductivity.

(b) p-type semiconductors: When Si or Ge is doped with IIIA or 13 th group elements (B, Al, Ga or In) is known as p-type semiconductors. ‘p’ stands for positive means that positively charged holes (electron vacancy) are responsible for electrical conductivity.

(B) When crystalline solids (insulators) are doped with suitable element, they conduct electricity even at ordinary temperatures

4. What do you mean by imperfection in solids? Write an essay on crystal defects. A: Imperfection in solids: The term imperfection (or) defect generally denotes departure from regularity in the

arrangement of the constituent particles (atoms or ions or molecules) in crystal. The defect may appear at a point, along a line or over a surface.

Following are the main defects in crystals: Schottky defect: * Schottky defect is a point defect. * This defect arises when an atom or an ion is missing from its normal site in the lattice.

* Ionic crystals maintain electrical neutrality. Hence no. of vacancies at cation sites are equal to no. of vacancies at the anion sites.

Ex: AB is ionic compound

A+ B- A+ B-

B- B- A+

A+ B- A+ B-

B- A+A+

Cation Vacancy

Anion Vacancy

* This defect occurs mainly in (a) Highly ionic compound where the cationic and anionic sizes are similar (b) Compounds with high coordination numbers. * The density of the crystal may be decreased or may become unpredictable. Ex: NaCl, CsCl Frenkel defect: * Frenkel defect is point defect.

* This defect arises when an atom or ion is missing from its normal site and occupies interstitial space in lattice.

* In ionic crystals, usually cation escapes from its site and occupies the space among anions because of small size.

Ex: AB is ionic compound.

A+ B- A+ B-

B- B- A+

A+ B- A+ B-

Interstitial site A+

* This defect occurs mainly in a) Ionic compound where the large difference exists in sizes between the cation and anion b) Compounds with low coordination numbers. * The density of the crystal is not changes. Ex: AgCl, AgI, ZnS


5. Derive Bragg’s equation. Or

Derive Bragg’s equation for X-rays of wavelength ( )l and a diffraction angle ( )q for an nth order reflection.

First layer

Second layer

Third layer

d q qqq

A

B

C D

1st ray 2nd ray

A: 1st ray is diffracted from point ‘A’ in the plane (I layer). 2nd ray is diffracted from point ‘B’ in the plane (II

layer). The second X-ray wave traveled more distance than the first X-ray. The extra distance traveled by second X-ray = CB + BD If the two waves are in constructive interference they exist in same phase i.e., the path difference is integral

multiple of the wave length.

CB BD n , here n 1,2,3,...\ + = l =

From the figure, in CB CB

ABC, sinAB d

D q = =

CB d sin\ = q

In BD BD

ABD, sinAB d

D q = =

BD dsin\ = q

CB BD dsin dsin 2dsin+ = q + q = q

n 2dsin\ l = q This is known as Bragg’s equation. Here ‘n’ is known as order of diffraction. n = 1, is called as first order diffraction; n = 2 is called as second order diffraction, n = 3 is called as third order diffraction.

6. Draw the differences between Schottky and Frenkel defects. A:

Schottky defect Frenkel defect

i) Schottky defect is due to absence of ions in the lattice sites

i) Frenkel defect is due to missing of an ion from lattice site and occupying in the interstitial space

ii) This defect arises in the compounds of high coordination number

ii) This defect arises in the compounds of low coordination number

iii) This defect arises in the compounds where the cation and anion sizes are similar

iii) This defect arises in the compound where cation size is small and anion size is large

iv) Due to this defect density of crystal decreases Ex: NaCl, CsCl

iv) Due to this defect density of crystal does not changes: Ex: AgCl, AgI

SAQ

1. Giving suitable examples, explain the following terms. a) Paramagnetic substances b) Feromagnetic substances c) Piezo electric effect

A: a) Paramagnetic substances: * The substances which are attracted into the applied magnetic field are called as paramagnetic

substances. * Paramagnetism is due to the presence of unpaired electrons in atoms, ions or molecules.

Ex: 2O , NO, Na atoms.

* They lose their magnetism when the applied magnetic fields are removed.


b) Ferro magnetic substances: * The substances which shows permanent magnetism even after the applied magnetic field is removed. * Ferro magnetism is due to presence of domains of magnetism. Ex: Fe, Co, Ni * A spontaneous alignment of magnetic moments in the same direction gives rise to ferromagnetism.

c) Piezo electric effect : The electrons in the insulators are closely bound to the individual atoms or ions. They do not migrate

under an applied field. But dipoles are created by shift in charges resulting in polarization. If these dipoles may align themselves in an ordered manner such that there is a net dipolement in the crystal, such crystals are deformed when mechanical stress or if an electric field is applied. The ions or atoms are displaced in this process, and hence electricity is produced. This is known as piezo electricity or pressure electricity.

Ex: Quarts, Lead Zirconate etc.

2. Write the differences between crystalline and amorphous solids. A:

Crystalline solids Amorphous solids

i) They have definite and regular geometrical shape

i) They do not have any definite geometrical shape

ii) They have sharp melting point ii) The do not have sharp melting point

iii) They are rigid and incompressible iii) They are not very rigid and compressibility is very less

iv) They have long range orders of crystals iv) They have short range orders

v) X-rays are diffracted v) X-rays diffraction bands are not given

VSAQ

1. What is metallic bond? A: “ The force that binds a metal ion to the mobile electrons within its sphere of influence is known as metallic

bond”

2. What is intrinsic region and extrinsic region? A: Intrinsic region:

The temperature zone where the conductivity depends on the thermal electrons and the holes in the lattice of the semiconductor is called as intrinsic region. Extrinsic region: At low temperatures the conductivity is mainly determined by the concentrations of the electron donors and the acceptors. This region is known as extrinsic region.

