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Grade XIChemistry

Exam Preparation Booklet

Chapter Wise - Important Questions and Solutions

Questions

Some Basic Concepts of Chemistry Q1. Differentiate between diffusion and osmosis.

(2 marks)

Q2. An organic compound contains 56.8 % carbon, 7.32 % hydrogen and the remaining oxygen. The vapour density of the compound is 68. Calculate the molecular formula of the compound.

(2 marks) Q3. Calculate the number of moles produced when 400 g of hydrogen reacts with nitrogen to

form ammonia. (2 marks)

Q4. What is the limiting reactant when 0.45 mole of Zn reacts with 0.7 mole of HCl to form

ZnCl 2 and H 2 ? (2 marks)

Structure of Atom

Q1. Explain the following:

(i) Dual behaviour of matter (ii) Heisenberg’s Uncertainty Principle and its significance

(2+3 marks) Q2. If the number of photons emitted by a 100 watt bulb is 4.01 × 10 20 , then what is the

wavelength of the electromagnetic radiation to be emitted? (2 marks)

Q3. Calculate the energy associated with the second orbit of Li + and the radius of this orbit.

(2 marks) Q4. What is the ratio of the first spectral line of Balmer and Paschen series of hydrogen

spectrum? (2 marks)

Classification of Elements and Periodicity in Properties

Q1. (i) Arrange the following in the order of increasing ionic size: Al 3+ , F ­ , O 2­ , Na +

(ii) How electronegativity affects the non­metallic character of elements?

(iii) Why does the ionic radius decreases on going from left to right in a periodic table? (3 marks)

Q2. The variation of ionisation enthalpy with atomic numbers for the elements of 2 nd period

of the Modern Periodic table is shown below. Which elements are likely to represent A and F?

(2 marks)

Q3. Use the following information to answer the question.

Element Ionisation energy (kJ/mol)

P 2080

Q 899

R 520

S 1681 (i) Which of the following elements has the highest metallic character? (ii) Which of the following elements belongs to the noble gas group? (iii) Which of the following elements is most likely to be a halogen? (iv) Which of the following elements acts as the strongest reducing agent? (v) Out of element P and R, which one has high hydration energy?

(5 marks)

Q4. Account for the following statement: Li shows anomalous behaviour compared to other group members.

(5 marks)

Chemical Bonding and Molecular Structure Q1. Draw the molecular orbital diagram of O 2 and calculate its bond order.

(3 marks) Q2. Consider the information given in column I and column II of the following table.

Column I Column II A. Trigonal planar P. NF3 and PH3

B. Tetrahedral Q. NO3− and CO32− C. Pyramidal R. SO42− and ClO4− D. Square pyramidal S. IF5 and BrF5

Match the given columns and draw the structures of the corresponding species.

(4 marks) Q3. What gives rise to the polarity in covalent bonds? What do you understand by dipole

moment and how will you explain that the resultant dipole moment in NF 3 is equal to 0.80 × 10 −30 C m, whereas the resultant dipole moment in NH 3 is 4.90 × 10 −30 C m.

(3 marks) Q4. Giving reasons, arrange the given species in order of their decreasing stabilities.

(5 marks)

States of Matter

Q1. A gas occupies a volume of 300 cm 3 at a temperature of 27°C and pressure of 760 mm.

What will be the volume of the gas at – 4°C and 760 mm pressure? (2 marks)

Q2 . (i) Define Boyle’s Law.

(ii) A gas occupies a volume of 900 cm 3 at a pressure P. If the pressure is altered to 3.5 atms, the volume of the gas is found to be 1000cm 3 . Calculate the value of initial pressure (P).

(2 marks) Q3. (i) State Dalton’s law of partial pressure.

(ii) A mixture of gases A (molecular mass = 23) and B (molecular mass = 30) contains 90 g of A and 119 g of B. If this mixture is present in a container at 10 bar, then what will be the partial pressure exerted by the two gases?

(1+2 marks)

Q4. Which two postulates of kinetic theory of gases do not hold true for real gases? How are these postulates corrected in van der Waals’ equation?

(4 marks) Q5. What gives rise to surface tension in liquids? What do you understand by viscosity of a

liquid? What is the effect of increase in temperature on surface tension and viscosity? (3 marks)

Thermodynamics

Q1 . Two moles of a gas are allowed to expand isothermally and reversibly at 30°C from a

volume of 10 m 3 to 80 m 3 . Calculate w, Δ U and q. (3 marks)

Q2 . The complete combustion of 5.6 g of benzene evolved 424 kJ of heat. This heat was

measured at constant volume and at a temperature of 30°C. What will be the heat of combustion of benzene at constant pressure?

