chemistry 20 - science with mr. danmrdancisabudhabi.weebly.com/.../notes_unit_3_bonding… · web...

CHEMISTRY 20

Unit 3 : Chemical Bonding

Prepared by Daniel VeraartCIS Abu Dhabi

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Chemistry 20Day 22

Date:______________________ Name: _____________________________

Electron representation for atomsIn nature, atoms of elements that have less than eight electrons in their outer energy levels nearly always form bonds with other atoms.As pointed out by the Bohr Model, electrons are located in certain specific energy levels around the nucleus. Scientists very quickly saw the very close relationship between the electron level representations for an atom and the atom's ability to bond with other atoms.

Electron energy level representation for atoms

1. The number of electrons in an atom equals the atomic number.2. The maximum number of electrons in each successive energy level equals the number of

elements in each successive period. There are 2, 8 and 8 elements in the first three periods of the periodic table.

3. The number of electrons in the outermost energy level are called the valence electrons.

Draw the energy level representation of the first 20 elements.

Hydrogen Helium Lithium

Carbon Nitrogen Oxygen

Sodium Magnesium Aluminum

Sulfur Chlorine Argon

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Electron Dot Diagrams

What are electron dot diagrams?

Electron dot diagrams are excellent models for visualizing valence electrons in atoms as they interact in chemical reactions.

An electron dot diagram consists of the chemical symbol of the element with the dots representing the valence electrons.

Example:

Rules to draw an electron-dot diagram

1. Write the element symbol and use dots to represent each valence electron.2. Place the first dot at the top of the symbol and then add dots moving around the symbol.3. For elements that have more than four valence electrons, you place the fifth dot beside

one of the single dot and continue if more dots are required. 4. No more than 8 electrons exist in the outer energy level of an atom.

Element Group number

Valence electrons

Electron-dot diagram # Bonding electrons

# Lone pairs

Hydrogen

Carbon

Nitrogen

Oxygen

3

Chlorine

Xenon

Magnesium

Aluminum

Forming Bonds

No more than 8 electrons exist in the outer energy level of any atom. When bonds form between atoms, the outer level usually becomes filled with its eight electrons.

The octet rule states that when bonds form between atoms, the atom gain, lose or share electrons in such a way that they create outer energy levels with eight electrons.

The octet rule does not apply to hydrogen since the first level of energy only have 2 electrons.

Ionic BondsIn ionic bonding, metals lose their valence electrons to become positively charged ions. Non-metals gain electrons to become negatively charged ions.

Covalent BondsA covalent bond forms when two non-metals atoms share one or more pairs of electrons. Including the share electrons, both atoms achieve the octet rule.

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Electron Dot Diagrams for ionic compounds

Draw the electron dot diagram for the following molecules

CaF2 K2S

Fe2O3 Mg3N2

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Chemistry 20Homework #22

Name: ____________________________ Date: __________________________

ExercisesDraw the electron dot diagram for the following molecules

a) Calcium phosphide b) Magnesium iodide

c) Sodium oxide d) Hydrogen chloride

e) Potassium chloride f) Aluminum oxide

g) Potassium sulfide h) Zinc oxide

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Chemistry 20Day 23

Date:______________________ Name: _____________________________

Electron-Dot Diagrams for molecular compounds

Molecular compounds = Covalent bonds

Covalent bonds: Covalent bonds are type of bonds formed between two nonmetals to form a molecular compound. Each atom of the molecule shares bonding electrons to a maximum of 8 electrons = octet rule.

Covalent bond = Sharing electrons between non-metals.

Electron dot diagram for molecular compound = Lewis structure

Electron dot diagrams of molecules are called Lewis structures. The examples below show the Lewis structures for methane (CH4) and water (H2O)

1. Draw the electron dot diagram of the atom that has the most bonding electrons (central atom).

2. Form sharing pairs of bonding electrons with the remaining atoms.3. If any bonding electrons remain, form a double or triple bond.

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4. In the finished diagram, all atoms (except hydrogen) should contain an octet.

Examples :

For each of the following molecular compound, draw an electron dot diagram.

HF OF2 CH2I2

HCOOH C2Cl2

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1. 2.

3. 4.