3. What are antiferromagnetic substances? Give an example. A: If the alignment of magnetic moments is in opposite directions with the equal number, the net magnetic

moment is zero. Such substances are termed as antiferromagnetic substances.

Ex: 2 3V O , NiO

4. What are Ferrimagnetic substance? Give an example. A: When unequal number of magnetic moments are aligned in opposite directions, the net magnetic moment is

not zero. Such substances are called as ferromagnetic substance.

Ex: 3 4Fe O

5. How the purity of the metal can be estimated? A: Purity of the metal can be estimated by a measurement of the resistance of the metal, as resistance ratio

300K

4.2K

rr

.

6. What will be the magnitude of the vapour pressure of these ionic crystals? A: At ordinary temperatures the vapour pressure is very low. Therefore, the heats of vapourization are very high.

Examples of ionic crystals: NaCl, MgO, 2CaCl , 3KNO , 2 4Na SO , 3CaCO


7. Which group and period of the periodic table does polonium occupy? A: VIA group or 16th group of the long form of the periodic table and 6th period.

8. Do amorphous solids have unit cells in them? A: Recent investigations report that the amorphous solids contains minute crystals but for all purposes

apparently they do not have unit cells or crystalline structures.

9. The diffracted X-rays from copper sulphate crystal are allowed to fall on a photographic plate. What happens to the photographic plate?

A: The photographic plate gets alternate bright and dark patched that result from the constructive and destructive interference of the diffracted rays.

10. Calculated contributions of lattice points in body centered cubic lattice arrangement.

A: Contributions from 8 corners = 1

88

´ =1 atom

Contributions from body centered point = 1 atom Total contribution per unit cell = 2 atoms or points.

11. A metal crystallizers into two cubic phases, f.c.c. and b.c.c. whose unit cell lengths are 3.6 0A and 2.7 0A ,

respectively. Calculate the ratio of densities of f.c.c. and b.c.c.

A: Density ( ) 3

1Z andr µ r µ

a

31 1 2

32 2 1

Z aZ a

r= ´

r

Here in f.c.c. density = 1r In b.c.c. density = 2r

Total number of atoms ( )1Z 4= Total number of atoms ( )2Z 2=

Length of the unit cell ( )1 0a 3.6 A= Length of the unit cell ( )2 0a 2.7 A=

By substitution the values

( )

( )31

32

2.74 272 323.6

r= ´ =

r \ The ratio of densities of f.c.c. and b.c.c. is

2732


ELECTROCHEMISTRY VERY SHORT ANSWER TYPE

1. What is electrolysis? Illustrate.

A. The decomposition of a chemical substance in the molten or in the solution state into its constituent elements

under influence of an applied EMF.

Ex: When molten potassium chloride is electrolysed potassium is obtained at cathode and 2Cl at anode.

2. What products are obtained in the electrolysis of fused KCl and aqueous KCl between platinum electrodes.

A. Fused KCl: 2KCl 2K 2Cl+ -® +

At cathode (reduction ): 2K 2e 2K+ -+ ®

At anode (oxidation): 22Cl Cl 2e- -® +

Aqueous KCl:

2KCl 2K 2Cl+ -® + ; 22H O 2H 2OH+ -® +

22H 2e H+ -+ ® ; 2K 2OH 2KOH+ -+ ®

At Anode: (oxidation)

22Cl Cl 2e- -® +

3. State Faraday’s First Law.

A. The mass of substance liberated or deposited or dissolved or underwent electrode reaction at an electrode

during the electrolysis of an electrolyte is directly proportional to the quantity of electricity passing through the

electrolyte.

4. Write the cell reaction for a : Ni|Ni ||Cu |Cu++ ++ , b: 21

Cu|Cu || Cl |Cl2

+ -

A. At LHS electrode oxidation half reaction ( )2aqNi Ni 2e+ -+ˆˆ†‡ˆˆ

At RHS electrode reduction half reaction aqCu 2e Cu++ -+ ˆˆ†‡ˆˆ

Cell Reaction ( )2

aqNi Cu Ni Cu++ ++ +ˆˆ†‡ˆˆ

5. Calculate the E of ( )Cu 0.1M |Cu++ electrode and ( ) 21

Cl 0.1M | Cl ,Pt2

- electrode

A. oCu /CuE 0.33V++ = ;

2

o1Cl /Cl2

E 1.36V-

= +

( )2

ocell 1 Cu /CuCl /Cl

2

E E E 1.36 0.33 1.03V++-= - = - + =

6. Write the equation for specific conductance.

A. Specific conductance K Ca

= ´l

where C = conductance, al

= cell constant

( ) ( )1

1 1 1 1 12

ohm cmunits ohm cm cgs ohm m SI Sm

cm

-- - - - -´

= = = =


7. Give the equation for equivalent conductance.

A. Equivalent conductance ( ) Specific conductance 1000 k 1000Normality N

´ ´L = =

Where k = specific conductance , N = Normality

( )1 3

1 2 12

ohm cm 1000 cmUnits ohm cm equiv in CGS

cm equiv

-- -´ ´ ´

= =´

( )1 2 1 2 1ohm m equiv SI Sm equiv- - -= =

8. What is fuel cell?

A. A fuel cell is a galvanic cell in which the chemical energy of fuel-oxidant system (combustion of fuel) is

converted directly into electrical energy.