(2 marks)

Q3 . What do you understand by the heat capacity of a substance? Derive the relation between

C p and C v . (3 marks)

Q4 . Calculate the heat of combustion of glucose (C 6 H 12 O 6 ) from the given data.

C (graphite) + O 2 (g) → CO 2 (g) (ΔH = − 465.0 kJ) H 2 (g) +(½)O 2 (g) → H 2 O (l) (ΔH = − 342.8 kJ) 6C (graphite) + 6H 2 (g) + 3O 2 (g) → C 6 H 12 O 6 (s) (ΔH = −1368.9 kJ)

(3 marks)

Q5 . Calculate for the reaction if the bond enthalpy of C−H bond = 413 kJ/mol, Cl−Cl bond = 243 kJ/mol, C−Cl bond = 328 kJ/mol and H−Cl bond = 432 kJ/mol.

(2 marks)

Equilibrium

Q1. Derive a relation between K p and K c for the given reaction.

(3 marks)

Q2. (i) Write the conjugate bases for each of the following acids:

HS − , H 2 SO 4 (ii) Write the conjugate acids for each of the following bases:

(2 marks)

Q3. Calculate the percent dissociation of PCl 5 , if 0.10 mol of PCl 5 is kept in 16.0 L flask at

600 K. The value of K c for the reaction is 4.2 × 10 −2 . (3 marks)

Q4. Calculate the value of ΔG ° at 50°C for the given reaction, for which the equilibrium

constant is 4.0 × 10 −7 . What will be the effect on equilibrium constant if the concentration of SO 2 is increased?

(2 marks)

Q5. Total pressure for the reaction at equilibrium was found to be 8.97 bar. If 15.8 g of N 2 O 4 was placed in a 1 L reaction vessel at 400 K, then calculate K c , K p and partial pressures at equilibrium.

(4 marks)

Redox Reactions Q1. Balance the following reaction in acidic medium using ion electron method.

(i)

(ii) Show all the steps involved.

(4 marks) Q2. Answer the following:

(i) What are disproportionation reactions? Explain giving suitable examples. (ii) What products are obtained when ClO ­ 3 and SF 6 undergo disproportionation

reaction? (4 marks)

Q3. Determine the oxidation number of each of the bromine atom in the given structure of Br 3 O 2 .

(2 marks)

Hydrogen Q1. What is the strength of a solution of H 2 O 2 marked as 15 V? Why is H 2 O 2 kept away from

dust? (2+1 marks)

The s­Block Elements

Q1. Why is an ammoniated solution of sodium metal paramagnetic and blue in colour?

(1 mark) Q2. Explain the following.

(i) Despite of having more negative reduction potential than sodium, Li does not react readily with water.

(ii) Hydrides and carbonates of Na and K are soluble, while that of Mg and Cs are insoluble in water.

(4 marks) Q3. Among BaCO 3 , CaCO 3 , SrCO 3 and BeCO 3 , which will require the

(i) highest energy to undergo decomposition? (ii) least energy to undergo decomposition?

(2 marks) Q4. Answer the following:

(i) How do the properties of beryllium differ from rest of the alkali earth metals? (ii) What are the similarities in the properties of beryllium and aluminium? (iii) Draw the structure of BeCl 2 in solid state.

(2+2+1 marks)

The p­Block Elements

Q1. Explain the inert pair effect. Tl(I) compounds are more common than Tl(III) compounds. Comment.

(3 marks) Q2. Justify the given statement:

(CH 3 ) 3 N is pyramidal while (SiH 3 ) 3 N is tetrahedral. (3 marks)

Q3. Answer the following:

(i) How is boric acid prepared and what are its physical properties? (ii) Why boric acid acts like a weak acid?

(2+1 marks)

Organic Chemistry ­ Some Basic Principles and Techniques Q1. Answer the following:

(i) Explain inductive effect. (ii) Comment on the stability of carbocations, carbanions and free radicals on the bases

of the inductive effect. (3+3 marks)

Q2. Write the IUPAC names of the following compounds .

(i)

(ii)

(iii)

(3 marks)

Q3. 0.4950 g of an organic compound gave on combustion 0.9900 g of carbon dioxide and

0.4050 g of water. Calculate the percentage of carbon and hydrogen in it. (2 marks)

Q4. How does the hyperconjugation effect contribute towards the stability of ethyl

carbocation? (2 marks)

Hydrocarbons Q1. What do you understand by conformations? Explain Sawhorse and Newman projections

using the example of ethane. (3 marks)

Q2. Which of the given compounds is not aromatic?

(3 marks)

Q3. The ozonolysis of an alkene forms methanol and propanone. What is the most likely

structure of the alkene? (2 marks)

Environmental Chemistry

Q1. What are the harmful effects of photochemical smog?