Structural Formulas

Lewis structures are very helpful for understanding bonding and making predictions about structures. They are however, very tedious to draw. To provide a clear image of a structure in a simpler form, you can draw a structural formula.

Structural formulas

A structural formula shows every individual atom in the molecule and shows the bond between atoms.

A structural formula uses a line to represent a bond, two lines to represent double bond and three lines to represent triple bonds.

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Name: ____________________________ Date: __________________________

Exercises1. Draw Lewis structures to represent each of the following molecules;

I. One carbon atom is bonded to two hydrogen atoms and two chlorine atoms. This compound is called methylene chloride and is sometimes used as a paint stripper.

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II. One hydrogen atom is bonded to a fluorine atom. The compound, hydrogen fluorine, is someimtes used to etch glass.

III. Three hydrogen atoms are bonded to one nitrogen atom in this compound that is found in some cleaning solutions.

IV. One carbon atom is bonded to two sulfur atoms. This compound, carbon disulfide, is sometimes found in gases of volcanic eruptions and in marsh gas.

V. A central carbon atom is bonded to a hydrogen atom and to a nitrogen atom. This is a toxic compound called hydrogen cyanide.

2. Draw the Lewis structure of the following compounds

a) NH2CH2COOH b) CH3OCH3 c) CH3COOH d) CF3CHF2

3. Draw structural formulas for each of the following compounds.

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a) BrNO b) HOCl c) H2S d) C2H4

e) H2O2 f) HCN g) NH2CH2COOH h) CH3OCH3

Lab 8

Evidence of Double Bonds

Problem: Which of the substances supplied contain molecules with a double bond?

Design: The samples are tested by adding a few drops of a solution of potassium permanganate.

Purple Potassium permanganate reacts with double bonds and changes color.

Materials:12

Test tubes Permanganate solution

Substances to be tested: Canola oil Methanol Vinegar Olive oilGlycerin Sulfuric Acid Cyclohexane Cyclohexene

Procedure:

1. Add a dozen drops of the substance in the test tube.2. Add 5 drops of permanganate solution. Mix.3. Observe color. 4. Dispose of all materials into the waste container.

Questions:

Make a table of your observations. Identify if a double bond is present or not

Substance Observation Double bond presentCanola Oil

Glycerin

Methanol

Sulfuric acid

Vinegar

Cyclohexane

Olive oil

Cyclohexene

Chemistry 20Day 24

Name: ____________________________ Date: __________________________

ElectronegativityThe nucleus of an atom is positively charged and therefore attracts electrons, which are negatively charged.

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What is the electronegativity?

The electronegativity (EN) of an element is the relative measure of the ability of that element's atom to attract the shared electrons in a chemical bond

Chemist have devised a scale from 0.7 to 4.0 to quantify the EN of the elements.

Higher EN indicate a greater attraction for the electrons.

Fluorine has the highest electronegativity.

Lower EN indicate a weaker attraction for the electrons.

Francium has the lowest electronegativity.

Bond type and Electronegativity

Ionic bonding

Ionic bonds form between metals (with low EN) and non-metals (with High EN).

Because metals do not attract electrons in a bond, they lose electrons.Because non-metals attract electrons in a bond, they gains electrons.

The resulting ions are attracted to each other, thus forming ionic bonds.

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Covalent bonding in diatomic molecules

Diatomic molecules are those elements: O2, H2, N2, F2, P4, Br2, Cl2, I2 .

Those molecules have bonds between two identical atoms (identical EN).

Because the EN are identical, the two nuclei attract the electrons of the bond with exactly the same strength. Neither atom loses or gains electrons.

They share electrons in covalent bonds and the molecules are symetrical.

Covalent bonding in molecular compounds

Some variations in bond structure begin to emerge when you consider covalent bonds between atoms of unlike elements.

For example, the bond between carbon and chlorine: EN for carbon = 2.6 and EN for Chlorine = 3.2

The EN for chlorine is higher indicating that chlorine attracts electrons more than carbon. As a result, the atoms do not share electrons equally. Such a bond creates an slightly positive end and a slightly negative end

Chemists call these partial charges and designate them with the Greek letter delta .