SHORT ANSWER TYPE

1. State Faraday’s Laws of electrolysis.

A. Faraday’s First Law:

The mass of substance liberated or deposited or dissolved or underwent electrode reaction at an electrode

during the electrolysis of an electrolyte is directly proportional to the quantity of electricity passing through the

electrolyte.

m q or m eqµ =

q C t= ´

Where q = quantity of electricity in coulombs, C = current in amperes, t = time (in seconds)

e = electrochemical equivalent of the metal deposited.

Faraday’s Second Law:

If same quantity of electricity is passed through different electrolytic cells connected in series containing

different electrolytic solutions the masses of different species deposited or liberated or dissolved at the

electrodes are directly proportional to the chemical equivalents of the substance.

1 1

2 2

m Em E

=

Where 1 2m and m are masses of the substances deposited at electrodes.

1 2E and E are chemical equivalents of the substances which are deposited at electrode.

2. What are galvanic cells? Give two examples.

A. Galvanic cells are the electrochemical cells. Electrochemical cell is a device which make use of spontaneous

redox reaction for the generation of electrical energy. In this device chemical energy is transformed into

electrical energy. Daniel cell is an exmple.

The spontaneous oxidation – reduction reaction in the Daniel cell is

(1) ( ) ( ) ( ) ( )s 4 aq 4 aq sZn CuSO ZnSO Cu+ ® +

Representation of the cell: ( ) ( )aq aqZn|Zn ||Cu |Cu++ ++

(2) ( ) ( ) ( ) ( )s 4 aq 4 aq sNi PdSO NiSO Pb+ ® +

( ) ( )aq aqNi|Ni ||Pb |Pb++ ++


3. What is single electrode potential and write the Nernst equation.

A. A metal rod dipped in aqueous solution of its salt or a gaseous non-metal in contact with solution containing

its anion in the form of salt, function as single electrode.

The value of this potential depends on the chemical identity of the metal or nonmetal. The concentration of

the concerned ion in aqueous solution and temperature.

Nernst equation for single electrode potential.

(1) Cation or metal ion electrodes:

0.059

E E log Cn

= ° + where n is number of electrons transferred in the electrode reaction

(2) For anion electrodes:

0.059

E E log Cn

= ° -

4. What is equivalent conductance?

A. The conductance of a volume of a solution containing one gram equivalent of the electrolyte placed between

two parallel electrodes separated by unit length of 1 m or 1 cm is called equivalent conductance ( )L .

Equivalent conductance ( ) Specific conductance 1000 k 1000Normality N

´ ´L = =

Where k = specific conductance , N = Normality

( )1 3

1 2 12

ohm cm 1000 cmUnits ohm cm equiv in CGS

cm equiv

-- -´ ´ ´

= =´

( )1 2 1 2 1ohm m equiv SI Sm equiv- - -= =

5. What is Kohlrausch law?

A. The law states that the equivalent conductance ( )aL at infinite dilution of an electrolyte is equal to the

algebraic sum total of the equivalent conductances or mobilities of the cation ( )0+l and the anion ( )0

-l of the

electrolyte at infinite dilution ( )0 0 0 ions+ -L = l + l

The equivalent conductance of the ions at infinite dilution ( )0 0,+ -l l are known as ionic conductance at infinite

dilution. These are proportional to ionic mobilities ( )u ,u+ -

0 0u+ +l µ ; 0 0u- -l µ

or

0 0k u+ +l = ; 0 0k u- -l =

Where k is proportionality constant. Its value is found to be 96,500 Faradays.

0 00 0u or u

F F

+ -+ -l l

= =

6. What is lead accumulator?

A. In a battery or storage battery or secondary cell chemical changes occur during the charging of the cell with

current. These changes are reversed during discharging.


Acid storage cell or lead accumulator is an example of reversible storage cell.

It consists of two lead electrodes, sponge lead anode, lead coated with 2PbO as cathode. The charged cell

is represented simply as 2 4 2Pb|H SO |PbO .

At LHS electrode:

( )2Pb Pb 2e oxidation+ -+ˆˆ†‡ˆˆ

2 24 4Pb SO PbSO+ -+ ˆˆ†‡ˆˆ

At RHS electrode:

42 2PbO 2H O Pb 4OH+ -+ +ˆˆ†‡ˆˆ

4 2Pb 2e Pb+ - ++ ˆˆ†‡ˆˆ

Both the electrodes are reversible. During passing of electricity (using a charger), the accumulator is

“charged”. During the use of accumulator, reverse reactions of the above occur and the accumulator is

discharged.

7. What is a fuel cell?

A. Fuel Cell :

A fuel cell is a galvanic cell in which the chemical energy of fuel-oxidant system (combustion of fuel) is

converted directly into electrical energy.

In fuel cell, the fuel is oxidised at the anode. The fuel cell has two electrodes and an electrolyte. Fuel cell has

two electrodes and an electrolyte. Fuel cell are liquid fuel cells and Gaseous fuel cells.

Liquid fuel cells use methanol, ethanol, hydrazine, formaldehyde as fules. Gaseous fuel cells use Hydrogen,

alkanes, carbon monoxide as fules.