(3 marks)

Solutions

Some Basic Concepts of Chemistry Ans 1.

Diffusion Osmosis

(i) It is the movement of molecules from a region of higher concentration to a

region of lower concentration.

(i) It is a special case of diffusion wherein the medium is water.

(ii) It does not require a membrane. (ii) It requires a semi permeable membrane.

Ans 2.

Element Percentage Atomic mass

Moles of atom Atomic ratio Simplest ratio

C 56.8 12 56.8/12=4.73 4.73/2.24=2.1 2

H 7.32 1 7.32/1=7.32 7.32/2.24=3.3 3

O 35.88 16 35.88/16=2.24 2.24/2.24=1 1

∴ Empirical formula = C 2 H 3 O Empirical formula mass = 12 × 2 + 1 × 3 + 16 × 1= 43 g Molecular mass = 2 × Vapour density = 2 × 68 = 136 Hence, the molecular formula = (E.F.) n = (C 2 H 3 O) 3 = C 6 H 9 O 3

Ans 3. The given chemical reaction can be represented as:

The number of moles of H 2 in 400 g

Now, 3 moles of hydrogen react with 1 mole of hydrogen to produce 2 moles of ammonia.

200 moles of hydrogen react with (200/3) moles of hydrogen to produce (2x200)/3 moles of NH 3. Therefore, moles of NH 3 produced = 133.33 moles

Ans 4. The balanced equation for the reaction is Zn + 2HCl → ZnCl2 + H 2 1 mole Zn reacts with 2 moles of HCl.

∴ 0.45 mole Zn will react with moles of HCl. The HCl used in the reaction is only 0.7 mole (given). Thus, Zn will not completely react with HCl.

∴0.7 mole of HCl will react with moles of Zn. HCl will completely react with Zn. Hence, HCl is the limiting reactant in the given reaction.

Structure of Atom Ans 1. Dual behaviour of matter:

de­Broglie proposed that like radiation, matter also exhibits dual nature, i.e., matter also exhibits properties of both a wave and a particle. Therefore, like a photon, an electron should also possess wave as well as particle nature. With regard to this, he derived an equation, known as de­Broglie equation.

Where, λ = Wavelength of the particle h = Planck’s constant p = Momentum of the particle m = Mass of the particle v = Velocity of the particle Heisenberg’s Uncertainty Principle : According to this rule, it is impossible to determine the exact position and the exact momentum (velocity) of an electron. To locate the position of an electron, it needs to be

illuminated with high­intensity photons. When these high­energy photons collide with electrons, they change their velocity by transferring energy to them. Hence, the position and the momentum of an electron can never be determined accurately at the same point of time. This can be shown mathematically as follows:

Where, Δ x = Uncertainty in position Δ p x = Uncertainty in momentum Δ v x = Uncertainty in velocity m = Mass of electron Significance of Heisenberg’s Uncertainty Principle : It rules out that electrons and other similar particles have specific paths or trajectories. This is because the trajectory of a substance is determined by its location and velocity at various moments/instances.

Ans 2. Number of photons emitted by the bulb = 4.01 × 10 20

Energy of 1 photon = 100/ (4.01 × 10 20 ) = 24.938 × 10­ 20 J

Ans 3. We know that,

Therefore, energy for Li + , n = 2, Z = 3

Ans 4. According to Rydberg equation,

For first line of Balmer series,

For first line of Paschen series, n 1 = 3; n 2 = 4

Hence, the ratio of Balmer and Paschen series is 7:20.

Classification of Elements and Periodicity in Properties Ans 1. (i) Al 3+ < Na + < F ­ < O 2­

(ii) Non­metallic character is directly related to electronegativity. As we move from left to right of a period, the outermost electrons feel greater pull of the nucleus. This is because the nuclear charge increases and atomic size decreases from left to right. Thus, the elements on the right of the periodic table do not lose electrons easily, so these are non­metallic in nature. In this way, electronegativity affects the non­metallic character of elements.

(iii) On going from left to right, the effective nuclear charge increases, as number of

electrons keep on increasing in the outermost shell. Due to this, electrostatic attraction between the nucleus and the electrons increases and ionic size decreases.

Ans 2. A represents the element lithium (Li) and F represents the element neon (Ne). This is

because the ionisation energy for A i s very less. So, A must lie at the extreme left of the 2 nd period. F on the other hand, has the maximum ionisation energy, so it must have a stable electronic configuration.

