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4.1 Exercises

1. Ionic bonding

a) Ionic bonding occurs between what types of elements?

A: non metal – non metal

B: metal – metal

C: metal – non metal

Answer: C

b) What is the octet rule?

Atoms have a tendency to gain or loose electrons in order to end up with a full outer shell of electrons. A full

outer shell consists of eight electrons (for all but the period 1 elements) and, therefore, the tendency to reach

this outer electron arrangement is called completing the octet.

2. Attraction

a) Define electronegativity

Electronegativity is the ability of an atom to attract electrons to itself in a bond. Electronegative elements tend to

gain electrons and form negative ions, e.g. the halogens.

There are various ways of assigning values for electronegativity. The IB Chemistry Data Booklet contains values

for the electronegativity of the elements using the Pauling scale, which is a dimensionless scale based on bond

dissociation energies in which fluorine (the most electronegative element) has a value 4.

b) What determines the electronegativity of an atom?

Electronegativity is determined by how easily an atom will either lose or gain electrons. This, in turn, is

determined by the number of protons in the nucleus and the size of the atom. Many protons in a small atom, e.g.

F, means a high electronegativity. It has been proposed that the average of an atoms electron affinity and first

ionisation energy is a good approximation of the atoms electronegativity. The electronegativity of an atom in a

bond is most easily predicted by its position in the periodic table. Atoms high in a RHS group have high

electronegativity. Atoms low in a LHS group have the lowest electronegativity.

c) What determines how many electrons and element will lose or gain?

An element will gain or lose electrons in order to satisfy the octet rule. By identifying an atom’s position of the

period table, it can quickly be determined how many electrons it will either lose or gain. For example, all atoms

in group 2 of the periodic table need to lose two electrons in order to have a full octet, while elements in group 6

must gain two electrons for a full octet.

3. Predicting Ionic Bonds

a) Predict, from their position on the periodic table, whether atoms of Rubidium, Rb and Oxygen,

O, will form ionic bonds when they react.

Rubidium is low in group 1 and is a metal. Oxygen is high in group 6 and is a non-metal. They will therefore form

an ionic bond due to the large difference in electronegativities between the elements.

b) If an ionic compound was formed by these elements, which element would lose electron(s) and

which would gain electron(s)?

Rubidium would lose one electron and oxygen would gain two in order to have a stable octet of electrons.

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c) With the help of Lewis dot diagrams, explain the bonding that occurs between rubidium and

oxygen and state the final formula of the compound that is formed.

O

RbRb

O

Rb+Rb+

2-

This electron transfer results in the formation of oppositely charged ions which bond via electrostatic attraction

to form an ionic lattice structure.

Since rubidium loses only one electron and oxygen requires two to obtain a stable octet, there must be two

rubidium atoms per oxygen atom; the ionic formula is therefore Rb2O.

d) Explain the nature of the forces that hold Rb and O atoms together in their crystal lattice?

Ionic bonding, i.e. the strong electrostatic force of attraction between oppositely charged species . The Rb+ and

O2-

ions are held together by electrostatic attraction to form a three-dimensional, continuous crystal lattice.

4. Deduce the formulas and names of the ionic compounds that arise due to the combination of:

Determine the charge on each ion then adjust with subscripts to balance charges.

a) Na and O

Na2O: Sodium oxide

b) Mg and P

Mg3P2: Magnesium phosphide

A phosphide compound is one in which phosphorous has an oxidation state of –3, that is it has gained three

electrons.

c) K and Br

KBr: Potassium bromide

d) Li and S

Li2S: Lithium sulphide

A sulfide compound is one in which sulfur has an oxidation state of –2, that is it has gained two electrons.

e) Ca and N

Ca3N2: Calcium nitride

A nitride compound is one in which nitrogen has an oxidation state of –3, that is it has gained three electrons.

f) Sr and F

SrF2: Strontium fluoride

5. Describe the formation and structure of a three-dimensional ionic lattice using KBr as an example.

Bromine is much more electronegative than potassium. Therefore, electrons are transferred from potassium to

bromine when the two collide. Potassium has one electron in its valence shell and can therefore lose one

electron easily to obtain a stable eight electron arrangement (full octet). Bromine has seven electrons in its

valence shell and readily gains one electron to obtain a stable eight electron arrangement. Potassium donates

an electron to bromine and the ions K+ and Br

- are formed. They are attracted to each other through electrostatic

forces to form an ionic crystal lattice.

2

6. Name, and give the formulas for, three compounds with ionic bonding, one with an ion of single

positive charge, the next with a double positive change, and then one with a triple positive change.

Sodium; Na+. Lithium; Li

+. Potassium; K

+ (Group 1 metals and transition metals) may form NaCl, LiCl, KCl.

Calcium; Ca2+

. Magnesium; Mg2+

. Strontium; Sr2+

(Group 2 metals and transition metals) may form CaCl2,

MgCl2, SrCl2.

Aluminium; Al3+

. Iron; Fe3+

. Chromium, Cr3+

(Group 3 metals and transition metals) may form AlCl3, FeCl3, CrCl3.

7. Some elements in the transition metals are able to form more than one ion, an example being iron,

Fe.

a) Write the formula for iron(II) sulfate and for iron(III) sulfate.

Iron(II) sulfate: FeSO4, iron(III) sulfate: Fe2(SO4)3

b) Copper can form copper(I) and copper(II) chlorides. Write formulas for these chlorides.

Copper(I) chloride: CuCl, copper(II) chloride CuCl2

8. What are some of the experimental tests you would carry out to decide if ionic bonding was present

in a compound?

Compounds containing ionic bonding have crystalline lattice structures. The crystalline lattice structures are

hard, brittle and not malleable (cannot be moulded into other shapes) and normally have very high melting and

boiling points. Compounds that contain ionic bonding do not conduct electricity in the solid state but will do so in

an aqueous environment or molten state because the free ions are then free to move. Experimental tests

therefore include: testing the melting and boiling points, testing for electrical conductivity in both the solid and

molten state.

9. Write the formula, and show the ions present, for the following:

a) iron(II) chloride

FeCl2, ions present: Fe2+

, Cl-

b) iron(III) chloride

FeCl3, ions present: Fe3+

, Cl-

c) magnesium nitride

Mg3N2, ions present: Mg2+

N3-

d) aluminium oxide

Al2O3, ions present: Al3+

, O2-

e) potassium nitrate

KNO3, ions present: K+, NO3

-

The IUPAC accepted systematic name for potassium nitrate is potassium trioxonitrate(V)

f) calcium fluoride

CaF2, ions present: Ca2+

, F-

g) magnesium hydroxide

Mg(OH)2, ions present: Mg2+

, OH-

h) sodium carbonate

Na2CO3, ions present: Na+, CO3

2-

3

i) ammonium sulfate

(NH4)2SO4, ions present NH4+, SO4

2-

j) sodium phosphate

Na3PO4, ions present: Na+, PO4

3-

k) sodium carbonate

Na2CO3, ions present: Na+, CO3

2-

l) sodium hydrogencarbonate

NaHCO3, ions present: Na+, HCO3

-

The IUPAC accepted common name for sodium hydrogencarbonate is sodium bicarbonate

Now use a circle to indicate in each pair above which species was originally the less electronegative, i.e.

which atomic species became the cation in the above pairs.