Oxygen, air, hydrogen peroxide, nitric acid are some oxidants used in fuel cells. The electrodes used are Pt,

porous PVC or PTFE (Teflon) coated with Ag to produce conducting surface.

8. What is corrosion?

A. The natural tendency of conversion of a metal into its mineral compound form on interaction with the

environment is known as corrosion.

The process of conversion may be chemical or electrochemical in nature. The electrochemical corrosion is

considered as the anodic dissolution of the metal undergoing corrosion. In anodic dissolution, the metal

getting corroded undergoes oxidation to the metal ion.

nM M ne+ -® +

Corrosion can be defined in electrochemical terms as anodic dissolution of the metal.

9. What is passivity?

A. A state of non-reactivity reached with time of action after an initial state of reactivity is passivity. Passivity of

the metal can be classified into

(1) Chemical passivity (2) Mechanical passivity (3) Electrochemical passivity

Passivity is explained on the basis of the formation of an invisible metal oxide layer on the metal. The oxide

layer is so thin that it is not visible to the naked eye. Its presence can be proved only through chemical

reactions.

10. What are electrolytes? Distinguish between electronic conductors (metallic) and electrolyte.

A. The substance which undergo dissociation to oppositively charged ions when electricity is passed through

them are called electrolytes. Ex: NaCl


Electronic conductors: The conductivity of them is due to movement of electrons. Ex: Metals

Increase in temperature increase the conductivity of electrolytes while decreases in electronic conductors.

Electrolytes conduct due to movement of ions while in electronic conductors it is due to movement of

electrons.

NUMERICALS

1. The resistance of 0.1 N KCl solution is found to be 702 W when measured in a conductivity cell. The specific

conductance of 0.1 N KCl is 1 10.14807 m- -W . Calculate the cell constant.

A. Ra

µl

R Sa

=l

Here R = 702 W ; k = 1 10.14807 m- -W

1 1 1R S

a

= ´ l where al

= cell constant

1

C k or C ka

a

= ´ ´ =l

l 1

0.14807702 a

= ´ =l

10.14807 702 103.9 ma

-= ´ =l

2. The specific conductance of 0.1 N KCl is 1 10.14870 m- -W . Calculate equivalent conductance.

A. Equivalent conductance ( ) Specific conductance 1000 k 1000N N

´ ´L = = (if k is given 1 1ohm cm- - )

If k is given in 1 1 1 2 1k 0.14870ohm m 1.487 ohm m eq

N 0.1- - - -L = = =

3. The electrode potentials of Cu |Cu++ and Ag |Ag++ electrode are +0.33 V and +0.8 V respectively. What

is the emf of the cell constructed from the electrodes.

A. oCu /CuE 0.33V++ =

oAg /AgE 0.8V++ =

Cell is ( ) ( )aq aqCu|Cu |Ag |Ag++ +

( )o oAg NiE E E 0.8 0.33 0.8 0.33 0.47° = - = - + = - =

4. The E° value of Zn |Zn++ electrode is 0.77V- . What is the E value of electrode containing 0.01 M 2Zn+

ions?

A. [ ]20.059 0.059E E log Zn 0.77 log 0.01 0.77 0.05 0.82V

n 2+é ù= ° + = - + = - - = -ë û .


THERMODYNAMICS VERY SHORT ANSWER TYPE

1. Explain system and surroundings.

A. System: System is that part of the universe which is under observation within a definite boundary.

Surrounding: Surrounding is that part of the universe other than the system. The surrounding and the

system are separated by the boundary which may be real or imaginary

2. Name three intensive properties.

A. Density, surface tension, viscocity.

3. Name three extensive properties.

A. Mass, volume, internal energy

4. Explain enthalpy.

A. Enthalpy is the amount of heat exchanged by a system with its surroundings at constant pressure and

temperature.

5. What is heat capacity? Explain.

A. Heat capacity (C) of a substance is defined as the amount of heat required to raise its temperature through

one degree.

6. What is thermochemical equation? Give one example.

A. The chemical equation in which heat change accompanying a reaction is also numerically specified with

proper sign by HD or ED by the side of the equation are known as thermochemical equation. In these

equations, the physical states of the reactants and the products are mentioned in brackets.

( ) ( ) ( )graphite 2 g 2 gC O CO+ ® ; H 393.5 kJD = -

7. State Hess’s law?

A. The total heat change in a reaction is the same whether the chemical reaction takes place in one single step

or in several steps.

8. What is Gibbs energy?

A. The thermodynamic function which involves both enthalpy (H) and entropy(S) functions.

G H TSD = D -

9. What is Gibbs equation?

A. G H T SD = D - D

Where G is Gibbs energy or Gibbs function, H is Enthalpy, S is entropy

SHORT ANSWER TYPE

1. State first law of thermodynamics or state and explain I law of thermodynamics.

A. The law is known as law of conservation of energy. The law states that the energy in a process may be

transformed from one form into the other but is neither created nor destroyed.