Polar covalent Bond and Non-polar covalent bond

Polar covalent bonds

Covalent bonds that have a separation of charge (or unequal sharing of electrons) are called polar covalent bonds.

Non-Polar covalent bonds

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Covalent bonds that have no EN difference (or equal sharing of electrons) are called non-polar covalent bonds.

Calculate Electronegativity Difference ( EN)

The difference in electronegativity (EN) of the elements involved in a bond determine some of the characteristics of the bond.

We can calculate the EN of two elements to predict the nature of the bond between atoms of those two elements.

Examples:

1. Find the EN for the bond between Potassium and Fluorine and classify the bond.

2. Find the EN for the bond between two oxygen atoms and classify the bond.

3. Find the EN for the bond between carbon and chlorine and classify the bond.


Name: ____________________________ Date: __________________________

Exercises1. For each of the following pairs of atoms, predict whether a bond between will be non-polar covalent, polar

covalent or ionic. For each polar bond, indicate the locations of the partial positive and partial negative charges;

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I. carbon and fluorine

II. oxygen and nitrogen

III. chlorine and chlorine

IV. copper and oxygen

V. silicon and hydrogen

VI. sodium and fluorine

VII. iron and oxygen

VIII. manganese and oxygen

2. Arrange the bonds below in order of increasing polarity;

hydrogen bonded to chlorine;oxygen bonded to nitrogen;

carbon bonded to sulfur;sodium bonded to chlorine

i. ____________________________________________________________

ii. ____________________________________________________________

iii. ____________________________________________________________

iv. ____________________________________________________________

Chemistry 20Day 25

Name: ____________________________ Date: __________________________

Three dimensional Structures for ionic crystals

Ionic compounds make crystal in their solid state. For example. Sodium chloride

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Because an attractive force exists between positive and negative charges, the ions pack together very tightly. This 3-D pattern is called a crystal lattice.

To visualize the attractive forces, examine the diagram below. The sticks represent the attractive forces between positive and negative ions.

Three Dimensional structures for molecular compoundsInteresting website http://www.worldofmolecules.com/

Molecular compounds also have a definite three-dimensional shape. Valence-Shell Electron Pair Repulsion Theory (VSEPR) allows you to predict the three dimensional shape of a molecule.

The number of bonding electron of the central atom(s) can be used to predict the shape of the molecule according to the VSEPR.

In this theory, we use the following symbols:

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http://www.worldofmolecules.com/

A = Central atom X = Any atom bonded to the central atom E = a lone pair

Bonding electrons and molecular shapes

VSEPR Class Type of electron

pairsExample Name of the

shapeShape

AX2 All BondingHCNCO2 Linear

AX3 All Bonding HCHO Trigonal plan

AX2E2

2 Bonding2 Lone pair

H2O Bent orV-Shaped

AX4 All Bonding CH4 Tetrahedral

AX3E 3 Bonding1 lone pair

NH3

Pyramidal

Steps for predicting Molecular shapes

1. Draw a Lewis structure for the molecule2. Determine the total number of electron groups around a central atom. A double or a triple bond

is counted as one electron group.3. Determine the type of electron groupings (bonding pairs or lone pairs)4. Determine which one of the shapes in the table above will accommodate this combination.

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Examples of molecule in 3-DMolecule Electron dot

diagramVSEPR Class Name of

shapeMolecular

shape

CH2BrF

COBr2

SBr2

COCl2

N

PF3

H2O


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Name: ____________________________ Date: __________________________

Exercises

Molecule Electron dot diagram

VSEPR Class Name of shape

Molecular shape

CH2Cl2

N2H4

CH3OH

OF2

COS

HCN

PIBr2

Lab #2Soap Bubble Molecule

Materials

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- Soap solution - Straw - Transparent ruler + Protractor

Procedure1. Obtain 25 mL of the soap solution. Use the soap solution to wet an area of about

10 cm x 10 cm .2. Dip a straw into the soap solution in the beaker and blow a bubble on the wet surface.

Then blow a second bubble of the same size so that it touches.3. Record the shape of this simulated molecule (diagram) and measure the bond length

between the centres of the two bubbles where the nuclei of the atoms would be located.4. Repeat procedure #3 but with three bubbles of the same size. Record the shape of this simulated

molecule (diagram) and all the bond angles and bond length.