Ans 3. (i) R

(ii) P (iii) S (iv) R (v) R

Ans 4. Li shows anomalous behaviour compared to other group members due to the following

reasons: • Li ion (Li + ) is very small. • Li has high electronegativity. • Li ion (Li + ) has large charge/radius ratio. • Li does not have any d ­electron in its valence shell. The positive values of the second electron gain enthalpy can be explained on the basis of the following reason:

After the addition of one electron, the atom acquires a negative charge. Now, the second electron has to be added to a negatively charged ion and this addition is opposed by coulombic repulsions. Thus, the energy has to be supplied to force the second electron into an ion.

Chemical Bonding and Molecular Structure Ans 1. Number of electrons in O 2 = 2 × 8 = 16

Electronic configuration of the molecule = σ1 s 2 , σ * 1 s 2 , σ2 s 2 , σ * 2 s 2 , σ2 p z 2 , π2 p x 2 = π2 p y 2 , π * 2 p x 1 = π * 2 p y 1

Bond order

Thus, oxygen molecule has two bonds (one sigma and one pi bond).

There are two unpaired electrons in the molecule. Thus, it is paramagnetic.

Ans 2. The structures of the given species are as under:

Ans 3. In the case of homoatomic molecules like H 2 and O 2 , the shared pairs of electrons are attracted by the two atoms in a similar manner. Hence, they are present at the centre of the bond.

However, in the case of heteronuclear molecules, the shared pairs of electrons are attracted more strongly by one atom depending on the electronegativity of the atoms. Thus, the shared pairs of electrons remain closer to the more electronegative atom. This gives rise to the polarity in covalent bonds.

Dipole moment is defined as the product of the magnitude of charge and the distance between the centres of the positive and the negative charges.

Dipole moment = Charge × Distance of separation We know that N is more electronegative than H, but less electronegative than F. Therefore, as shown in the figure, the dipole moment in NF 3 due to N­F bonds opposes the dipole moment due to the lone pair present on N. However, in NH 3 , the dipole moment due to N­H bonds enhances the dipole moment due to the lone pair of electrons, making the value of the resultant dipole moment extremely high.

Ans 4. The respective electronic configurations of and can be written as follows:

In , there are 10 electrons in the bonding molecular orbitals, while there are 8 electrons in the anti­bonding molecular orbitals.

Hence, bond order of = = = 1

In , there are 10 electrons in the bonding molecular orbitals, while there are 7 electrons in the anti­bonding molecular orbitals.

Hence, bond order of = = 1.5

In O 2 , there are 10 electrons in the bonding molecular orbitals, while there are 6 electrons in the anti­bonding molecular orbitals.

Hence, bond order of O 2 = = = 2

In , there are 10 electrons in the bonding molecular orbitals, while there are 5 electrons in the anti­bonding molecular orbitals.

Hence, bond order of = = = 2.5 Greater the bond order, more stable is the molecule. Therefore, the decreasing order of the stability of given species are as follows:

States of Matter Ans 1. Given that,

Initial volume of gas (V 1 ) = 300 cm 3 Initial temperature of gas (T 1 ) = 27°C = (273 + 27) K = 300 K Final temperature of gas (T 2 ) = –4°C = (273 – 4) = 269 K Final volume of gas (V 2 ) = ?

According to Charles’ Law :

∴

⇒ V 2 = = 269 cm 3 Ans 2. (i) Boyle’s Law states that the absolute pressure and volume of a given mass of gas

within the closed system, are inversely proportional to each other, if the temperature remains unchanged.

(ii) Given that, P 2 = 3.5 atm, V 1 = 900 cm 3 , V 2 = 1000 cm 3 According to Boyle’s Law, P 1 V 1 = P 2 V 2 P × 900 = 3.5 × 1000

⇒ P = Therefore, P = 3.88 atm

Ans 3. (i) Dalton’s law of partial pressure: It states that the total pressure exerted by the

mixture of non­reactive gases is equal to the sum of the partial pressures of the individual gases. p total = p 1 + p 2 + p 3 +…….

(ii)

Ans 4. The two postulates of kinetic theory of gases that do not hold true for real gases are as

follows: • There is no force of attraction between the molecules of a gas. • The volume of the molecules of gas is negligibly small in comparison to the space

occupied by the gas. Correction because of van der Waals’ equation of state: Pressure correction: Attractive forces exist between molecules of real gas. When a molecule strikes the wall of a container, it experiences an inward pull because of the attractive force exerted by the neighbouring molecules inside the vessel. Thus, the molecules strike the wall with a lesser force because of the presence of attractive forces.