Mathematical formula of I law

E Q WD = - or Q E W= D +

For infinitesimally small changes q dE W= +

ED ® change in internal energy

Q ®heat content of the system

W ® work done by the system


2. Give the mathematical formulation of I law of thermodynamics.

A. Let a system in state A, has internal energy AE absorb from the surroundings a certain amount of heat ( )q Q

and undergo a change in its state to B. Let the internal energy of B, be BE

The increase in internal energy ED of the system is given by the equation B AE E ED = -

If ‘W’ is the workdone by the system in the process the net gain of energy ( )Q W- must be equal to

aED from first law. The increase in the internal energy of the system therefore is

( )B AE E E Q WD = - = - or Q E W= D +

For infinitesimally small changes q dE W= +

As per IUPAC conventions; heat absorbed by a system is given ‘+sign’, heat given out by a system is given ‘-

sign’. Work done by a system is given ‘-sign’ and work done on a system is given ‘+sign’.

3. State second law of thermodynamics.

A. II law of thermodynamics is stated in different forms:

(1) All spontaneous process are thermodynamically irreversible and entropy of the system increases

in all spontaneous process.

(2) Heat cannot be converted into work completely without causing some permanent changes in the

system involved or in the surroundings.

(3) Heat cannot flow from a colder body to a hotter body on its own.

4. What is entropy?

A. Entropy is taken as a measure of disorder of molecules or randomness of a system. The greater the disorder

of molecules in a system, the higher is the entropy. The entropy change SD between any two states is given

by the equation revqS

TD =

revq = Heat absorbed by the system isothermally and reversibly at ‘T’ during the state change.

5. State third law of thermodynamics.

A. The entropy of a pure and perfectly crystalline substance is zero at the absolute zero temperature ( )273 C- ° .

This is known as the third law of thermodynamics. Some times it is referred to as Nernst heat theorem.

III law of thermodynamics imposes a limitation on the value of entropy.

Tp

T0

CS .dT

T= ò

Entropy(S) of a substance at any temperature is calculated if the temperature dependence of PC is known in

evaluating the absolute value of entropy of any substance.

6. State and explain the significance of second law.

A. II law of thermodynamics is stated in different forms:

(1) All spontaneous process are thermodynamically irreversible and entropy of the system increases

in all spontaneous process.

(2) Heat cannot be converted into work completely without causing some permanent changes in the

system involved or in the surroundings.

(3) Heat cannot flow from a colder body to a hotter body on its own.

The II law of thermodynamics deals with the draw backs like


(1) Whether a process (or transformation) occurs in the specified direction on its own or not without

the intervention of any external agency.

(2) If a transformation or a process occurs, what fraction of one form of energy is converted into

another form of energy in this transformation or process.

7. Explain the terms entropy. Gibbs energy.

A. Entropy is taken as a measure of disorder of molecules or randomness of a system. The greater the disorder

of molecules in a system, the higher is the entropy. The entropy change SD between any two states is given

by the equation revqS

TD =

revq = Heat absorbed by the system isothermally and reversibly at ‘T’ during the state change.

Gibbs Energy

The thermodynamic function which involves both enthalpy (H) and entropy(S) functions.

G H TS= -

Gibbs Equation: G H T SD = D - D

Where G is Gibbs energy or Gibbs function, H is Enthalpy, S is entropy.

8. What are the criteria for the spontaneous nature of a chemical reaction? Give examples.

A. All natural process are spontaneous. When the disorderness increases the process will be spontaneous. A

process is said to be spontaneous if it occurs on its own without the intervention of any external agency.

Entropy increases in all spontaneous process. Entropy is thus taken as a measure of disorder of molecules or

randomness of a system. For a spontaneous process in an isolated system the entropy change is positive

( )S positiveD = .

total system surroundingS S S 0D = D + D >

Ex: Crystalline solid state is the state with lowest entropy and the gaseous or vapour state is the state with

highest entropy. In liquid state the entropy lies between the values for solids and the gaseous state.

9. State and explain enthalpy and internal energy terms of substance.

A. Enthalpy is the amount of heat exchanged by a system with its surroundings at constant pressure and

temperature.

The absolute value of H cannot be determined directly by experiment. But the change in enthalpy ( )HD

during a chemical change can be experimentally and directly determined.

products reactantsH H Hé ùD = -ë û

Reaction in which HD value is negative are called exothermic reactions and those which have HD positive

are endothermic reactions.

Internal energy:

The energy stored in the substances at constant temperature and pressure is called the internal energy and is

denoted by E.

The value of E depends on state conditions.

10. How are HD and ED related for a gaseous equilibrium reaction? What is ( )H ED - D for

( ) ( ) ( )3 s s 2 gCaCO CaO CO® +

A. H E P VD = D + D or H E nRTD = D + D

HD is change in enthalpy, ED is change in internal energy


( ) ( ) ( )3 s s 2 gCaCO CaO CO® +

nD = Gaseous products – Gaseous reactants = 1

H E nRTD - D = D or H E RTD = D +

11. Define and explain “heat of formation” of a compound.

A. The amount of heat evolved or absorbed during the formation of one mole of the compound at constant

temperature from the constituent elements in the standard state.

Compound which liberate heat in their formation reaction from the constituent elements are called exothermic

compounds. Compounds which absorb heat in the formation from the constituent elements are called

endothermic compounds.

fH veD = - for exothermic compounds , fH veD = + for endothermic compounds.

12. What is heat of neutralization? Explain.

A. Heat evolved when one gram equivalent of an acid is neutralized completely with one gram equivalent of an alkali, at constant temperature.

Neutralization process is always exothermic. But the amount of heat liberated depends on the fact whether the acid or base or both are strong or weak.