5. On top of the group of three equal-sized bubbles, blow a fourt bubble of the same size that touchesthe center. Record the shape of this simulated molecule (diagram) and all the bond angles andbond length.

6. Repeat Procedure step 3, but this time, to simulate a lone pair, make one bubble slightly largerthan the other two. Record the shape of this simulated molecule (diagram) and all the bond anglesand bond length.

7. To simulate two lone pairs, make a fourth bubble that is the same size as the large bubble and restit on top of the three bubbles you made. Record the shape of this simulated molecule (diagram)and all the bond angles and bond length.

Analysis1. What shapes and bond angles were associated with two, three, and four of the same sized

bubbles?2. Give an example of a molecule that matches each of these shapes.

Chemistry 20Day 26

Name: ____________________________ Date: __________________________

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Polar Bonds and Polar Molecules

Polar bond: Bond formed between unlike atoms with different electronegativity.

The word polar implies an overall charge separation. One end of the bond is positive and the other end is negative.

We represent a polar bond with an arrow pointing to the element with the highest EN

O - H C = O

Polar molecule: Molecule, which has an overall charge separation; the polarity vector do not cancel each other.

Polar molecules vs Non-Polar Molecules

The following molecules are polar: The following molecules are non-polar:

►Bent (V-shaped) molecule►All non-symmetrical molecules►If there are one or more lone pairs on the molecule

► All symmetrical molecules around the central element

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Example :

1. Are the following molecules polar or non-polar. Explain.

Molecule VSEPR notation 3-D Shape Polarity

HCl:

BF3

H2O

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CBr4

PCl3

CH3OH

Properties of polar and non polar molecules Properties of polar molecules:

Property #1: Polar molecules attract other polar molecules but do not attract non polar molecules

Property #2: Polar molecules dissolve well in other polar molecules but do not dissolve in non polar molecules.

Properties of non polar molecules:

Property #1: Non polar molecules attract other non polar molecules but do not attract polar molecules

Property #2: Non polar molecules dissolve well in other non polar molecules but do not dissolve in polar molecules.

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Chemistry 20Homework 26

Name: ____________________________ Date: __________________________

Exercises

1. Indicate if the following substances dissolve well in water?

A. CH4 : ________

B. C2H5OH : ________

C. C4H10 : ________

D. CH3COOH : ________

2. Are the following compounds polar or non-polar.

Molecule Explanation of polarity

C2Cl2

CH3OH

OF2

CO2

H2O2

CF4

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CH3Br

CH3OCH3

C2H5OH

Chemistry 20Day 27

Name: ____________________________ Date: __________________________

Intermolecular forces

Forces binding atoms in a molecule are due to chemical bonding. They are known as covalent and ionic bonds.

The forces involved in these bonds act only between atoms within thesame molecule and are called intramolecular forces.

What then holds molecules together in a solid or a liquid.

The forces holding molecules together are called intermolecular forces.

The Nature of Intermolecular Forces

Important definition

Intermolecular forces : An intermolecular force is the force between molecules.

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It’s the force that holds the molecules together in a solid or a liquid

Note : In general, the greater intermolecular force between the molecules of a compound, the higher the boiling point of the compound.

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Three types of intermolecular forces

London (Dispersion) Forces:

are very weak forces of attraction between all molecules resulting from the weak attraction of the nuclei in one molecule for the electrons in a neighbouring molecule.

The more electrons that are present in the molecule, the stronger the dispersion forces will be.

Dispersion forces are the only type of intermolecular force operating between non-polar molecules.

Example: Hydrogen (H2) moleculeschlorine (Cl2) moleculescarbon dioxide (CO2) moleculesdinitrogen tetraoxide (N2O4) moleculesmethane (CH4) molecules.

Dipole-Dipole Forces:

are stronger intermolecular forces than Dispersion forces

occur between polar molecules.

Examples: SCl2 moleculesPCl3 moleculesCH3Cl molecules.

Hydrogen Bonds:

occur between polar molecules that have a hydrogen being covalently bonded to either fluorine, oxygen or nitrogen.