In other words, the observed pressure is lesser than the ideal pressure. Hence, Van der Waals added the correction term ‘ a ’ to the observed pressure ' p' , which is equal to (a n 2 / V 2 ) . Then, p corrected for n moles of gas. Volume correction: The volume occupied by the molecules cannot be neglected in comparison to the total volume. Hence, the ideal volume of a gas is less than its observed volume. van der Waals, thus, subtracted the correction term ‘ b ’ from the observed volume i.e., V corrected = ( V − nb ) for n moles of gas

Ans 5. In liquids, the molecules present in the bulk experience equal force from all directions.

However, the molecules present on the surface experience force only from below, as shown in the figure.

Therefore, a molecule present on the surface experiences a net inward force. Due to these net forces, the surface of a liquid behaves like a stretched membrane. This is known as surface tension. It is defined as the force acting per unit length perpendicular to the line drawn on the surface of the liquid. The viscosity of a liquid is defined as its measure of resistance to flow. It arises due to the internal friction between the layers of a fluid. Stronger the forces of attraction of a liquid, more viscous it will be. We know that the kinetic energy of particles increases on increasing temperature. Therefore, intermolecular forces of attraction become less effective when temperature is increased. As intermolecular forces of attraction decreases, both the surface tension and viscosity decrease.

Thermodynamics Ans 1. It is given that,

V 1 = 10 m 3 , V 2 = 80 m 3 , n = 2 moles R = 8.314 JK −1 mol −1 T = 30 + 273 = 303 K (i) Work done, w can be calculated as follows:

(ii) Δ U = 0 for an isothermal expansion. (iii) Heat, q can be calculated as follows:

Δ U = q + w Now, for isothermal expansion, Δ U = 0 ∴ q = − w Or , q = 5239.35 J mol −1

Ans 2. The combustion of benzene is represented as follows:

C 6 H 6 ( l ) + (15/2)O 2 ( g ) → 6CO 2 ( g ) + 3H 2 O ( l )

Heat evolved in the complete combustion of 5.6 g of benzene = 424 kJ ∴ Heat evolved in the complete combustion of 78 g (1 mole) of benzene = (78/5.6) × 424 = 5 905.71 kJ Thus, heat liberated at constant volume = −5 905.71 kJ Here, Δ n = (6 − 7.5) = −1.5 T = 273 + 30 = 303 K Δ H = Δ E + Δ n R T = −5 905.71 + (−1.5) × 8.3 × 10−3 × 303 = −5 909.48 kJ mol−1 Hence, the heat of combustion of benzene at constant pressure is −5 909.48 kJ mol−1.

Ans 3. The increase in the temperature of a substance is directly proportional to the heat supplied to it. q ∝ Δ T q = C Δ T Here, C is known as the heat capacity of the substance and depends upon its size, nature and composition. At constant volume, heat capacity is denoted as C v and at constant pressure, heat capacity is denoted as C p . Therefore, at constant volume

q v = C v Δ T = Δ U At constant pressure, q p = C p Δ T = Δ H Now, we know that for 1 mole of an ideal gas, Δ H = Δ U + Δ( pV ) = Δ U + Δ(R T ) = Δ U + RΔ T Now, substituting the values of Δ H and Δ U in the equation, we get C p Δ T = C v Δ T + RΔ T C p = C v + R C p − C v = R

Ans 4. The required equation is represented as follows:

C 6 H 12 O 6 ( s ) + 6O 2 ( g ) → 6CO 2 ( g ) + 6H 2 O ( l ) ...(i) C ( graphite ) + O 2 ( g ) → CO 2 ( g ) (Δ H = − 465.0 kJ) ...(ii) H 2 ( g ) +(½)O 2 ( g ) → H 2 O ( l ) (Δ H = − 342.8 kJ) ...(iii) On multiplying equations (ii) and (iii) by 6 and adding them up, the reaction that is obtained is represented as follows: 6C ( graphite ) + 6H 2 ( g ) + 9O 2 ( g ) → 6CO 2 ( g ) + 6H 2 O ( l ) ........ (iv) Δ H = −4 846.8 kJ On subtracting equation (i) from equation (iv), the reaction obtained is as follows: C 6 H 12 O 6 ( s ) + 6O 2 ( g ) → 6CO 2 ( g ) + 6H 2 O ( l ) (Δ H = −3 477.9 kJ) Hence, the heat of combustion of glucose is −3 477.9 kJ.

Ans 5.

= Bond energy of reactants − Bond energy of products Bond energy of reactants = (4 × C − H) + (4 × Cl − Cl) = (4 × 413) + (4 × 243) = 1652 + 972 = 2624 kJ Bond energy of products = 4 × C − Cl + 4 × H–Cl = (4 × 328) + (4 × 432)

= 1312 + 1728 = 3040 kJ

= 2624 − 3040 = −416 kJ mol−1

Equilibrium

Ans 1.

