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Quick Revision/Bonding and Structure/P.1

Physical Chemistry: Bonding and Structure

I. Ionic Bonds

A. Definition

The formation of ionic bond in terms of electrostatic attraction between ions of opposite charge. These ions

are formed by complete transfer of electrons between atoms.

B. Energetics of Formation of Ionic Compounds

a. Lattice energy of an ionic compound is the standard heat of formation one mole of the crystal lattice from its

constituent ions in the gaseous state.

M+(g) + X-(g) M+X-(s) H298 = lattice energy

b. Taking the formation of NaCl(s) as an example,

Na(s) + 1/2 Cl2(g) NaCl(s)H

f[NaCl(s)]

Na(g) Cl(g)

Na+(g) Cl-(g)

Hatom

[Na(s)] Hatom[Cl2(g)]

HIE HEA

Hlattice

By Hesss Law,

Hlattice = Hf[NaCl(s)] {Hatom[Na(s)] + Hatom[Cl2(g)] + HIE + HEA}

c. Theoretical lattice energy of ionic compounds could be found by calculation, basing of some assumptions:

(i) ions are spherical, separate entities;

(ii) charges are distributed uniformly around ions;

(iii) transfer of electrons is complete.

d. Comparison between the experimental lattice energy and theoretical lattice energy can illustrate the degree of ionic

character of the ionic bond. Large difference between the experimental lattice energy and theoretical lattice energy

shows that the bond is not purely ionic.

C. Ionic crystals

A unit cell is the smallest basic portion of the space lattice that, when repeatedly stacked together at various

directions, can reproduce the entire crystal structure.

a. NaCl lattice

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two interpenetrating face centred cubic lattices of Na+ and Cl-.

Number of Na+ = (1/4)12 + 1 = 4

Number of Cl- = (1/8)8 + (1/2)6 = 4

Co-ordination number of Na+ (and Cl-) = 6

[Co-ordination number is the number of nearest neighbours.]

b. CsCl lattice

two interpenetrating simple cubic lattices of caesium and chloride ions.

Number of Cs+ = 1

Number of Cl- = (1/8)8 = 1

Co-ordination number of Cs+ (and Cl-) = 8

c. CaF2 lattice

Calcium fluoride has the fluorite structure, which is based on the

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(i) face-centred cubic lattice of calcium ions.

(ii) fluoride ions are located at the tetrahedral holes of the lattice and each fluoride ion is tetrahedrally

surrounded by four calcium ions.

Number of Ca2+ = (1/8)8 + (1/2)6 = 4

Number of F- = 1(8) = 8

Co-ordination number of Ca2+ = 8; co-orindation number of F- = 4

D. Ionic Radii

a. Cations are smaller than the atom from which they are formed, while anions are larger.

1. Cation has a smaller radius than the corresponding atom because

(i) after losing the outermost shell electron(s) in forming the positive ion, the number of electron shells

decreases.

(ii) the number of protons is greater than the number of electrons in a cation, the attraction of the nucleus for

the remaining electrons (effective nuclear charge) must be stronger than that in the atom.

2. An anion has a larger radius than the corresponding atom because

(i) the repulsion between the newly added electron(s) and the electrons already present.

(ii) the number of protons is smaller than the number of electrons in an anion, the effective nuclear attraction for

the electron cloud becomes weaker.

b.

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For a series of ions with the same arrangement of electrons as in neon, e.g. F-, Na+, Mg2+, Al3+, the size of the

ion decreases as the atomic number increases. Increasing atomic number means increasing nuclear charge and a

correspondingly greater inwards attraction on the electrons.

II. Covalent bonds

Atoms may also achieve the noble gas electronic structures by sharing of electrons (overlapping of orbitals) to

form covalent bonds.

A. Dative covalent bond

When one atom provides both of the electrons necessary for the formation of a single covalent bond. Once the

dative bond is formed it is indistinguishable from other covalent bonds of the same type.

e.g. H N

H

H

B

F

F

F |

|

|

|

H H H+ ++ : N

H

H

[ H N

H

H

|

|

|

|

]

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B. Bond energy

Bond energy is the enthalpy change when one mole of a particular bond is broken under standard conditions,

producing free atoms or radicals.

e.g. H2(g) 2H(g)

a. Example

Find the heat of reaction of the following reactions

CH2=CH2(g) + H2(g) CH3CH3(g)

Solution

CH2=CH2(g)+ H2(g) CH3CH3(g)

2C(g) + 6H(g)H1

H

H2

H1 = E(C=C) + 4E(C-H) + E(H-H) = +611 + 4(413) + 435=+2698

H2 = E(C-C) + 6E(C-H) = +346 + 6(413) = +2824

H = H1 - H2 = +2698 2824 = -126 kJmol-1

b. Breakdown the Additive Rule of Bond Energies on Calculating Heats of Reaction

Some compounds containing conjugated double bonds such as benzene, cannot be represented by simple

structures. The additive rule of bond energies will be broken down when it is used to calculate the enthalpies of

some reactions involving these compounds.