The positive ion (cation) forms when a species (often a metal) loses electrons to a more electronegative

species. In the majority of the examples above, the metal loses electrons and is, therefore, the less

electronegative species. In part i) the ionic compound, NH4SO4, has formed as the result of an acid-base

reaction rather than as the result of differing atomic electronegativities.

10. Ionic compounds have many properties in common. Outline the general properties associated with

compounds that have ionic bonding.

• Crystalline structures

• Hard, brittle, high melting solids

• Do not conduct electricity in solid state but conduct when molten or in solution

• Examples include common table salt. Our bodies need these and other salts as a sort of “electricity”

conductor in our bodies. In solution ionic salts are called electrolytes, because they conduct.

11. Crystals of sodium chloride consist of a lattice of Na+ and Cl

-, in which each ion is surrounded by six

of the other kind.

a) Draw a diagram showing how the 6 Cl- surrounding a Na

+ are arranged.

Please note: These diagrams only to show part of the structure of Na+Cl

-(s). In reality there is not only one

Na+ surrounded by six Cl

− nor one Cl

− surrounded by six Na

+. Instead there is always one Na

+ for every one

Cl−, in a repeating fashion as shown in the diagram on page 118.

b) Draw a diagram showing how the 6 Na+ surrounding a Cl

- are arranged.

Please see above diagram and note.

c) Give the name for the structure of sodium chloride

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Continuous ionic lattice.

d) Name the forces holding these ions in place.

The electrostatic attraction between positive and negatively charged ions

e) Explain how this type of bonding is formed.

Sodium needs to lose one electron in order to end up with a full outer shell (octet). Chlorine needs to gain one

more electron to obtain a full outer shell. Sodium transfers one electron completely to chlorine, and the ions Na+

and Cl- are formed. These are then held together by electrostatic attraction to form the ionic crystal lattice.

12. Why does chlorine exist as Cl2 and not just as Cl like elemental argon, Ar?

Chloride is a group 7 non-metal that contains 7 valance electrons. In order to satisfy the octet rule, two chlorine

atoms share one valence electron each with each other through a single bond. Argon on the other hand is a

noble gas and has complete octet in its valence shell. Therefore, it does not need to form additional bonds to

complete its octet.

4.2 Exercises

1. Draw Lewis (electron dot) structures for:

Lewis structures, also known as electron dot structures, show all outer (valence) electrons, plus any charge.

a) Cl-

Cl

b) H+

H

c) O2-

O2

d) CH4

CH

H

H

H

e) NH3

NH H

H

f) H2O

5

OH H

g) NH4+

NH

H

H

H

h) C2H4

CH

H

C H

H

or

CH

HC

H

H

Both are correct Lewis structures. However, the second structure shows the arrangement of the electron pairs due to their repelling affect, known as the VSEPR

(Valence Shell Electron Pair Repulsion) theory.

i) CO32-

CO O

O

j) CO2

CO O

k) BF3

BF F

F

l) SO2

SO O

m) H3O+

OH H

H

n) C2H2

CH C H

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2. Why are only valence (or outer shell) electrons drawn in Lewis dot structures?

Only valence electrons participate in the formation of chemical bonds.

3. Use your Lewis structures to predict the shape of the molecules d) to n) above from question 1

above and explain your reasoning using the VSEPR Theory.

Parts a) – c) are all individual atoms, not molecules, so shape not applicable here.

d) The central atom is carbon; it has four regions of electron density around it, these are the C-H single

bonds. These are all bonding regions, and the electrons in each bond repel the others equally, so

according to VSEPR the structure is tetrahedral.

e) The central atom is nitrogen; it has four regions of electron density around it: three are bonding regions

(N-H single bonds) and one is a non-bonding region (the nitrogen lone pair). According to VSEPR the

lone pair electrons will repel the electrons in the three bonds to a slightly greater extent, so the three

bonding regions are pushed into a trigonal pyramidal shape.

f) The central atom is oxygen; it has four regions of electron density around it: two are bonding regions (O-

H single bonds) and two are non-bonding regions (the two oxygen lone pairs). According to VSEPR the

lone pair electrons will repel the electrons in the two bonds to a slightly greater extent, so the two

bonding regions are pushed into a bent shape.

g) The central atom is nitrogen; it has four regions of electron density around it: all are bonding regions (N-

H single bonds), and the electrons in each bond repel the others equally, so according to VSEPR the

structure is tetrahedral.

h) There are two central atoms in this molecule, both are carbon. Each carbon atom has three regions of

electron density; all are bonding regions (2 x C-H bonds and 1 x C=C double bond on each carbon). By

VSEPR each of these three bonding regions will repel each other equally, therefore the shape around

each carbon is trigonal planar.

i) The central atom is carbon , it has three regions of electron density (2 x C-O single bonds and 1 x C=O

double bond). By VSEPR each of these three bonding regions will repel each other equally, therefore

the shape is trigonal planar.

j) The central atom is carbon, it has two regions of electron density (2 x C=O double bonds). By VSEPR

each of these two bonding regions will repel each other equally therefore the shape is linear.

k) The central atom is boron, it has three regions of electron density (3 x B-F single bonds). By VSEPR

each of these three bonding regions will repel each other equally, therefore the shape is trigonal planar.

l) The central atom is sulfur, it has two regions of electron density (2 x S=O double bonds). By VSEPR

each of these two bonding regions will repel each other equally therefore the shape is linear.

m) The central atom is oxygen, it has four regions of electron density around it: three are bonding regions

(O-H single bonds) and one is a non-bonding region (the oxygen lone pair). According to VSEPR the

lone pair electrons will repel the electrons in the three bonds to a slightly greater extent, so the three

bonding regions are pushed into a trigonal pyramidal shape.

n) The central atom is carbon, it has two regions of electron density (2 x CO triple bonds). By VSEPR each

of these two bonding regions will repel each other equally therefore the shape is linear.

4. Covalent bonding in CO2

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a) Predict, with reasoning, whether a molecule consisting of carbon and oxygen, CO2, will contain

covalent bonding given the positions of the atoms in the periodic table.

Both carbon and oxygen are non-metals, close to each other on the periodic table. Their electronegativities are

therefore not different enough to enable a complete transfer of electrons between the two elements. Carbon

requires 4 electrons to complete its octet and oxygen requires two. Covalent bonding in the form of two C=O

double bonds ensures that both carbon and oxygen complete their octet.

b) Deduce the Lewis structure of the CO2 molecule.

CO O

c) Indicate the bond dipoles on the structure by considering the electronegativity of the atoms

present. Show which atom(s) has the partial positive charge and which atom(s) has the partial

negative charge.

C OOδ

+δ

−δ

−

Oxygen is more electronegative therefore it will withdraw electron density away from carbon, leaving carbon with

a partial positive charge and oxygen with a partial negative charge.

Bond dipole moments are indicated by using

d) Predict the shape and bond angles of CO2 using VSEPR theory. How many regions of electron

density are around the central atom? How many of these are bonding regions of electron

density?

The central atom is carbon, it has two regions of electron density (2 x C=O double bonds). By VSEPR each of

these two bonding regions will repel each other equally at 180 degrees, therefore the shape is linear.

Note that a double bond does not count as two regions of electron density despite the fact that there are two bonds present. A single region of electron density refers to any type of bond – whether it’s a single, double or triple bond – or a lone pair of electrons.

e) From (c) it is shown that the bonds are polar. Is the molecule polar overall? Give your reasons.