Using the relation, PV = n R T

Or, P = C R T P NO = [NO] R T ...(i)

= [Cl 2 ] R T ...(ii) P NOCl = [NOCl] R T ...(iii) Substituting the values from equations (i), (ii) and (iii) in the expression of K p, , we obtain

Or, K p = K c RT

Ans 2. (i) Following are the conjugate bases: HS − : S 2−

H 2 SO 4 : HSO ­ 4

(ii) Following are the conjugate acids:

Ans 3.

Hence, the percent dissociation is 2.6 × 10 −1 × 100 = 26%

Ans 4. Using the relation,

ΔG ° = 2.303 RT log K It is given that, K = 4.0 × 10 −7 , R = 8.314 J K −1 mol −1 , T = 323 K ΔG ° = 2.303 × 8.314 × 323 log (4.0 × 10 −7 ) = − 39.56 kJ Hence, the value of ΔG° at 50°C for the given reaction is − 39.56 kJ. If the concentration of SO 2 is increased, then according to Le­Chatelier’s principle, more and more of SO 2 will react with O 2 to form SO 3 . Hence, the reaction moves in the forward direction.

Ans 5. Total volume (V) = 1 L

Molecular mass of N 2 O 4 is 92 g Number of moles of N 2 O 4 = 15.8/92 = 0.172

Now,

Or,

Initial pressure

Equilibrium pressure P total at equilibrium P N2O4 + P NO2

Partial pressures of individual components at equilibrium is

Redox Reactions Ans 1. (i)

I . Divide the equation in two halves. Cr 2 O 7 2­ →Cr 3+ (reduction half reaction) SO 2 →SO 4 2­ (oxidation half reaction) II. Balance the atoms other than O and H. Cr 2 O 7 2­ →2Cr 3+ SO 2 →SO 4 2­ III. Then, O and H are balanced using H and H 2 O. Cr 2 O 7 2­ +14H + →2Cr 3+ +7H 2 O SO 2 + 2H 2 O→SO 4 2­ +4H + IV. Electrons need to be added to balance the charge. Cr 2 O 7 2­ +14H + +6e ­ →2Cr 3+ +7H 2 O .........................(1) SO 2 + 2H 2 O→SO 4 2­ +4H + +2e ­ ...............................(2) V. Loss and gain of electrons need to be made equal. For this, we need to multiply equation (2) by 3 VI. Adding both equations Cr 2 O 7 2­ +2H + +3SO 2 →2Cr 3+ +7H 2 O

(ii)

I. Separate the equation into two half­reactions.

Oxidation half­reaction:

Reduction half­reaction:

Reduction half­reaction II. Balance the number of oxygen atoms in the reduction half­reaction by adding water molecule. Here, 4 water molecules are added on the right side.

• Balance the number of hydrogen atoms by adding 8 H + ions on the left.

III. Add electrons to balance the charges on both sides of the reactions.

...........(1) Oxidation half­reaction: IV. Balance the atoms other than O and H in each of the oxidation half­reactions.

V. Balance the number of oxygen atoms in the oxidation half­reaction by adding water molecule. Here, 5 water molecules are added on the left side

VI. Balance the number of hydrogen atoms by adding 10 H + ions on the left

VII. Add 4 electrons to balance the charges of the half reactions

.........(2) To equalize the number of electrons, multiply the oxidation half­reaction (2) by 5 and reduction half­reaction (1) by 4.

VIII. Add the two half­reactions to achieve the overall reaction and cancel the electrons on each side.

Ans 2. (i) The reaction in which the same element is oxidised and reduced simultaneously is

called disproportionation reaction. For this reaction to take place, the element should exist in at least 3 oxidation states. The reaction takes place when the element is in the intermediate oxidation state and both the lower and the higher oxidation states of the element are obtained as products.

For example, oxygen can display three oxidation states, 0, −1, −2. The oxidation

state of oxygen in H2O2 is −1(intermediate oxidation state). When H2O2 undergoes disproportionation reaction, water and molecular oxygen are produced. The oxidation number of oxygen in water is −2 and in molecular oxygen is 0.

It can be observed from the reaction that the oxidation number of hydrogen remains the same in the reaction.

(ii) Cl − and ClO ­ 4 are produced as products when ClO ­ 3 undergoes disproportionation

reaction.

SF 6 , on the other hand, cannot undergo disproportionation reaction as the oxidation state of S in SF 6 is +6, which is the highest oxidation state displayed by sulphur.

Ans 3. The oxidation numbers of bromine atoms (1) and (3) are +6 each and that of (2) is +4.