C. The Shapes of Covalent Molecules and Polyatomic ions

No. of

electron

pairs

Basic arrangement of the

electron pairs

Type and no. of

electron pairs

Examples Shape

2 Linear

AB B

180o2 bond pairs BeCl2, CO2,

[Ag(CN)2]-Linear

AB B

180o

3 Trigonal planar 3 bond pairs BF3, BCl3, BI3, SO3 Trigonal Planar

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A

B

B

120o

B

A

B

B

120o

B

2 bond pairs

1 lone pair

SO2, SeO2, PbCl2 V-shaped

A

B B

xx

less than

120o

No. of

electron

pairs


electron pairs

Type and no. of

electron pairs

Examples Shape

4 Tetrahedral

AB

B

B

B

B

4 bond pairs CH4, CHCl3, CCl4,

SiCl4, NH4+,

PCl5+

Tetrahedral

A

B

B

B

B

109.5o

3 bond pairs

1 lond pair

NH3, PCl3, NF3, H3O+ Pyramidal

A

B B

xx

104.5o

xx

2 bond pairs2 lone pairs

H2O, H2S, H2Se

Se: Selenium, Gp.6

V-shaped

A

B B

xx

104.5o

xx

5 Trigonal Bipyramidal

AB

B

B

B

B

5 bond pairs PCl5, SbCl5

Sb: Antimony, Gp.5

Trigonal Bipyramidal

AB

B

B

B

B

4 bond pairs

1 lone pair

ICl4+ Distorted tetrahedral

A

B

B

B

B

xx

3 bond pairs

2 lone pair

ClF3, BrF3 T-shaped

AB

B

B

xx

xx

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2 bond pairs

3 lone pair

I3-, Icl2

- Linear

A

B

B

xx

xx

xx

No. of

electron

pairs


electron pairs

Type and no. of

electron pairs

Examples Shape

6 Octahedral

A

B

B

BB

B

B

6 bond pairs SF6, PCl6- Octahedral

A

B

B

BB

B

B

5 bond pairs

1 lone pair

SbF52- Square pyramidal

A

B

BB

B

B xx

4 bond pairs2 lone pairs

XeF4, SeBr42-

Square Planar

A

B

BB

B xx

xx

D. Sigma ( ) and Pi ( ) Bonds

(i) Sigma ( ) bond

For all single covalent bonds, orbitals overlap directly so as to form sigma () bonds.

(ii) Pi ( ) bond

The lateral overlap of two parallel p orbitals of adjacent atoms results in the formation of an entirely different

type of covalent bond, which is called a Pi () bond.

E. Hybridization Theory

Rules for hybridization:

(1) Hybridization is a process of mixing atomic orbitals on a single atom (ion).

(2) Only orbitals of similar energies can be mixed to form hybrids.

(3) The number of orbitals mixed equals the number of hybrids obtained.

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(4) All hybrids are similar, but differ from one another in orientation in space.

a. sp hybridizationExample

BeF2

1s 2s 2p

ground state of Be

1s 2s 2p

excited state of Be

1s sp 2p

sp hybridization

The molecule is considered to be formed by overlapping of each of two hybrid orbitals of Be atom with the singly

occupied orbital of each of two chlorine atoms, 3p orbital.

Two bonds results. Hence it is predicted that BeCl2 has a linear structure.

b. sp2 hybridization

e.g. BF3

1s 2s 2p

ground state of B

1s 2s 2p

excited state of B

1s sp2 2p

sp2 hybridization

Each sp2 hybrid orbitals of boron overlaps with the 2p orbital from each of three F atoms. 3 bonds give a planar

trigonal molecule.

c. sp3 hybridization

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e.g. CH4

1s 2s 2p

ground state of C1s 2s 2p

excited state of C

1s sp3

sp3 hybridization

Overlapping of sp3 hybrids of C with four 1s orbitals of four H atoms give rise to four equivalent bonds. The

molecule takes a shape of a tetrahedron. Each bond angle is 109.5o.