Overall the molecule is not polar because the bond moment vectors of the C=O bonds are equal and opposite to

one another. They are equal in magnitude (size) and opposite in direction and, therefore, cancel each other out.

This makes the compound overall non-polar despite its polar bonds.

f) What force is actually holding the atoms within this molecule together?

Electrostatic forces, i.e. the attraction between the positive nuclei and the negatively charged electrons within

the bonds.

5. Explain, with the help of an example, the following terms:

a) single bond

A single bond is one pair of electrons shared between two atoms, for example the C-H bonds in CH4.

b) double bond

A double bond is two pairs of electrons shared between two atoms, e.g. O=O, in O2.

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c) triple bond

A triple bond is three pairs of electrons shared between two atoms, e.g. N≡N, in N2.

d) polar bond

A polar bond arises from the difference in electronegativity between two atoms covalently bonded. The electron

pair is more localised around the atom that has the greatest electronegativity, creating a partial negative charge,

δ-, on that atom and a partial positive charge, δ+, on the less electronegative atom in the bond. An example

H(δ+) – Cl(δ-)

e) coordinate covalent bond (dative bond)

A coordinate covalent bond (or dative bond) is a type of covalent bond in which one atom donates both

electrons. For example, ammonia can react with a proton (hydrogen ion) to form the positively charged

ammonium ion. The bond is formed by the nitrogen using both the electrons in its lone pair in order to form the

covalent bond with the proton:

NH

HH

NH

HHH

H

The proton, H+, has no electrons which it can donate in order to form a bond.

f) bond moment (dipole)

A bond moment is a vector that indicates the magnitude and size of the polarity of a covalent bond, i.e. over

which of the two atoms the electrons spend the majority of their time and, therefore, which atom is partially

positive and which is partially negative.

For example, in the molecules of dichloroethene shown below, there are bond moments indicated for the C-Cl

bonds:

C

H

C

Cl

H

Cl

C

H

C

H

Cl

Cl

bond moment

The chlorine atom is much more electronegative than carbon which results in a polar bond with partial charges

distributed in this way: δ

−δ

+

Cl C . The C−H and C=C bonds are non-polar bonds, so have no bond moments

associated with them.

Carbon and hydrogen are different elements and so their electronegativities are not identical. However, according to the Pauling scale, they are very close. Therefore, despite being different elements, the pull each atom has on the electrons within the

bond is similar and therefore the bond is non-polar.

g) dipole moment

The dipole moment of a molecule is the sum of its individual bond moments. If the molecule has an overall

dipole moment it is polar. If the dipole moment = 0 the molecule is non-polar, even if there are some polar bonds

within the molecule.

For example again considering the molecules of dichloroethene from the previous question:

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C

H

C

Cl

H

Cl

C

H

C

H

Cl

Cl

no overall dipole momentnon-polar molecule

polar molecule

dipole moment

The molecule on the left had two bond moments which, by vector addition, give the overall dipole moment

shown. Therefore the molecule is polar. The molecule on the right has two bond moments in equal and opposite

directions so they cancel each other out and there is no net dipole moment. The molecule is non-polar.

h) polar molecule

A polar molecule has an overall dipole moment, the individual bond dipole vectors do not cancel each other out,

e.g. H−Cl or CH3Cl.

i) bond length

A measurement of the distance between two nuclei. It varies in length depending on the nature of the bond.

Single bonds are longer than double bonds, and double bonds are longer than triple bonds.

j) bond strength

A measurement of the amount of energy required to break the bond apart and, therefore, a reflection of how

tightly the bonds hold the connected atoms together. Strong bonds, like primary bonds (ionic, metallic, covalent)

require large amounts of energy to break. Weak bonds, like secondary bonds (dispersion forces) require small

amounts of energy to break.

k) bond angle

The measurement of the angle at which two atoms sit apart from each other in space. For a tetrahedral

molecule the bond angles are 109.5º.

6. Explain how Valence Shell Electron Pair Repulsion theory can be used to predict the shape and

bond angles in molecules. Address how electron density affects the bond positions.

The basis of the VSEPR theory considers how repulsive forces between valence electrons produces a

geometric shape that results in the repulsive forces being as far apart from each other as possible. The negative

charge density of electrons – whether they are in bonds or lone pairs – affects repulsion and hence the bond

position. Lone pairs of electrons have a higher negative charge density than electrons in a single bond and

therefore repel other regions to a greater extent. This is reflected in the differing bond positions when molecules

contain lone pairs instead of all single bonds. For example, the electrons in the four single bonds around the

carbon atom in methane (CH4) all repel each other equally producing bond angles of 109.5°. However, the

electrons in the three single bonds around the nitrogen atom in ammonia (NH3) do not repel the lone pair on the

nitrogen atom equally. The electrons in the lone pair have a higher charge density (their double negative charge

is more concentrated) and they repel the bonding electrons to a greater extent. This results in angles >109.5°

between the lone pair and the single bonds and < 109.5° between the single bonds.

7. How is the type of bonding in the chlorides of potassium (KCl) and phosphorus (PCl3) related to the

position of K and P in the periodic table?

K is a group 1 metal and can readily give up an electron to a chlorine atom (group 7 non-metal) to obtain a

stable octet. This complete transfer of one electron results in ionic bonding. P is a group 5 non-metal and forms

covalent bonds in order to share electrons with chlorine atoms and so obtain a stable octet.

8. Both N2 and H2O contain covalent bonds, yet only one molecule is polar. Which is this and why?

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H2O is a polar molecule and N2 is a non-polar molecule. N2 is a diatomic molecule that contains two atoms of

the same element which consequently have the same electronegativity. Therefore the electrons are equally

shared between both N atoms. There is no bond moment and therefore no overall dipole moment. N2 is non-

polar. H2O has O−H bonds. Oxygen and hydrogen have different electronegativities and the electrons spend

more time around the O atom than the H atom, creating a bond moment. The dipole vectors do not cancel each

other out, so creating an overall dipole moment in the molecule. Therefore H2O is polar.

9. A, B and C represent elements with atomic numbers 9, 16 and 19 respectively.

a) Give the electron arrangement for each and name the elements

A 2.7 fluorine

B 2.8.6 sulfur

C 2.8.8.1 potassium

b) What kind of bonding would you expect from:

A and B covalent

A and C ionic

B and C iIonic

c) Draw Lewis structures for the compounds formed in (b)

A and B

SF F

A and C

An ionic compound – potassium fluoride, KF: continuous lattice of F- and K

+ ions.

Ions: FK

B and C

An ionic compound – potassium sufide K2S: continuous lattice of S2-

and K+ ions.

Ions: SK

2

10. Using an example, e.g. CH4, outline the formation of covalent bonds.

Covalent bonds are a result of electron sharing. Each atom must share enough electrons with other atoms in

order to complete its octet. In CH4, the carbon atom needs four electrons to complete its octet, so it forms four

bonds with four hydrogen atoms sharing one electron with each. Each hydrogen atom needs one electron to fill

its valence shell (which for hydrogen has only two electrons) so each gets one electron by forming a single bond

with carbon.