Example: water (H2O) molecules ammonia (NH3) molecules hydrogen fluoride (HF) molecules hydrogen peroxide (H2O2) molecules alcohols such as methanol (CH3OH) molecules caboxylic acids such as ethanoic (acetic) acid (CH3COOH).

are a stronger intermolecular force than either Dispersion forces or dipole-dipole interactions.

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http://www.ausetute.com.au/molpolar.html

Effect of intermolecular forces on boiling point

Since boiling result in a weakening of the forces between the molecules, the stronger the intermolecular force is, the more energy is required to boil a liquid. If only dispersion forces are present, then the more electrons the molecule has, the stronger the dispersion forces will be, so the higher the melting and boiling points will be.

Consider the hydrides of Group 14, all of which are non-polar molecules, so only dispersion forces act between the molecules.

CH4 (molecular mass = 16)SiH4 (molecular mass = 32)GeH4 (molecular mass = 77) SnH4 (molecular mass = 123)

As the mass of the molecules increases, so does the strength of the dispersion force acting between the molecules, so more energy is required to break the attraction between the molecules resulting in higher boiling points.

Boiling Points of Group 14 Hydrides

If a molecule has a dipole-dipole force, then the attraction between these molecules will be stronger than if only dispersion forces were present between the molecules. As a consequence, this substance will have a higher melting or boiling point than similar molecules that are non-polar in nature.

Consider the boiling points of the hydrides of Group 17 elements.

HF (molecular mass ~ 20 + Polar + Hydrogen bond) HCl (molecular mass ~ 37 + Polar)HBr (molecular mass ~ 81 + Polar)) HI (molecular mass ~ 128 + Polar))

As a consequence, the stronger dipole-interactions acting between the hydride molecules of Group 17 elements results in higher boiling points than for the hydrides of Group IV elements as seen above.

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With the exception of HF, as the molecular mass increases, the boiling point of the hydrides increase.HF is an exception because of the stronger force of attraction between HF molecules resulting from hydrogen bonds acting bewteen the HF molecules. Weaker dipole-dipole interactions act between the molecules of HCl, HBr and HI. So HF has a higher boiling point than the other molecules in this series.

Boiling Points of Group 17 hydrides

Substance Polarity # electrons Dipole-dipole London Hydrogen bond Boiling point

F2

Cl2

Br2

I2

CH4

C2H6

C3H8

C4H10

C5H12

CH3F

CH3Cl

CH3Br

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http://www.ausetute.com.au/mmcalcul.html

CH3I

CH3OH

C2H5F

C2H5Cl

C2H5Br

C2H5I

C2H5OH


Name: ____________________________ Date: __________________________

Exercises

1. Which types of intermolecular attractive forces operate between polar molecules?

A. London-dispersion and hydrogen bonding B. London-dispersionC. dipole-dipole attraction D. London dispersion and dipole-dipole attractionE. hydrogen bonding

2. Identify the types of intermolecular forces that are present in C3H8 and CH3OCH3 and select the substance that has the highest boiling point.

A. C3H8, London, dipole; CH3OCH3, London, dipolar; C3H8, highest boiling pointB. C3H8, London, dipole; CH3OCH3, London, dipolar; C3H8, highest boiling pointC. C3H8, London; CH3OCH3, London; CH3OCH3, highest boiling pointD. C3H8, London; CH3OCH3, London, dipole; CH3OCH3, highest boiling pointE. C3H8, London; CH3OCH3, London; C3H8, highest boiling point

3. Predict the boiling point of H2Se if the following is true for the boiling points:

H2O : 100°C H2Te: -2°C H2S: -61°C

A. It will have a boiling point of about -100°C.

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B. It will have a boiling point about 110°C.C. It is impossible to estimate.D. It will have a boiling point about -32°CE. It will have a boiling point about 10°C.