d. sp3d hybridization

e.g. phosphorus pentachloride PCl5

1s 2s 2p 3s 3p 3d

ground state of P

1s 2s 2p 3s 3p 3d

excited state of P

1s 2s 2p sp3d 3d

sp3d hybridization

Shape of the sp3d hybrids: trigonal bipyramidal

e. sp3d2 hybridization

e.g. SF6 molecules

1s 2s 2p 3s 3p 3d

ground state of S

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1s 2s 2p 3s 3p 3d

excited state of S

1s 2s 2p sp3d2 3d

sp3d2 hybridization

Shape of sp3d2 hybridized orbitals - octahedral (all bond angles = 90o)

F. Covalent crystals

a. Diamond

all the carbon atoms are sp3

hybridized, joining into a three dimensional network by strong, rigid and directionalC-C covalent bonds. Each carbon is tetrahedrally bonded to four neighbours by strong covalent bonds.

b. Graphite

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carbon atoms are all sp2 hybridized, so that each is bonded planar trigonally to three other carbon atoms by

single, rigid covalent bonds, forming a hexagonal layer lattice.

The unhybridized p electrons in the bonding carbon atoms overlap with each other, and are delocalized above and

below the layer lattice, thus holding the lattices together by weak van der Waals' forces.

c. Quartz (Silicon(IV) oxide, SiO2)

In silicon(IV) oxide structure, each silicon atom is covalently bonded to 4 oxygen atoms tetrahedrally. Each

oxygen atom is bonded to 2 silicon atoms.

III. Intermediate Bonding Between Ionic and Covalent

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A. Polarization Of Ionic Bonds/Ions (i.e. ionic bond with covalent character)

Factors affecting the polarization of ions

a. Size

Cation the smaller the cation the greater is the distortion of electron clouds of the neighbour anions

Anion - the larger the anion the easier it will be distorted.

b. Charge

The larger the charge of the cation the better is the effective it will be in causing distortion.

Conclusion, cations of small size and high charge (i.e. higher charge density) will cause the polarization of ionic

bonds easily.

e.g. AlCl3, FeCl3, ZnS are the common examples of ionic compounds with covalent character.

B. Polarization of covalent bonds (i.e, covalent bond with ionic character)

Polarization of covalent bond is due to the large difference in electronegativity of the bonding atoms.

e.g. HCl

Cl is very electronegative, hence the bonding electrons are attracted to the side of the Cl atom and then the Cl

carries partial negative charge while the H carries partial positive charge.

IV. Metallic Bond

A. Atoms/ions in metals are attracted by the delocalized electrons in the metallic structure.

B. Metallic structure

a. Close packed structure

(i) hexagonal close packed structure (hcp); with ababab. structure

Co-ordination number = 12

Packing efficiency = 74%

(ii) cubic close packed structure (ccp) or face-centred cubic (fcc); with abcabc. structure

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b. open structure / Body centred cubic strucuture (b.c.c.)



c. Hexagonal close packed (h.c.p.): Be, Mg, Sc, Ti, Co and Zn

Cubic close packed (c.c.p.): Ca, Fe, Cu and Pb

Body centred cubic (b.c.c.): alkali metals

C. Alloys

a. When compared with pure metals

(i) is stronger and harder

(ii) is less malleable and ductile

(iii) has a lower electrical and thermal conductivity and

(iv) has a lower melting point

b.

Alloy Composition In compare with pure metal Uses (in making)

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Brass 70% copper

30% zinc

cheaper than copper, golden

appearance, malleable, harder and

more corrosion resistant

musical instrument,

ornament, electrical

fitting, buttons,

screw, door knobs

Bronze 90-95% copper

5-10% tin

more expensive than brass, yellowish

brown appearance, malleable,

stronger and more resistant to wear

than brass

ancient weapons,

ornaments, statues,

church bell, ship

propellers

Coinage

Metal

Silver coin:

75% copper

25% nickel

Copper coin:

97% copper

2.5% nickel

0.5% tin

harder than copper, malleable and

corrosion resistant coins

Duralumin 95% aluminium

3% copper

1% magnesium

1% manganese

more corrosion resistant than

aluminium, has strength similar to

steel, but lower density than steel,

expensive

aircraft, spacecraft

Mild steel 99.5-99.9% iron

0.1-0.5% carbon

strong, malleable, cheap thin sheets, car bodies.

cans

Stainless

steel

73% iron

18% chromium

8% nickel

1% carbon

hard and strong, very corrosion

resistant

cutlery, surgical

instruments, cooking

utensils, bearing,

razor bladesSolder 33-50% tin melts at a temperature lower than the

pure metals

welding

N-carat

gold

n/24 gold

(24-n)/24 silver

or copper

harder than pure gold, lustrous

jewellery, gold coins

(since pure fold is too

soft in many cases)

V. Intermolecular forces

2 types of intermolecular forces, namely

a. van der Waals forcesb. hydrogen bonding.