Ex: ( ) ( ) ( ) ( )2aq aq aqStrongacid Strong base

HCl NaOH NaCl H O ; H 57.3 kJ+ ® + D = -l

( ) ( ) ( ) ( )4 4 2aq aq aqStrongacid Strong base

HCl NH OH NH Cl H O ; H 51.46 kJ+ ® + D = -l

13. State Hess’s law of constant heat summation and explain it with an example.

A. The total heat change in a reaction is the same whether the chemical reaction takes place in one single step or in several steps.

Ex: 2CO is obtained from graphite and oxygen in two different path ways.

I path : ( ) ( ) ( )graphite 2 g 2 gC O CO+ ® ; H 393.5 kJD = -

II path: ( ) ( ) ( )graphite 2 g g1

C O CO2

+ ® ; H 110.5 kJD = -

( ) ( )2 g 2 g1

CO O CO2

+ ® ; H 283.02D = -

Total ( )( )H 110.5 283.02 393.52 kJD = - - + - = -

The two HD values are same.

14. What are pC and vC ?

A. Heat capacity (C) of a substance is defined as the amount of heat required to raise its temperature through one degree.

Heat capacity at constant volume vC gives the measure of the change of internal energy (E) of a system with temperature.

vv

EC

T¶æ ö= ç ÷¶è ø

Heat capacity at constant pressure pC gives the measure of change of enthalpy (H) of a system with

temperature.

pp

HC

T¶æ ö= ç ÷¶è ø


SURFACE CHEMISTRY VERY SHORT ANSWER TYPE

1. What is adsorption?

A. Concentration or accumulation of a substance on the surface of a solid or a liquid is known as adsorption.

2. What is absorption?

A. The process of concentration of molecules of gas or liquid on the surface of the solid and throughout the solid

is called absorption.

3. Give the mathematical equation relating the pressure (P) and the extent of adsorption (x/m) of gas on a metal.

A. 1/nxP

mµ

1/nxKP

m= at constant temperature

Where x is amount of gas adsorbed, m is mass of adsorbent, P is adsorption equilibrium pressure K and n are

adsorption constants.

4. What is physisorption?

A. If the molecules of the adsorbate are held on the surface of adsorbent with physical forces or van der Waal’s

forces, the process is called physisoaption or physical adsorption.

Ex: Adsorption of 2H or 2O on charcoal.

5. What is the chemisorption?

A. If the molecules of the adsorbate are held on the surface of adsorbent with chemical forces, the process is

called chemisorption or chemical adsorption.

Ex: 2H gas passed over Ni surface

6. What is catalyst?

A. The substance which increases rate of a reaction to which it is added without itself undergoing chemical

change is called catalyst.

7. What is negative catalyst?

A. A catalyst which decreases the rate of reaction is called negative catalyst.

8. What is homogenous catalysis? Give example.

A. If the catalyst and the reactants are in the same phase it is called homogeneous catalysis.

Ex: ( ) ( )( )

( )gNO

2 g 2 g 3 g2SO O 2SO+ ¾¾¾¾®

The reactants and the catalyst are in same phase.

9. What is heterogeneous catalysis?

A. If the catalyst and the reactants are in different phase it is called heterogeneous catalysis.

Ex: ( ) ( )( )

( )sPt

2 g 2 g 3 g2SO O 2SO+ ¾¾¾®

10. What are the catalyst used in the reactions?

( ) ( ) ( )2 g 2 g 3 gN 3H 2NH+ ® and ( ) ( ) ( )2 g 2 g 3 g2SO O 2SO+ ®

A. The catalyst used in the formation of ammonia is Fe.

The catalyst used in the formation of sulphur trioxide is NO or Pt.


11. What is auto catalysis? Give two examples.

A. If the product of the reaction acts as a catalyst, for the reaction itself, the process is called autocatalysis.

Ex: Oxidation of oxalic acid by 4KMnO in acidic medium. The 2Mn+ ion formed in the reaction is acting as

catalyst.

12. What is lyophilic colloid? Give example.

A. The colloidal solution in which the dispersed phase has great affinity with the dispersion medium is called

lyophilic collidal solution.

Ex: Starch solution.

13. What is lyophobic colloid? Give example.

A. The colloidal solution in which dispersed phase has not much affinity with the dispersion medium is called

lyophobic colloid.

Ex: Gold solution

14. What is emulsion? Give example.

A. Colloidal solution in which both the dispersed phase and dispersion medium are liquids is called emulsion.

Ex: Milk – fat in water emulsion; cold cream – water in fat

15. What is micelle?

A. A colloidal sized particle formed in water by the association of simple molecules each having a hydrophobic

end and hydrophobic end.

Ex: Higher concentration of soap solution.

16. What is gold solution?

A. It is an aqua solution or hydro solution. It is colloidal suspension of gold particles in water.

Dispersed phase: Gold particles (solid)

Dispersion medium: Water (liquid)

17. What is milk?

A. This is liquid in liquid type of colloid. Liquid fat is dispersed in water.

Dispersed Phase: Liquid fat ; Dispersion medium: Water

18. What is emulsifying agent?

A. The third substance added in small amounts to an emulsion to keep the emulsion stable is emulsifier or

emulsifying agent.

19. What is cloud?

A. It is an aerosol. Water (drops) disperses in air and forms the colloidal solution.

Dispersed phase: Water drops (liquid)

Dispersion medium: Air (gas)

20. What is smoke?

A. This is an aerosol. It is formed by the dispersion of carbon particles in air.

Dispersion phase: Carbon particles (solid)

Dispersion medium: Air (gas)

21. What is tyndall effect?

A. When light passes through a solution we will be able to see the path of light as a luminous beam. This is

called tyndall effect.