11. Using an example, eg. H2O, outline the general physical properties of covalent compounds.

Covalent molecules, like H2O, interact with each other through intermolecular forces. These forces are not very

strong compared to covalent or ionic and metallic primary bonding and this is the reason why most covalent

molecules have relatively low boiling and melting points. They therefore exist mostly as liquids and gases.

Molecules that have covalent bonds do not conduct electricity in the solid or molten state as their outer electrons

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are shared between the atoms and are not free to move between molecules. Water, H2O, is a covalent molecule

which exists as a liquid despite having the strongest of all intermolecular forces between its molecules –

hydrogen-bonding. There are some exceptions to these generalisations of course; continuous covalent lattices

of Si, SiO2 and some allotropes of carbon for example are high melting solids. This is because rather than

existing as single molecules they form three dimensional covalent lattices which require the breaking of strong

covalent bonds to melt.

Section 4.3 covers intermolecular forces in detail.

12. Which of the following is not an electrical conductor?

A: KBr(aq)

B: octane, C8H18

C: HCl(aq)

D: NaCl(l)

Answer: B

Octane is a covalent liquid. There is no conduction because there are no free electrons nor free ions.

KBr and NaCl are ionic compounds and will conduct when molten or in solution. HCl(aq) contains H+(aq) and

Cl−(aq) and so conducts when it ionizes.

13. Which of these group Vll hydrides has the smallest bond dipole?

A: HF

B: HCl

C: HBr

D: HI

Answer: D

Iodine is the lowest in group 7 and it therefore has the lowest electronegativity. Therefore iodine and hydrogen

will have the smallest electronegativity difference of the group 7 hydrides, resulting in the smallest bond dipole.

14. The table below relates to the some chlorides of period 3 in the periodic table:

NaCl MgCl2 SiCl4 PCl3

Type of bonding

Ionic Ionic Covalent Covalent

Structure Lattice of Na

+

and Cl- ions

Lattice of Cl- and

Mg2+

ions SiCl

ClCl

Cl

PCl

ClCl

Boiling Point ºC 1413 1412 58 76

a) Fill in the Type of Bonding row with either ‘ionic’ or ‘covalent’.

b) Name and draw the structure for each. NaCl and MgCl2 ionic lattices; silicon tetrachloride is

tetrahedral; phosphorus trichloride is trigonal pyramidal.

c) Explain, in terms of forces, the difference between the boiling points of NaCl and PCl3.

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NaCl is a compound that contains ionic bonding. This type of primary bonding requires vast amounts of thermal

energy in order to break the ionic lattice and cause the compound to boil. PCl3 is a compound that contains

covalent bonding and the individual molecules are held together by intermolecular interactions: dispersion and

dipole-dipole forces. Intermolecular interactions do not require as much thermal energy to break and therefore

PCl3 has a much lower boiling point.

15. The sequence that shows these molecules in increasing order of carbon-carbon bond strengths is:

A: C2H2, C2H4, C2H6

B: C2H4, C2H6, C2H2

C: C2H2, C2H6, C2H4

D: C2H6, C2H4, C2H2

Answer: D

If you draw out the structural formulas of these molecules, you will find that C2H2 contains a carbon-carbon triple

bond, C2H4 contains a carbon-carbon double bond, and C2H6 contains a carbon-carbon single bond.

16. Bond strengths.

a) Draw a structural formula for ethanoic acid, CH3COOH.

C

O

OH

H3C

b) Comment on the bond lengths and strengths of the C―O bond and the C=O bond present in

ethanoic acid.

The C–O bond is weaker than the C=O double bond. There are more electrons in the double bond, and the

atoms are drawn closer together because of this, making the C=O bond shorter and stronger than the C–O

single bond.

17. What is a difference between the bond formed between H+ and NH3 and the bond formed between H

and Br? Include an explanation as to how each bond is formed.

The bond between H+ and NH3 is a coordinate (or dative) covalent bond, whereas the H and Br bond is a

covalent bond. In the coordinate covalent bond, an electron pair from the N is used to form the bond with H+.

One atom donates both of the two electrons required to form the single bond. The H-Br bond is a “regular”

covalent bond, which is formed when the two atoms both donate an electron to complete their valence shells.

Each atoms donates one of the two electrons required to form the single bond.

18. Is the C―H bond considered to be a polar bond? Again explain your answer.

Though the C−H bond is made up of two different atoms, the bond is said to be non-polar because there is a

small difference in electronegativity between these two elements. This small electronegativity difference means

that the electrons are shared fairly equally between both the carbon and hydrogen atoms, resulting in a non-

polar bond.

19. Which statement correctly describes polar molecules?

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A All molecules with polar bonds are polar.

B Some molecules without polar bonds are polar.

C Polar molecules have bond moments which cancel out.

D Polar molecules have bond moments that do not cancel each other out.

Answer: D (Because in these molecules there is overall polarity)

A: Not all molecules with polar bonds are polar, as the individual bond moments may cancel out, to give a non-

polar molecule overall.

B: Only molecules with polar bonds can have a dipole moment and therefore be classified as being polar.

C: If the bond moments in a molecule cancel out, the molecule is non-polar.

20. Consider the following un-named elements:

Element A B C D E

Electronegativity 0.9 1.0 2.8 3.0 4.0

a) Which two elements would combine to form the most polar compound (which can include the

extreme – an ionic compound)?

A and E would combine to form the most polar compound as they have the greatest difference in

electronegativity. They may undergo a complete transfer of electron(s) to form an ionic bond.

b) Which two elements would combine to form the least polar compound?

A and B would combine to form the least polar compound because they have the smallest electronegativity

difference.

c) Suggest the most likely group and period for element E.

Group 7, period 2, (the element is fluorine).

21. Allotropes

a) What is an allotrope?

An allotrope is the same pure element differing in the geometric structure, i.e. having a different physical forms.

For example, oxygen has two allotropic forms, dioxygen (O2) and ozone or trioxygen (O3), while carbon has

three.

b) Describe and compare the structure and bonding present in the three carbon allotropes:

diamond, graphite and fullerene (C60) to account for properties such as hardness and electrical

conductivity.

Diamond is the hardest mineral known. This hardness results from the covalent crystal structure in which each

carbon atom is linked to four others by covalent bonds to form a continuous tetrahedral structure. The carbon

atoms are all sp3 hybridised. The C−C bond length is 0.154 nm and the bond angel 109.5º and these properties

are constant throughout the entire crystal lattice which results in no point of weakness in the structure. There are

no free electrons in the structure since all four outer electrons of carbon are used in bonding, therefore diamond

does not conduct electricity.

Graphite is a soft black slippery substance. These properties are derived from the structure in which carbon

atoms are bonded covalently to three others to form a flat sheet. The carbon atoms are sp2 hybridised and

sheets of carbon hexagons are are stacked and held together by weak van der Waals’ forces using delocalised

electrons. The delocalised electrons can move between the layers, which explains the electrical conductivity of

14

graphite. The C-C length in the layers of graphite is 0.142 nm and the distance between the layers is 0.34 nm.

Longer bonds are weaker than shorter bonds, therefore there is a weakness between the layers and they slip

over each other, resulting in the soft, slippery properties of graphite.

C60 fullerene is a form of carbon that a bonded to form a cluster which resembles a sphere composed of sp2

hybridised carbons joined to form hexagons and pentagons, to look like a soccer ball. Fullerene does not

conduct electricity. However the properties of fullerenes are still the subject of much scientific investigation.