The oxidation number of Br (1) and Br (2) can be calculated as follows: x + 3 (−2) = 0 x − 6 = 0 x = 6

The oxidation number of Br (2) is calculated as follows: x + 2 (−2) = 0 x − 4 = 0 x = 4

Hydrogen Ans 1. (i) 15 V H 2 O 2 solution means that 1 L of this H 2 O 2 solution will give 15 L of oxygen

at STP. H 2 O 2 gives O 2 according to the following reaction.

That is, at STP, 22.4 L of O 2 is produced from 68 g of H 2 O 2 .

Therefore, at STP, 15 L of O 2 is produced from of H 2 O 2 . = 45.54 g of H 2 O 2

(ii) H 2 O 2 is kept away from dust because dust can induce explosive decomposition of H 2 O 2 .

The s­Block Elements Ans 1. Sodium metal dissolves in liquid ammonia forming sodium ammoniated cation and

ammoniated electron.

The presence of ammoniated electron is responsible for the blue colour and the paramagnetic nature of ammoniated solution.

Ans 2. (i) When Na reacts with water, it releases hydrogen gas with a lot of heat. This

released heat is sufficient to burn the produced hydrogen. This makes Na more reactive for water. In case of reaction of Li with water, Li + ion is totally surrounded by the water due to its smaller size and high effective nuclear charge. This provides high hydration enthalpy, which makes reduction electrode potential highly negative. However, this is not sufficient to burn Li. Hence, Lithium having more E ­ reacts less vigorously with water than sodium.

(ii) Hydration enthalpy increases solubility, while lattice enthalpy decreases solubility. Smaller radius of cation increases lattice enthalpy. The cationic radii of Mg and Ca hydroxides are smaller than that of Na and K. Due to this, they have high lattice enthalpy and low hydration enthalpy as compared to that of Na and K and hence lesser solubility. Due to the same reason, the carbonates of Mg and Cs have lower solubility.

Ans 3. (i) Among the given carbonates, BaCO 3 needs the highest energy to undergo

decomposition. This is because Ba ion is the largest in size and is stabilised by the

large anion . Hence, highest amount of energy is required to break the crystal lattice to undergo decomposition.

(ii) Among the given carbonates, BeCO 3 needs the least energy to undergo

decomposition. This is because Be ion being smallest in size, is not much

stabilised by the ion. Hence, it requires the least energy to undergo decomposition.

Ans 4. (i) Anomalous Properties of Beryllium

• Exceptionally small size of the atom and the cation • Forms covalent compounds (due to small size and high ionisation enthalpy) • Highest coordination number is four while other members exhibit six. • Reason: Be has four orbitals in its valence shell, while other members can

use d­orbitals. • Unlike other elements, the oxides and hydroxides of Be are amphoteric in

nature.

(ii) Similarities Between Be and Al • Like Al, Be is not readily attacked by acids (Reason − presence of an oxide

film on the surface of the metal) • Both dissolve in excess of alkali, giving beryllate ion [Be(OH) 4 ] 2− and

aluminate ion [Al(OH) 4 ] − • Like Al 2 Cl 6 , BeCl 2 has a Cl − bridged polymeric structure. Both these

compounds are Lewis acids, and are soluble in organic solvents. • Both form fluoro complex ions, [BeF 4 ] 2− , [AlF 6 ] 3−

(iii) Structure of BeCl 2 in solid state:

The p­Block Elements Ans 1. Group 13 elements have ns 2 np 1 electronic configuration. Hence, they would be

expected to be trivalent. In most of their compounds, this is the case; however, for the heavy elements, lower oxidation states are more stable. This is explained by the s electrons that remain paired and not participating in bond formation. This inertness of s­subshell electrons towards the bond formation is called inert pair effect. This happens because the s orbitals are held closer to the nucleus. The electrons present in s orbitals are held strongly by the nucleus because of large electrostatic forces. Hence, the energy required to unpair the s­electrons is high because of which they remain paired.

For example: (1) In 13 th group, thallium can exhibit +1 and+3 oxidation states, but it is stable in +1

oxidation state only due to inert pair effect (2) In 14 th group, lead shows both +2 and +4 oxidation states, but it is stable in +2

oxidation state due to inert pair effect. Tl(I) compounds are more common due to the inert pair effect. The valence shell electronic configuration of Tl is 6 s 2 6 p 1 . After the removal of one p ­electron, the 6 s ­electrons do not take part in compound formation.

Ans 2. In (CH 3 ) 3 N, lone pair of electrons are present in nitrogen atom resulting in sp 3

hybridisation. Hence, it is pyramidal.

In case of (SiH 3 ) 3 N, the lone pair of electrons in nitrogen atom overlaps with the empty d ­orbital of silicon atom. This results in dπ­ p π back­bonding.