22. What is gold number?

A. The minimum mass in milligrams of a dry lyophilic solution which is required to be added to 10 mL of standard

gold to just prevent its coagulation by addition of 1 mL of 10% NaCl solution or red gold solution to become

blue gold solution.

23. What is protective colloid?

A. A lyophobic solution can be protected from coagulation by adding electrolyte, which is lyophilic colloid. This

lyophilic solution added is called protective colloid.

24. What is Brownian movement?

A. The zigzag movement of colloidal particles in colloidal solution is called Brownian movement.

25. What is coagulation?

A. Precipitation of lyophobic colloids on addition of electrolytes is called coagulation of flocculation.

SHORT ANSWER TYPE

1. What is adsorption? Explain with two examples.

A. Accumulation or concentration of a substance on the surface of a solid or a liquid is known as adsorption.

The molecules of gases or liquid or the solutes in solutions get adhered to the surface of solids. When the

solids are kept in contact with them for sufficiently long time.

(Adsorption occurs more effectively if the surface of the solid is clean and free from the surface impurities)

Ex. 1: Activated charcoal adsorbs gases like 2CO , 2SO , 2Cl etc.

Ex. 2: Pt or Ni metal kept in contact with a gas - Hydrogenation of oils

In the process of adsorption, two substances are involved one is solid or liquid on which adsorption occurs.

The second is the gas or liquid or solute from a solution which gets adsorbed on the surface. These two

substances are called adsorbent and adsorbate.

2. What is absorption? Explain with two examples?

A. The process of concentration of molecules of gas or liquid on the surface of the solid and throughout the solid

is called absorption.

In absorption the molecules of a gas, liquid or a solute present in a solution are not only present on the

surface but also pass through the surface of the solid/liquid into the bulk of the solid or liquid. The substance

that passes into the bulk of the solid or liquid is uniformly distributed. Absorption is a bulk phenomenon.

Ex: 1. If a chalk piece dipped into a solution of coloured ink and kept for some time, the chalk piece absorbs

the coloured substance.

2. A sponge placed in water absorbs water into it.

3. What is physical adsorption? Give example.

A. Physical adsorption is also called as physisorption. If the forces responsible for the adsorption of adsorbate

molecules on the surface of the adsorbed are physical forces or van der Waal’s forces, the adsorption is

referred as physical adsorption or physisorption. It is a weak force and is reversible.

Ex: Adsorption of 2H or 2O on charcoal.

4. What is chemical adsorption? Give example.

A. Chemical adsorption is also called as chemisorption (or) activated adsorption.

If the forces that are responsible for the adsorption of adsorbate molecules on the surface of the adsorbent

are chemical forces (i.e. chemical bond formation). The adsorption is called chemical adsorption or

chemisorption. It is a strong force and irreversible.

Ex: Adsorption on Ni metal surface, 2H molecules are held by valence forces on the surface.


5. Write any forces differences between physical and chemical adsorption.

A. Property Physical adsorption Chemical adsorption

(1) Nature of adsorption Held by Physical forces or van der

Waal’s forces hence weak

Held by chemical bond

formation, hence strong

(2) Reversibility of adsorption

process

Reversible and occurs rapidly Irreversible and occurs slowly

(3) Temperature at which

adsorption is more

pronounced

Low temperature (below the boiling

point of the adsorbate gas)

High temperature (generally

above the boiling point of the

adsorbate gas

(4) Effect of change in

temperature

Decreases with rise in temperature Increases with rise in

temperature

6. Give the factors that effect adsorption of gases on metals.

A. The extent of adsorption of gases depend on the following factors.

(1) Surface area of adsorbent

The extent of adsorption is directly proportional to the surface area. Greater the surface area, greater is the

extent of adsorption. The amount of gas adsorbed by unit mass of the adsorbent xm

increases with increase

in surface area of adsorbent. Gases like 2 2 3 2H ,N ,CO,NH ,SO are adsorbed on the surface of finely

divided transition metals such as Ni, Co, Pt and Pd.

(2) Nature of adsorbate (gas)

Easily liquefiable gases are generally adsorbed to great extent. 2 3 2SO ,NH ,HCl,CO are easily liquefied.

Hence they are easily adsorbed.

(3) Pressure of the adsorbate (gas)

In case of physisorption the gas forms monolayer at low pressure and multilayer at high pressure. Increase in

pressure leads to increase in extent of adsorption.

(4) Temperature

Physisorption takes place at low temperature while chemisorption takes place at high temperature.

7. What is colloidal solution? Give example.

A. Colloidal solution contains two phases dispersed phase and dispersion medium. The continuous medium of

the colloidal solution in which disperse phase particles are dispersed is dispersion medium.

Colloidal solution is a binary system in which the particle size of the dispersed phase (solute) is of the order

1m 1m - m . It is heterogeneous binary system. (1 millimicon = 910- m)

Colloidal solutions are classified into lyophilic solutions and lyophobic solutions.

Lyophilic solution - Ex: Starch solution

Lyophobic solution – Ex: Gold solution

8. Give any four differences between true solution and colloidal solutions.

A. Property True solution Colloidal solution

(1) Particle size 1m< m 1m 1m - m

(2) External appearance Very clear Generally clear or opaque.