For diagrams see page 132 or refer to other sources.

22. Continuous covalent network structures

a) Explain the structures of silicon and silicon dioxide.

Silicon can occur in a diamond like continuous covalent network structure, where each silicon atom is covalently

bonded to another silicon atom:

SiSi

SiSi

Si

Silicon dioxide also forms a continuous covalent network structure, with the molecular formula SiO2

Si

Si

SiO

Si

Si

SiO

Si

Si

Si

Because silicon is a larger atom than carbon, it is more stable with single bonds in continuous covalent networks

rather than in forming double and triple bonds like carbon does.

b) Use the bonding and structures to explain why silicon and silicon dioxide (quartz) both have

melting points in excess of 1400º C.

Silicon and silicon dioxide both occur as crystalline structures made from a continuous covalent network of

atoms. It is the continuous network structure made up of strong covalent bonds that gives these compounds

their high melting points. Unlike other covalent molecules that have weak intermolecular forces between

individual molecules, there are no separate molecules in silicon or silicon dioxide. Therefore, in order to convert

between the solid and liquid states, the strong covalent bonds between atoms must be broken and this requires

a lot of energy, reflected in the melting points in excess of 1400º C.

4.3 Exercises

1. Explain how the English meaning of the word “disperse” relates to the use of the word in the term

“dispersion forces” in chemistry.

Disperse means to spread in different directions.

Dispersion forces arise when a molecule collides or comes into close proximity to another molecule and their

associated electron clouds are ‘dispersed’ (displaced or spread) due to electron-electron repulsion. This creates

areas of low and high electron density associated with the molecule, i.e. temporary dipoles within the molecules

that may then become instantaneously attracted to one another. This attraction is only instantaneous. However,

as the electron cloud quickly disperses back, the molecules repel each other once more. This is why dispersion

forces are such a weak intermolecular interaction.

15

See also: diagram question 3 answer below.

2. What is the difference between a permanent electric dipole and temporary electric dipole?

Permanent electric dipole refers to a permanent partial positive and negative charge between molecules, for

example a bond moment or overall dipole moment found in polar molecules. Induced electric dipole refers to

one molecule inducing a partial negative or positive charge into another molecule when in close proximity, such

as the case in a dispersion force or an ion-dipole interaction. Once the molecules separate, the partial charge is

no longer present as the electron density redisperses.

3. Define the following terms with diagram to illustrate your answer:

a) Dispersion force

A weak force of interaction between temporary and induced dipoles.

See question one for an explanation of dispersion forces as well as the following diagram:

1. Collision of two non-polar molecules

2. Associated electron clouds are dispersed due to electron-electron repulsions, resulting in induced dipoles

3. The molecules are then instantaneously attracted to each other due to their induced dipoles

b) Dipole-dipole interaction

Molecules which have a permanent dipole are attracted to one another

For example molecules of HCl interact through dipole-dipole forces:

δ−

δ+

Cl H

δ−

δ+

Cl H

There is electrostatic attraction through oppositely charged dipoles.

c) Ion-dipole interaction

The solvation (dissolving) of salt, NaCl, is an example of ion-dipole interaction. The ion, either Na+ or Cl

-, is

“solvated” by water molecules as shown below:

16

δ−δ

+

H

OH

δ+

δ−

δ+

HO

Hδ

+

δ−

δ+

H

O

H

δ+

δ−

δ+

HO

H

δ+

δ−

δ+

HO

Hδ

+

Na+

δ−

δ+

H O

Hδ

+

δ−

δ+

HO

H

δ+

δ−

δ+

HO

H

δ+

δ− δ

+

H

OH

δ+

δ−

δ+

H

O H

δ+

Cl-

d) Hydrogen bond

Hydrogen bonding is a very strong type of dipole-dipole interaction. The strength of this electrostatic interaction

is a result of the large bond moment that exists when hydrogen is bonded to the highly electronegative and

small atoms, nitrogen, oxygen and fluorine (N, O, F). Hydrogen bonding occurring between molecules of water

is shown:

δ−

δ+

H

OHδ

+

δ−

δ+ H

OH δ

+

When drawing hydrogen bonds make them longer than the covalent bonds. They are an intermolecular interaction and therefore the "bond" is longer and weaker than a

intramolecular interaction.

4. Comment of the relative electronegativities of N and Cl and explain why molecules of NH3 can

hydrogen bond to one another, but HCl molecules cannot.

Nitrogen and Chlorine have the same electronegativity value on the Pauling scale (see IB Chemistry date

booklet). However, nitrogen (period 2) is smaller than chlorine (period 3). This smaller size allows the nitrogen

atom to get closer to hydrogen atoms in adjacent NH3 molecules, so the electrostatic interaction between the δ+

hydrogen and δ- nitrogen atoms is much stronger and hydrogen bonding can occur.

5. Draw a diagram to show the hydrogen bonding that occurs between molecules of ethanol,

CH3CH2OH.

δ−

δ+

H3CH2CO

H

δ−

δ+ CH2CH3

OH

There is no hydrogen bonding with hydrogen atoms that are attached to carbon

atoms. The C–H bond is not polar enough to cause the δ+ charge on the hydrogen

atom in order for it to take part in hydrogen bonding.

6. Hydrogen bonding has been described as nature’s glue. Can you give some examples to support

this statement?

17

The high Tb of water, the open structure of ice and the stickiness of honey depend upon hydrogen bonds.

Hydrogen bonding is important is maintaining biological structures for proteins and DNA fragments. DNA relies

on hydrogen bonding between base pairs to maintain the double helical structure, and the unique three

dimensional structure and folding of proteins is heavily reliant on the individual hydrogen bonds that can occur

between amino acid side chains and water molecules.

Life depends upon the reactions that complex structures, like DNA and proteins, undergo. Certain bonds in these structures must be able not only hold the units

together strongly but also be able to break and reform readily. Two such types of interactions that occur in proteins and DNA are hydrogen bonding, and pi-stacking (an

aromatic interaction-learn about aromatic/benzene rings in Topic 10).

7. Explain how intermolecular forces account for the following properties, use a diagram to illustrate

your answer where possible.

a) Water has a higher boiling point than ammonia

Oxygen has two lone pairs, it can form two hydrogen bonds in the water (H2O) molecule, whereas nitrogen only

has one lone pair and can only form one hydrogen bond per ammonia (NH3) molecule.

b) Hexane C6H14 is immiscible with water

Hexane is a non-polar molecule, water is polar. The two do not mix as the strong polar interactions between

water molecules (hydrogen bonding) cannot be overcome by hexane as it can only have weak dispersion and

induced dipole interactions with water.

c) Solid water (ice) less dense than liquid water

When water freezes an open structure is formed as the molecules are held apart and in an ordered network

structure by hydrogen bonding. Each ice molecule is hydrogen bonded to four other ice molecules

δ−

δ+

H

OHδ

+

δ−

δ+ H

OH δ

+

This structure is less dense than liquid water, which although it has hydrogen bonding, has too much energy for

the molecules to be fixed in a rigid structure. Hence the molecules are a lot closer to one another are therefore

more dense in liquid water. When the kinetic energy of the water molecules is removed by freezing, they are

able to resume the highly ordered, more open structure of ice. When ice melts some of the hydrogen bonds

break and the water molecules come closer to each other.

d) NH3 has a higher boiling point than CH4

Ammonia, NH3 is capable of hydrogen bonding which is the strongest intermolecular attraction. This is a strong

force that requires more energy to break, hence the boiling point of NH3 is higher than that of CH4. CH4 is a non-

polar molecule and only has weak dispersion forces. Therefore less energy is required to interrupt these forces

so making the boiling point lower.

e) KCl has a higher melting point than I2

KCl is an ionic species, it is held together by electrostatic attraction between ions which results in a strong

continuous crystalline lattice. Iodine, I2, on the other hand, is a non polar molecule with only weak dispersion

forces between molecules of I2. More energy is required to break the strong electrostatic interactions of the

continuous lattice of KCl than the weak dispersion forces of I2. Therefore KCl has a higher melting point than I2.