4. Arrange the following in terms of decreasing boiling points: N2 , O2 , NO

A. NO > N2 > O2 B. N2 > O2 > NO C. NO > O2 > N2

D. O2 > N2 > NO E. N2 > NO > O2

5. Predict the order of the boiling points (from highest to lowest) of NH3, CH4, SiH4, and GeH4.

A. NH3 > SiH4 > CH4 > GeH4 B. NH3 > CH4 > SiH4 > GeH4

C. CH4 > SiH4 > GeH4 > NH3 D. GeH4 > SiH4 > CH4 > NH3

E. NH3 > GeH4 > SiH4 > CH4

6. Which one of the following substances would have hydrogen bonding as one of its intermolecular forces?

A. HCHO B. CH3F C. NH2CHO D. none of the above

7. Which of these molecules can form hydrogen bonds?

A. BH3 B. CH4 C. PH3 D. CH3F E. HOCH3

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8. Which of these molecules do not form hydrogen bonds?

A. ether, CH3OCH3 B. water, H2OC. ethyl alcohol, C2H5OH D. ammonia, NH3

9. Identify which forces are present in the following compounds by putting a checkmark in the box.

Substance Dipole-dipole forces

Dispersion forces

Hydrogen bonds

H2O

CHBrCl2

CH4

CH3CF3

C2H5COOH

Cl2

CO2

C2H2

CH3Br

10. True or False. Justify the false statements.

a) ______ H2O has a higher boiling point than CH4.

b) ______ C2H5COOH has a lower boiling point than CH3COOH.

c) _______ Cl2 has a higher boiling point than CO2.

d) ______ H2O2 has all three forces.

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Chemistry 20Day 28

Name: ____________________________ Date: __________________________

Formation of ionic compounds = Redox reaction

Properties of ionic compounds

1. The simplest ionic compounds are formed in the reaction of a metal with a non-metal.

2. Ionic compounds are all solid at room temperature.

3. Ionic compounds are soluble in water (some have a high solubility and some other have a low solubility).

4. Chemists have used ionic compounds to extract elements from naturally occurring compounds.

Example: Hematite (Iron one) is changed into pure iron.Fe2O3(s) 2 Fe(s) + O2(g)

Example: Sodium and chlorine can be produced from sodium chloride.

2 NaCl(s) 2 Na(s) + Cl2(g)

Formation of ionic compounds

When ionic compounds are formed, electrons are transferred from metal to non-metal.

Example: Fe(s) + S(s) FeS(s)

Scientists explain the formation of an ionic compound as a collision between a metal and a non-metal atom that results in a transfer of electrons, forming positive ion = cation and negative ion = anion.

This electron transfer from the metal to the non-metal is known as an oxidation-reduction reaction.

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This chapter examines the oxidation of a metal and the reduction of a nonmetal. These reactions are explained as an exchange of electrons.

Oxidation-reduction reactions are also called REDOX reactions.

Redox reactions have been used since ancient times in primitive metallurgy. More recent examples of redox reactions include batteries, electroplating and refining elements.

The term oxidation originally meant the combining of oxygen with other substances, such as occurs in combustion or corrosion of metals.

The term reduction originally applied to the process of reducing the metallic compounds found in ores to pure metals.

Today, chemists have extended the concepts of oxidation and reduction to include all transfer of electrons.

One example of an oxidation reduction reaction is the formation of ionic compounds from the reaction between a metal and a non-metal.

For example: The formation of ZnS(s) when Zinc reacts with Sulfur.

Zn(s) + S(s) ZnS(s)

In this reaction, the zinc loses 2 electrons and becomes Zn2+. At the same time, the non-metal sulfur gains two electrons and becomes S2-.

We can represent these by two half reactions: Zn(s) Zn2+(s) + 2e- and S(s) + 2e- S2-

(s)

Important definitions

Oxidation : Oxidation refers to the Loss of Electrons in a half reaction.

LEO : Lost of electrons = Oxidation

Example: Zn(s) Zn2+(s) + 2e-

Reduction : Reduction refers to the Gain of Electrons in a half reaction

GER : Gain of electrons = Reduction

Example: S(s) + 2e- S2-(s)

When you combine an oxidation half reaction and a reduction half reaction, you have what we call an oxidation-reduction reaction (or a redox reaction).

Oxidation-reduction reaction: Reaction that involves both processes, the loss and gain, or the transfer of electrons

Example: Zn(s) + S(s) ZnS(S)

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Decomposition of ionic compounds = Redox reaction

Decomposition of ionic compounds

Elements can be produced when ionic compounds decomposes, electrons are transferred from the anion to the cation.