(3) Nature of system Homogeneous Heterogeneous

(4) Tyndall effect Not shown Shown


9. What are lyophilic colloids and lyophobic colloids?

A. The colloidal solution in which the dispersed phase has great affinity with the dispersion medium is called

lyophilic collidal solution.

The colloidal solution in which dispersed phase has not much affinity with the dispersion medium is called

lyophobic colloid.

10. What is emulsion? Give two examples.

A. Colloidal solution in which both the dispersed phase and dispersion medium are liquids is called emulsion.

Ex: Milk – fat in water emulsion; cold cream – water in fat

Both milk and cold cream are liquid in liquid type of colloid. Liquid fat is dispersed in water.

Dispersed Phase: Liquid fat

Dispersion medium: Water

Gold solution:

It is an aqua solution or hydro solution. It is colloidal suspension of gold particles in water.

Dispersed phase: Gold particles (solid)

Dispersion medium: Water (liquid)

11. How are emulsions classified? Give examples.

A. Emulsions are classified into two classes. They are (1) Oil in water (2) water in oil

(1) Oil in water (o/w) type solutions:-

In this type of emulsion the dispersed phase is oil (immiscible liquid) and the dispersion medium is water.

Ex: Milk, Vanishing cream

(2) Water in oil (w/o) type emulsion:- In this type of emulsion the dispersed phase is water and the

dispersion medium is oil (immiscible liquid)

Ex: Stiff greases: Water in lubrication oils ; Cold liver oil: Water in cod liver oil

12. Explain the cleaning action of soap.

A. Clothes which are to be washed contain greeze or fat adhered to cloth called dirt.

In cleaning process dirt forms an emulsion with water which is used for cleaning. This emulsion is not stable,

it is stabilized by converting the dirt into micelle by the soap. Soap dissolves in water and gives Na+ and

stearate ions containing hydrophobic end (alkyl group end). This is the tail part of anion. The stearate ion

also contain hydrophilic end ( )COO- . This is called head part of anion. The tail part dissolves the dirt and

forms micelle and this micelle is removed by water in the cleaning process.

13. What is catalyst? Give two examples.

A. A catalyst is the substance that increase the rate of a chemical reaction to which it is added without itself

being consumed in the reaction.

Ni

2 4 2CatalystCO 3H CH H O+ ¾¾¾¾® +

Cu

2 CatalystCO H HCHO+ ¾¾¾¾®

14. How is catalysis classified? Give examples.

A. Catalysis is classified into two types based on the physical state (phase) of the catalyst and the reactant.

They are homogeneous and heterogeneous catalyst.

Homogeneous catalysis: In both the reactants and catalyst are present in same phase then it is called

homogeneous catalysis.


Ex: ( ) ( )( )

( )gNO

2 g 2 g 3 g2SO O 2SO+ ¾¾¾¾®

( ) ( )( )

( )gNO

g 2 g 2 g2CO O 2CO+ ¾¾¾¾®

Heterogeneous catalysis: If the reactant and catalyst are present in different phases it is called

heterogeneous catalysis.

( ) ( )( )

( )sPt

2 g 2 g 3 g2SO O 2SO+ ˆˆˆ †̂‡ˆˆˆ̂

15. What is autocatalysis? Give two examples.

A. In a reaction if one of the intermediate product formed itself functions as catalyst it is called autocatalysis.

Ex: (1) 4 2 2 4 2 4 4 2 4 2 22KMnO 5H C O 3H SO 2MnSO K SO 8H O 10CO+ + ® + + +

4MnSO formed acts as autocatalyst.

(2) 3 22AsH 2As 3H® +

As formed functions as autocatalyst.

16. What is tyndall effect? Explain.

A. It is an optical property. When light enters a colloidal solution, it is scattered by the large sized colloidal

particles. When light passes through a solution we will be able to see the path of the light as luminous beam.

This is called tyndal effect. This luminescent path can be viewed through a microscope placed at right angles

to the direction of the path of the light. The bright come of light is tyndall cone. Tyndall effect is observed

when

(1) The diameter of dispersed particles is not much smaller than the wavelength of light used and

(2) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

17. What is coagulation? Write about Hardy-Schulze law.

A. The phenomenon of colloidal substance loosing charge and coming down as a precipitate on addition of an

electrolyte is called coagulation or flocculation.

The effectiveness of an ion or electrolyte in causing coagulation is dependent on the sign of the charge and

magnitude of charge. This fact is explained by Hardy and Schulze. Hence the law or rule is known as Hardy-

Schulze rule.

Hardy Schulze Law

The ion with charge opposite to the charge of the colloid particle is effective in coagulating the colloid.

Greater the charge of the ion greater is the coagulating ability of the ion.

Positive colloids are coagulated by negative ions or anions of the salt added.

2 34 4Cl SO PO- - -< <

Negative colloide are coagulated by positive ions or cations of the salt added.

2 3K Ba Al+ + +< <

18. What is Brownian movement?

A. Robert Brown observed that pollen grains when suspended in a liquid and absorbed under microscope show

random motion of pollen grains. This type of motion is known as Brownian movement. All colloidal particles

in colloidal solution exhibit this type of motion. The motion becomes less vigorous as the size of the particle

increases and also with the viscocity of the dispersion medium but increases with rise in temperature.

Wwish you all the bestW

ipe second year - fiitjee · ipe second year `kukatpally centre ... inorganic chemistry va group...

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