18

8. List the types of intermolecular forces that exist between molecules of each of the following

species:

a) CH3Cl

Dipole-dipole interactions

b) NaCl

Ionic bond - electrostatic attraction.

c) CS2

Dispersion forces

d) PF3

Dipole-dipole interactions

e) CH3OH

Hydrogen bonding

f) CH3(CH2)4CH3

Dispersion forces

9. The following graph illustrates the boiling point (K) against period number for the following

hydrides, H2O, H2S, H2Se and H2Te. Explain the deviation shown by water.

-1000100123456

2 3 4 5

0

100

-100

Period

Bo

ilin

g p

oin

t (o

C)

Boiling points of group 6 hydrides

Since oxygen, sulfur, selenium and tellurium are all in group 6, one may expect that the properties of hydrides of

this group would be similar to one another, or at least follow a trend. The graph shows that H2O, the hydride of

oxygen (period 2) deviates quite significantly from the increasing boiling point trend of the other group 6

hydrides. This is because water is capable of hydrogen bonding interactions, whereas all the other hydrides can

only interact through weaker dipole-dipole interactions.

You may be wondering why hydrides of sulfur, selenium and tellurium follow a slow steady increase in boiling point. This is because the size of the atoms of the group 6 elements increases down the group and this increases the dispersion forces at work between their molecules. The larger molecules are, the more electron density they

19

have and the stronger the dispersion forces become. These forces become significant, for example, in plastics. Plastics are made up of very long carbon chains that exist as solids at room temperature due to the presence of extensive dispersion forces between the very large molecules.

10. Nitrogen and phosphorus are in the same group on the periodic table. Account for the large

difference in boiling points of the hydrides, NH3 -33ºC and PH3 -90ºC

The difference in boiling points between NH3 and PH3 is due to the types of intermolecular interactions present

between the different molecules. PH3 has dipole-dipole interactions, whereas NH3 is able to form hydrogen

bonds, which is a much stronger interaction. Hence more thermal energy is required to break intermolecular

forces between NH3 molecules and therefore it has a higher boiling point than PH3.

11. Despite their similar molecular masses, propane, C3H8 and ethanol, C2H5OH have boiling points of

231 K and 351 K respectively. Account for this seemingly large discrepancy.

The difference in the boiling points between ethanol and propane is due to the types of intermolecular

interactions present between the different molecules. Propane is a non-polar molecule and only contains

dispersion forces, the weakest type of secondary interactions. Ethanol is a polar molecule and has hydrogen

bonding as well as dispersion forces. Hydrogen bonding is much stronger than dispersion forces, and hence

requires more thermal energy to break and therefore ethanol has a higher boiling point.

12. The smell associated with decaying animal matter is, in part, due to the smelly gas mixture of

ammonia, NH3, methane, CH4, and hydrogen sulfide, H2S.

a) Draw Lewis structures for each, showing all valence electrons.

SH HCH

H

H

H

NH H

H

b) Describe the geometry of the electron pairs and the atoms around the central atom in each

molecule and predict the bond angles around the central atom in each molecule, using the

valence shell electron pair repulsion (VSEPR) theory.

In NH3 there are four areas of electron density around the central nitrogen atom, three bonding (N-H bonds),

and one non-bonding (nitrogen lone pair). The nitrogen lone pair repels the electrons in the N-H bonds more

than they repel each other, which results in the angles being slightly more or less than 109.5º as displayed on

the diagram below. The structure is a trigonal pyramidal molecular shape.

>109.5º

<109.5º

NH

HH

In CH4 there are four regions of electron density around the central carbon atom, all of which are bonding

regions (4 x C-H bonds). The electrons in these bonds all repel each other equally so the result is a tetrahedral

molecular shape.

20

all bond angles 109.5º

CH

HH

H

In H2S there are four regions of electron density around the central sulfur atom, two bonding (S-H bonds) two

non bonding (sulfur lone pairs). The lone pairs repel the bonding electrons more, which results in the angles

being slightly less or more than 109.5º. The structure is a bent molecular shape.

>109.5º

<109.5º

SH

H

c) The relative molecular masses, Mr, and boiling points, Tb/K, of these three substances are given

below. Identify the intermolecular forces present in each substance and account for the boiling

points in terms of these forces.

Compound Tb/K, Mr Intermolecular forces

ammonia 239.2 17.03

Hydrogen bonding, highest boiling point as

it is the strongest type of intermolecular

interaction

methane 111.6 16.04 Dispersion forces, lowest boiling point

weakest type of intermolecular interaction

hydrogen sulfide 212.9 34.08

Dipole-dipole interactions, moderate boiling

point due to moderate strength

intermolecular interaction

13. Water is the most abundant compound on Earth. Much of the chemistry of water is influenced by its

polarity and its ability to form hydrogen bonds.

a) Why are water molecules polar? Use a diagram to illustrate your answer.

OH

H

overall dipole moment

Water molecules contain two polar O-H bonds. The vector addition of these bond moments results in an overall

dipole moment for the molecule, meaning it is polar.

Although water is abundant on earth, only a very small amount (about 3%) is fresh water. Much of the water on Earth is trapped in glaciers or found as seawater.

b) The high polarity of the O-H bond in water results in the formation of hydrogen bonds. Draw a

diagram including dipoles and lone pairs of electrons, to illustrate hydrogen bonding between

water molecules.

21

δ−

δ+

H

OHδ

+

δ−

δ+ H

OH δ

+

c) Why must many people living in Canada drain their swimming pools before winter?

Water expands when it freezes, for reasons outlined in question 7 c). The people in Canada need to drain their

pools to stop damage to the structure that the large volume of water expanding as it freezes would cause.

14. Consider the following hydrides: NaH, NH3, H2O and HCl. Explain how the difference in the polar

nature of these hydrides can lead to a difference in their acid character, and their ability to form

hydrogen bonding. Give equations to support your answer.

NaH is an ionic material composed of Na+ and H

- ions. It is extremely reactive and used as a base in organic

syntheses as it readily removes a proton from even weak acids:

NaH + HA � Na+A

- + H2

NH3 is a covalent compound with polar N-H bonds. The nitrogen is electronegative and has a lone pair. Thus

ammonium acts as a base (accepts a proton) and reacts with water by the following equation:

NH3 + H2O � NH4+ + OH

-

Water may act as an acid or a base:

H2O + HA � H3O + A_ (water acts as BASE)

H2O + A_ � OH

_ + HA (water acts as ACID)

HCl is a polar covalent molecule. Due to its large polarity the HCl bond is quite weak and HCl is therefore acidic:

HCl + H2O � Cl- + H3O

+

Only water and ammonia are capable or H-bonding. NaH is ionic so will not hydrogen bond (the species are

already ionised, the electrostatic interactions by far overcome any intermolecular interactions). HCl cannot be

involved in hydrogen bonding as the Cl atom is too large (period 3), the atoms simply cannot get close enough

to one another for the hydrogen bonds to form.