Example: 2 NaCl(s) 2 Na(s) + Cl2(g)

In this reaction, the sodium ion gains 1 electron and becomes Na(s). At the same time, the chloride ion loses two electrons and becomes Cl2(g).

We can represent these by: Na+(s) + e- Na(s) (reduction)

2Cl-(s) Cl2(g) + 2e- (Oxidation)

Examples

1. Classify the following half reactions as oxidation or reduction.

a) Mg(s) Mg2+(s) + 2e- _________________

b) Br2(l) + 2e- 2 Br-(l) ________________

c) Al3+(s) + 3e- Al (s) _________________

d) Hg(l) Hg2+(s) + 2e- _________________

e) 2 H+(aq) + NO3

-(aq) + e- NO2(g) + H2O(l) ________________

2. Complete the following half reactions by adding the electrons required to balance the reaction.Indicate if the half reaction is an example of reduction or oxidation.

a) Ca2+(s) Ca (s) _________________

b) As3-(s) As(s) _________________

c) Cr(s) Cr2+(s) _________________

d) H2(g) H+1(g) _________________

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3. Write the reduction and oxidation half-reaction equations for the formation of the following ionic compounds.

a) Formation of CoI2(s) : _______________________________________________

Oxidation ½ reaction: ___________________________

Reduction ½ reaction: ___________________________

b) Formation of Ni3N2(s) : _______________________________________________



c) Formation of FeCl3(s) : _______________________________________________



4. Write the reduction and oxidation half-reaction equations for the decomposition of the following ionic compounds.

a) Decomposition of Cr2O3(s) : _______________________________________________



b) Decomposition of Li2S(s) : _______________________________________________



c) Decomposition of CaF2(s) : _______________________________________________



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Chemistry 20Homework 28

Name: ____________________________ Date: __________________________

Exerci ses

1. Classify the following half reactions as oxidation or reduction.

a) Ca(s) Ca2+(s) + 2e- _________________

b) 2 F- (s) F2(l) + 2e- ________________

c) Na+(s) + e- Na(s) _________________

d) Ag(l) Ag+(s) + e- _________________

e) MnO4-(aq) + 8 H+

(aq) + 5 e- Mn+2(aq) + 4 H2O(l) ________________

2. Complete the following half reactions by adding the electrons required to balance the reaction.Indicate if the half reaction is an example of reduction or oxidation.

a) Ba(s) Ba2+(s) ___________________

b) Al (s) Al3+(s) ___________________

c) Cl2(g) Cl-(s) ___________________d) F2(g) 2 F-

(s) ___________________e) Mn(s) Mn4+

(s) ___________________f) N2(g) N3-

(s) ___________________g) Zn(s) Zn2+

(s) ___________________h) Fe3+

(s) Fe(s) ___________________

Use the following incomplete half-reactions to answer the next question

Cl2(g) 2 Cl-(aq) Fe3+

(aq) Fe2+(aq)

Cr2+(aq) Cr3+

(aq) Zn2+

(aq) Zn(s)

3. Which species loses electrons?

A. Cl2(g) B. Fe3+(aq) C. Zn2+

(aq) D. Cr2+(aq)

4. In the reaction 2 Fe2+(aq) + Br2(l) 2 Fe+3

(aq) + 2 Br-(aq) , the Fe2+

(aq)

A. gains electrons and undergoes reduction B. gains electrons and undergoes oxidation C. loses electrons and undergoes reduction D. loses electrons and undergoes oxidation

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5 The species most likely to undergo oxidation is

A. Li+(aq) B. H+(aq) C. Ag(s) D. F2(g)

6. For the following reactions:- Write a balanced equation and find the charge of each element.- Write the balanced half reduction and half oxidation reactions.

d) Formation of Al2O3(s) : _______________________________________________



e) Decomposition of KBr(s) : _______________________________________________



f) Formation of MgCl2(s) : _______________________________________________



g) Decomposition of CaF2(s) : _______________________________________________Oxidation ½ reaction: ___________________________


h) Formation of Iron (III) oxide : _______________________________________________



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