A. van der Waals forces

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a. Dipole Dipole Interactions

for polar molecules e.g. HCl, CHCl3 etc

+

+

H Cl

H

Cl

H

Cl

+

b Dipole Induced Dipole Interactions

For a mixture of a polar substance and a non-polar substance

e.g. mixture of CH3COCH3 and CCl4

CH3COCH3 CCl4

+ +

permanent dipole induced dipole

c. Instantaneous Dipole Induced Dipole Interactions

(i) the momentary dipoles arise because the electrons in a molecule are in a continual motion and at some moment

that more electronic charge is one side of the molecule than the other, and then a momentary dipole is set up. Thisdipole induces a dipole in neighbouring molecules, and attractive forces result.

(ii) the greater the molecular mass of the the molecules, the more the number of electrons and hence the stronger is the

instantaneous dipole induced dipole interactions

d. van der Waals' radii and covalent radii

induced dipole

+

--

----

+

++

++++

++

++

+

--

--

--

+

instantaneousdipole

Ne Ne

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covalent radius0.128 nm

van der Waals' radius0.215 nm

Iodine molecules

The covalent radius of an atom is half the distance between the nuclei in a homonuclear diatomic molecule of the

element (bond length).

The van der Waals' radius is one half of the distance between the nuclei of two atoms in adjacent molecules.

Molecules Covalent radius (nm) van der Waals' radius (nm)

H2 0.037 0.120

N2 0.074 0.150

The van der Waals radius is significantly greater than the corresponding covalent radius. It is due to the fact that

van der Waals forces are much weaker than covalent bonds.

B. Hydrogen bonds

a. It is therefore a special kind of intermolecular forces in which the link is due to a proton lying between two very

electronegative elements.

X-H ........ :X'

where X or X' is the electronegative atoms (N, O or F)

........ is hydrogen bond

: is lone pair from electronegative atom

b. Hydrogen bond is much stronger than van der Waals forces

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c. Intermolecular and intramolecular hydrogen bonding

2-nitrophenol boils at 489K whereas the 4-nitrophenol, which is hydrogen bonded to other molecules, boils

at 532K.

d. Hydrogen bonds in water

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Each water molecule can donate and accept 2 protons.

In the ice lattice, each oxygen atom is tetrahedrally surrounded by four other hydrogen atoms. The result is an

open, cage like lattice with very inefficient packing of molecules, thus accounting for the low density of ice .

Therefore ice floats on water

C.Phase diagram

(i) Phase diagram of carbon dioxide

(i) The line TA is called the sublimation curve.

Solid carbon dioxide and carbon dioxide vapour are in equilibrium at any point of this curve.

(ii) The line TB is the melting point curve for solid carbon dioxide

Solid carbon dioxide and liquid carbon dioxide are in equilibrium at any point of this curve.

(iii) The line TC is the vaporization curve

Liquid carbon dioxide and carbon dioxide vapour are in equilibrium at any point of this curve.

(iv) At point T is called the triple point at which solid carbon dioxide, liquid carbon dioxide and carbon dioxide

vapour are in equilibrium.

(v) Point C is the critical point.

When the temperature of the system is higher than the critical temperature, the vapour cannot be changed

into liquid by increasing pressure alone.

Two important characteristics:

a. The triple point is above atmospheric pressure

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Solid carbon dioxide will pass directly into carbon dioxide gas without ever becoming liquid, i.e. it sublimes

b. The second characteristics is about the fusion curve, which slopes from left to right in an upward direction

(i.e. the slope is positive). This is typical behaviour and means that an increase in pressure raises the

melting-point of solid carbon dioxide.

(ii) Phase diagram of water

Two differences when compared with the pressure-temperature diagram of carbon dioxide:

a. The line TB is the melting point curve (often called the fusion curve). This line represents the effect of

pressure on the melting point of ice.

It slopes slightly from right to left in an upward direction(i.e. the slope is negative), indicating that the

melting-point of ice is lowered by an increase in pressure.

b. The triple point is lower than 1 atm, therefore, ice will melt instead of sublime when heated.

D. Molecular crystals

Molecular crystals consists of discrete molecules which are held in lattice sites by weak intermolecular forces such

as van der Waals forces or hydrogen bonds.

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Buckminsterfullerene (C60)

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