15. Cabon dioxide CO2 and silicon dioxide SiO2 are oxides of group 4 elements. CO2 is a gas at room

temperature, while SiO2 is a hard, high melting solid.

a) Draw a Lewis structure for CO2.

CO O

b) Predict the shape and polarity of CO2 using VSEPR theory.

There are two regions of electron density around the central carbon atom, and according to VSEPR theory the

electrons in these bonds will repel one another equally at 180º. Therefore CO2 will have a linear molecular

shape. The bond moments of the C=O bonds are in equal and opposite so the CO2 molecule is non-polar.

c) Draw the structure of SiO2.

22

Si

Si

SiO

Si

Si

SiO

Si

Si

Si

d) Why do CO2 and SiO2 have such different structures?

Because silicon is bigger than carbon, it is more stable with single bonds in continuous covalent networks rather

than forming double and triple bonds like carbon does. Silicon cannot form double bonds with oxygen like

carbon so it cannot form a discrete SiO2 molecule. Instead its octet must be satisfied by forming four individual

covalent bonds in a continuous network structure.

e) Explain how their structures relate to the large difference in boiling points and other physical

properties.

Silicon is a continuous covalent network of silicon and oxygen, which is very strong and gives silicon dioxide its

hard, high melting properties. Carbon dioxide is a small non-polar molecule that is held together by weak

dispersion forces. At room temperature there is enough thermal energy to break the dispersion forces apart;

hence carbon dioxide is a gas at room temperature.

16. Consider the following molecules, ethane CH3CH3 (bp 184.5 K), CH3CHO (bp 293 K) and CH3CH2OH

(bp 352 K).

a) Calculate the Mr of each.

CH3CH3 Mr = 30.08, CH3CHO Mr = 44.06, CH3CH2OH Mr = 46.08.

b) Explain the differences in their boiling points in terms of their intermolecular forces.

The difference in the boiling point of ethane, ethanal and ethanol is due to the type of intermolecular interactions

present between the different molecules. Ethane is a non-polar molecule with only weak dispersion forces, it

also has a smaller mass than the other two, so the dispersion forces will be much less. Ethanal and ethanol

have quite similar masses, so there must be differences in their intermolecular interactions to account for the

large difference in their boiling points. Ethanal is a polar molecule (due to polar C=O bond) and has dipole-

dipole interactions. Ethanol is a polar molecule and is capable of hydrogen bonding. Dipole-dipole interactions

are stronger than dispersion forces and hydrogen bonding is much stronger than dipole-dipole interactions.

More thermal energy is required to overcome the stronger interactions, therefore the highest boiling point is

ethanol (CH3CH2OH) followed by ethanal (CH3CHO) and ethane (CH3CH3).

23

4.4 Exercises

1. Explain the term: delocalised electrons.

The spreading of valence electrons over two or more bonds or atoms. In certain compounds, such as benzenes

and the nitrates, and in metals, the valence electrons are not restricted to definite bonds between atoms but

rather are spread out over several atoms and are free to move.

2. Illustrate metallic bonding using a diagram.

Na+ Na+ Na+ Na+ Na+ Na+ Na+ Na+

In metallic bonding the metal (in this case sodium) loses its valence electron (becomes ionised) and they are

spread out and able to move virtually freely through the lattice forming an electron cloud as shown in the

diagram.

The metal bond is the attraction the positive ions have for the delocalised electrons around them.

3. Properties of metals include ductility, malleability, electrical conductivity, high density.

a) Explain the terms ductile and malleable.

Ductile means able to be shaped into sheets or wires; malleable means able to be hammered out or pressed

into shape, e.g. wire or jewellery.

b) How does the model used for metallic bonding explain the above four properties of metals?

The existence of these “free” electrons accounts for the good thermal and electrical conductivities of metals.

The electrons can move freely so as to transfer heat and electricity. They also allow the structure to be moulded

or bent without snapping or breaking as they can resume a new shape easily, hence the ductility/malleability of

metals. Metals have high melting points as large amounts of thermal energy is required to break apart the

cations from the delocalised electrons; they are held together by strong electrostatic interactions

4. Explain why it is possible to bend metals but not ionic crystals.

Metals can be ‘bent’ into shapes as metallic bonding involves cations surrounded by delocalised electrons,

which are free to move, which allows the structure to be bent and shaped. Ionic crystals do not contain these

“free” delocalised electrons, the lattice is fixed as it contains both cations and anions which are attracted to each

other and cannot be disrupted or bent without breaking the lattice altogether.

24

4.5 Exercises

1. Write the formula for each of the following substances and using table 1, classify each of the

according to the type of bonding present. For parts b) – h), indicate any intermolecular forces present,

and explain the link between bonding and the physical properties of the substance.

Table 1 is on page124 and gives electronegativities of some common elements.

a) Iron

Fe, metallic bonding. The delocalisation of electrons and strong electrostatic attraction between these electrons

and metal cations results in most metals existing as high melting solids at room temperature.

b) iodine

I2: molecular covalent structure. Intermolecular forces holding I2 molecules together are weak dispersion forces.

Due to the size of Iodine, would have a relatively high boiling point when compared to the other group 7

elements. This explains why iodine is a solid at room temperature and the halogens higher in the group are

gases.

c) diamond

An allotrope of carbon, continuous covalent network structure. It is hard, crystalline and a solid at room

temperature. The lattice requires a large amount of thermal energy to break it apart.

d) sugar (sucrose)

C12H22O11: molecular covalent structure. It contains OH groups that are available for hydrogen bonding, it is also

a relatively large molecule so there are a significant amount of dispersion forces present hence sucrose is a

solid at room temperature.

e) hydrogen chloride gas

HCl: molecular covalent structure with dipole-dipole interactions due to the large difference in electronegativity.

As HCl is a small molecule, and is not capable of hydrogen bonding; it is a gas at room temperature.

Hydrochloric acid is actually HCl gas dissolved in water. The gas reacts with water to form H

+(aq) and Cl

−(aq). The reaction of HCl(g) as it dissolves in water is an acid-base

reaction. Topic 8 discusses this type of chemistry further.

f) sodium chloride

NaCl: ionic bonding, large electronegativity difference, a continuous lattice of ions which requires large amounts

of thermal energy to break; hence a hard, high melting solid at room temperature.

g) ice

H2O(s): molecular covalent structure, hydrogen bonding interactions are formed in a regular repeating pattern of

H2O molecules that are evenly spaced apart. At room temperature the water molecules have too much kinetic

energy to resume this highly ordered, open crystalline structure so water is a liquid at room temperature. It is

also denser than ice.

h) quartz

Silicon dioxide, SiO2: continuous covalent network. It is an extremely high melting solid at room temperature, as

the network requires a large amount of thermal energy to break apart.

i) sulphur

Sulfur forms more than 30 solid allotropes, more than any other element. It exists mostly as S8 in covalently

bonded rings.

j) copper(II) sulphate

25

CuSO4: Ionic bonding, strong electrostatic interactions between Cu2+

ions and SO42-

requires large amount of

thermal energy to break, solid at room temperature.

k) bronze

Cu and Sn alloy, metallic bonding. Hard, shiny.

