Chemical Bonding

b) Describe, including the use of ‘dot-and-cross’ diagrams,

i) covalent bonding in H2, O2, Cl2, HCl, CO2, CH4, C2H6

Covalent bond is formed from sharing of electrons between atoms. A covalent bond is the electrostatic force of attraction between the nuclei of 2 atoms for the shared pair of electron.

Molecule
‘dot-and-cross’ diagram
Description(to achieve stability)
H2




2 hydrogen atoms shared 1 pair of e- in the valence shell
O2




2 oxygen atoms shared 2 pairs of e- in the valence shell
Cl2




2 chlorine atoms shared 1 pair of e- in the valence shell
HCl


1 hydrogen & 1 chlorine atoms shared 1 pair of e- in the valence shell
CO2




1 carbon & 2 oxygen atoms shared 2 pairs of e- in the valence shell
CH4




1 carbon & 4 hydrogen atoms shared 4 pairs of e- in the valence shell
C2H6




2 carbon & 6 hydrogen atoms shared 7 pair of e- in the valence shell







ii) dative bond, in BF3->NH3

Dative bond is a type of covalent bond where the shared paired of electrons comes from one donor only. The donor must possess 1 lone pair of electron in its valence shell. The acceptor must have an empty orbital in its outer quantum shell to accommodate the lone pair.
NH3 + BF3----à BF3->NH3






NH3’s 1 lone pair donates to BF3. Now both N and B have 8e- in valence shell.

c) Explain shapes and bond angles of molecules by, using qualitative model of electron pair repulsion, using as simple examples: BF3 (trigonal), CO2 (linear), CH4 (tetrahedral), NH3 (pyramidal), H2O(V-shape), SF6(octahedral), by using VSEPR Theory.

Valence Shell Electron Pair Repulsion Theory can be use to predict the shape f simple covalent molecules.
I. These b.p & l.p. electrons around central atom determine the shapes of the molecule.
II. The electrons repel each other as far as possible.
III. Degree of repulsion
l.p.-l.p > l.p-b.p. > b.p.-b.p.
105’ 107’ 109’
Molecule
Structural formula
Kind of repulsion
Bond angle
Shape
Description
BF3




l.p.-l.p.
360’/3=120’
Trigonal planar
(because it is a triangle in 2D.)
B is in Grp III F is in Grp VII. There are 3F atoms. The strength of repulsion is equal because all are l.p.
CO2




l.p.-l.p.
360’/2=180’
Linear (because it is in a straight line)
There is repulsion between 2 O atoms, so the repulsion is the largest.
CH4




b.p-b.p.
109’
Tetrahedral (because there are 4 faces)
There is repulsion between H atoms.
NH3




l.p.-b.p.
107’
Trigonal pyramidal (because it is a triangle in 3D.
There is repulsion between N and H atom. So the angle will be pushed smaller.
H2O




l.p.-l.p.
105’
V-shape
There is repulsion causing the H atoms to be pushed closer to each other.
SF6




b.p-b.p
90’
Octahedral (because it has two 4 sided pyramid)
This is a molecule with more than 8e-.
Such case can be seen in period 3 and 4. There is no l.p. repulsion, so there is equal bond angle.

Factors affecting Bond Size:

I. Total number of e- pairs.
The more the number of e- pairs the central atom has, the smaller the bond angle.
II. Number of lone pairs.
L.p. electrons cause more repulsion than the b.p. electrons. Bond pairs are pushed closer to one another.
III. Size of central atom.
As size of central atom increases, the l.p. region increases, resulting in greater repulsion between the l.p.-l.p. & b.p-l.p.. The b.p. are pushed closer to one another, bond angle decrease.
IV. Electronegativity of central atom
The greater the electronegativity of a central atom, the closer the b.p. are towards the central atom, increasing b.p.-b.p. repulsion, hence bond angle increases.

d) Describe covalent bonding in terms of orbital overlap, giving σ and π bonds

A sigma bond is formed from the head-on overlapping of orbitals.

A Pi bond is formed from the side-on overlapping of p orbitals.

e) Predict the shapes and bond angles of molecules analogous to those specified in c).

1) Identify the central atom.
2) Determine the no. of electrons surrounding the central atom.
· For molecules with no charge,
No. of e- surrounding central atom = No. of valence e- of central atom + No. of surrounding atoms
· For charged particles,
No. surrounding atom = No. of valence e- of central atom ± magnitude of charge + No. of surrounding atoms
3) Determine the no. of electron pairs.
· No. of electron pairs = ½ x No. of e- surrounding central atom
4) Determine no. of b.p. electron and l.p. electron.
· No. of b.p. = No. of atoms around the central atoms
· No. of l.p. = No. of electron pairs – No. of b.p.
5) Identify shape of the molecule using the table below.
No. of Electron pairs
No. of Bond pairs
No. of Lone pairs
Geometrical shape
Bond angle
2
2
0
Linear
180’
3
3
0
Trigonal planar
equal or near 120’
2
1
V-shape/bent
4
4
0
Tetrahedral

equal or near 108’
3
1
Trigonal pyramidal
2
2
V-shape/bent
5
5
0
Trigonal bipyramidal
90’, 120’
4
1
Distorted tetrahedron
-
3
2
T-shape
90’
2
3
Linear
180’
6
6
0
Octahedral

90’
5
1
Square pyramidal
4
2
Square planar