Hence, the lone pairs are no longer available in nitrogen atom. Thus, it involves sp 2 hybridisation and has trigonal planar geometry.

Ans 3. (i) Boric acid is prepared by acidifying an aqueous solution of borax. The chemical

equation for the process is:

Boric acid exists as a white crystalline solid and is soapy to touch. It is sparingly soluble in cold water and highly soluble in hot water.

(ii) Boric acid is a weak acid as it does not release H + ions on its own. When added to

water, it accepts OH­ ions from it thereby furnishing H + ions.

Organic Chemistry ­ Some Basic Principles and Techniques Ans 1. (i) Inductive effect is an electronic effect due to the polarisation of σ bonds within a

molecule or ion. In a covalent bond, formed by the atoms of different electronegativity, the electron density moves towards the more electronegative atom of the bond.

There are two types of inductive effect: +I effect: The groups that push the electron density away from themselves exhibit positive inductive effect (+I). Example: The groups exhibiting + I effect are −CH3,−NH2. −I effect: Atoms or groups that pull the electron density towards themselves exhibit negative inductive effect (−I). Example: The groups exhibiting − I effect are −COOH,−NO2.

(ii) Stability of carbonium ions: The stability of carbonium ions increases with

increase in number of alkyl groups, due to their +I effect. The alkyl groups release electrons to carbon, bearing positive charge and hence stabilize the ion. The order of stability of carbonium ions is:

Stability of free radicals: In the same way, the stability of free radicals increases with increase in the number of alkyl groups. Thus, the stability of different free radicals is:

Stability of carbanions: Stability of carbanions decreases with increase in the number of alkyl groups, since the electron donating alkyl groups destabilise the carbanions by increasing the electron density. Thus, the order of stability of carbanions is:

Ans 2. (i)

1 − Ethoxypropan − 2 − ol

(ii)

Ethyl 2 − bromo − 2 − (3 −nitrophenyl) propanoate

(iii)

2 − (2− bromophenyl) ethanal

Ans 3. It is given that, Weight of organic compound = 0.4950 g Weight of carbon dioxide = 0.9900 g Weight of water = 0.4050 g

Percentage of carbon

Percentage of water ∴ The percentages of carbon and hydrogen in the organic compound are 54.54% and 9.09% respectively.

Ans 4. The hyperconjugation in ethyl carbocation can be represented as follows:

The electron density from the adjacent sigma bond helps in dispersing the positive charge, hence stabilising the cation.

Hydrocarbons Ans 1. Carbon atoms joined by only a sigma bond can rotate freely about the bond. This

rotation gives rise to different spatial arrangements of atoms. These spatial arrangements are interchangeable, and are known as conformations. For example, in a molecule of ethane, the two carbon bonds are connected by a single bond, and are thus, free to rotate. This gives rise to infinite number of conformations. However, there are two extreme cases − staggered and eclipsed. In the former, the hydrogen atoms joined to the carbon atoms are at the maximum distance, and in the latter, the hydrogen atoms are at the minimum distance. There are two ways in which these conformations can be represented. These representations are called Sawhorse and Newman projections.

Sawhorse projection: In this projection, the molecule is drawn in such a way that its molecular axis is clearly visible. The staggered and eclipsed conformations for ethane are as follows:

Newman projection: In this projection, the molecule is viewed from the side in such a way that only one carbon atom is visible. Therefore, in this representation, the carbon atom present nearer to the eye is represented by a dot and the atom away from the eye is represented by a circle. The staggered and eclipsed conformations for ethane are as follows:

Ans 2. For Aromaticity , a compound must obey Huckel’s rule. According to this rule, there

should be (4 n + 2) π electrons in the ring, where n is an integer ( n = 0, 1, 2…).

In structure (I) , there are (4×1 + 2) i.e. 6π electrons in the ring. This means that it obeys Huckel’s rule. Hence, it is aromatic. In structure (II), there are (4×1 + 2) i.e., 6π electrons in the ring. This means that it obeys Huckel’s rule. Hence, it is aromatic. Structure (III) has 8π electrons in its ring. Hence, it does not obey Huckel’s rule. Hence, it is not aromatic.

Ans 3. The products methanol and propanone are obtained by the ozonolysis of 2­methyl propene. The reaction takes place as follows:

Environmental Chemistry Ans 1. Harmful effects of photochemical smog:

1. Ozone and PAN (Peroxyacetyl nitrate) − the components of photochemical smog cause irritation to the eyes. Ozone and nitric acid cause irritation to the nose and throat. When these chemicals are present in higher concentrations, they can cause headaches, chest pain and breathing problems.

2. Photochemical smog causes cracking of rubber. It also corrodes metals, painted surfaces and building materials.

3. It causes extensive damage to plants.