2. Explain the difference between electronic and ionic conduction.

Electronic conduction is the movement of electrons; ionic conduction is the movement of ions, charged

molecules. Conduction by ions results in chemical reactions (electrolysis).

3. Account for the following observations;

a) ionic solids do not conduct electricity, but metals do.

Ionic solid cannot conduct electricity as there is no way for the electrons to pass through the lattice. The ions are

held fixed in place. Solid metals can conduct electricity as the ‘sea’ of delocalised electrons is able to move

freely throughout the lattice.

Metals conduct electricity without decomposing. As electrons are forced in, the same number are forced out accompanied by a temperature increase.

b) molten ionic compounds do conduct electricity.

In the molten state, the ions are mobile and can conduct electricity by the movement of the anions from the

cathode to the anode. However, as they give up their charge a chemical reaction occurs.

c) compounds which are gases at room temperature are normally small molecules.

Compounds that are gases at room temperature generally are small molecules, most non-polar. These

compounds are held together by dispersion forces or dipole-dipole interactions that do not require larger

amounts of thermal energy to break. The dispersion forces are weaker than in larger molecules due to the

reduced surface area of small molecules. At room temperature, there is enough energy to break apart these

secondary interactions so the compounds exist in the gaseous state.

d) Aluminium oxide is ionic, but aluminium chloride is polar covalent.

Oxygen is more electronegative than chlorine (Pauling scale values 3.5 and 3 respectively). Aluminium has a

Pauling scale value of 1.5. As a general rule, compounds that differ by 2.0 or more points on the Pauling scale

form ionic compounds. Therefore, aluminium oxide is ionic, but aluminium chloride is polar covalent.

e) Graphite is used in pencils, while diamond is used in glass cutting instruments.

Graphite and diamond are allotropes of solid carbon. Graphite is arranged in planar sheets of covalently bonded

hexagonal covalent lattices which are held together by weak dispersion forces. These sheets are therefore able

to slip over each other easily. The result is a soft slippery compound. Diamond exists in a three dimensional

covalent network structure consisting of all strong covalent bonds connecting each carbon atom. This highly

ordered, repetitive structure is very hard to break, hence diamond is the hardest mineral known as is often used

in glass cutting instruments.

f) petrol is highly volatile, water has low volatility

Volatility is a measure of a compounds ability to evaporate, i.e. change from liquid to a vapour (gas). Petrol (i.e.

octane CH3(CH2)6CH3) is non polar and has weaker intermolecular interactions than water. Petrol only has

dispersion forces holding it in the liquid phase, only a small amount of thermal energy is required to break apart

these dispersion forces and allow some molecules to escape into the vapour phase (gaseous phase). Water is

capable of hydrogen bonding, the strongest type of intermolecular interactions. Water molecules require a

26

relatively large amount of energy to break the hydrogen bonding present in order to` release water molecules

into the vapour phase.

g) adding salt to water lowers the boiling point of the water

Does it? No! Adding salt to water increases the boiling point of the water, contrary to popular belief about why

we add salt to a pot of pasta! The reasons for this are beyond the scope of this course but are derived from a

concept called vapour pressure. However, adding salt to water lowers the freezing point of the water.

All liquids and solids have a tendency to evaporate into their gaseous (or vapour) forms, and all gases have a tendency to condense back into their liquid or solid forms. The vapour pressure of a substance is the pressure exerted by its vapour form. Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution of salt in water has a higher boiling point than pure water. A liquid will boil when its vapour pressure equals the surrounding pressure. If the water is diluted by adding salt for example, the presence of this solute decreases the vapour pressure of the water, and more energy is required to boil it. Freezing point depression describes the phenomenon that the freezing point of a liquid such as water will be lowered when another compound (such as salt) is added. This can be observed for seawater, when remains unfrozen at 0°C.

4. Consider the changes given below and for each state the type of bonds or forces being broken and

the types of particles being separated.

a) water boiling

Change of state, requires the breaking of hydrogen bonds between the water molecules.

b) iodine subliming (solid ���� gas)

Change of state, requires the breaking of dispersion forces between the I2 molecules

c) H2(g) ���� 2H

Converting a diatomic molecule into its atoms, requires the breaking of a molecular covalent bond. This requires

much more energy than breaking intermolecular interactions, such as hydrogen bonding or dispersion forces!

d) NaCl dissolving in water

In forming an aqueous solution the Na+ and Cl

- ions in the crystal lattice are solvated by water molecules. This

ion-dipole interaction produces Na+(aq) and Cl

-(aq).

27

5. The following unknown substances were tested in order to classify them. Classify the following

substances as ionic solids, metallic solids, molecular/covalent solids or network compounds

(continuous covalent lattice)

Substance Properties Melting

point, ºC Conducts

electricity? Soluble in

water? Classification

A hard,

colourless 700

only if melted or dissolved

soluble Ionic

B lustrous, malleable

1600 High insoluble Metallic

C hard,

colourless 130 None soluble

Polar molecular covalent

D soft, yellow 110 None insoluble Non-polar molecular covalent

E very hard, colourless

1650 None insoluble Continuous

covalent network

F hard, orange 398 yes, if melted or

dissolved in water

soluble Ionic

6. As general rule is ‘likes dissolve like’, enlarge upon this statement by giving explaining the

following phenomena:

a) ionic compounds dissolving in water

Water is a polar molecular covalent molecule. When ionic compounds are dissolved in water the ions are

“solvated” by water molecules through ion-dipole interactions. See diagrams from question 4.3, 3 (c).

b) petrol floating on water, not dissolving in water

Hydrogen bonding between water molecules is strong and petrol is capable of only weak dispersion forces, so

cannot interrupt the strong interactions between water molecules.

c) ethanol and water being miscible (i.e. completely dissolve)

Ethanol is able to form hydrogen bonds with the water molecules through the common ―OH groups.

7. Magnesium oxide and hydrogen oxide are both compounds containing the element oxygen.

a) By referring to their positions in the periodic table predict the type of bonding likely to be found

in both

MgO, the electronegativity difference is large so ionic bonding occurs. Mg is a group 2 metal and can donate 2

electrons to become Mg2+

and O accepts 2 electrons to become O2-

. In hydrogen oxide (water H2O!) the

electronegativity difference between H and O is not large enough to form ionic bonding; instead a covalent

molecule is formed. Oxygen requires two electrons to fill its octet, hydrogen requires one, two atoms of

hydrogen combine with one oxygen atom via two single bonds to satisfy the octet of both atoms.

b) Explain how this knowledge of bonding can explain the fact that magnesium oxide is a solid with

a high melting point while hydrogen oxide is normally a liquid and, as ice, has a low melting

point

Ionic compounds have a high melting point as it requires a large amount of thermal energy to break apart the

ionic lattice. Water is a molecular covalent molecule that is held together by strong hydrogen bonding. However,

while these are the strongest type of intermolecular interactions (because water is a small molecule and O has a

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high electronegativity) there is still enough thermal energy at room temperature to ensure that water will not

exist as a solid. It is the hydrogen bonding that ensures that it is not a gas at room temperature!

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