KPK 11th Class Chemistry Chapter 3 Theories of covalent bond Short Questions Answers

Let us consider the two possible structures of O. Eitherwater have a bent structure or a linear structure.

In structure (A) the dipole moment of two O-H bonds are equal and in opposite direction. They will cancel out each other. The net dipole moment will be zero. In structure (B) the bonds moments add and give a definite net dipole moment. Water has (1.) dipole moment. So the structure of water molecule is like (B) and structure (A) is not possible.

Nitrogen molecule (N2):  in nitrogen molecules each nitrogen atom contributes four atomic orbitals i.e. 2S², 2P_x1 , 2P_y1, 2P_z1. Thus eight molecular orbitals are formed. Four of them are bonding molecules orbitals i.e. σ2S, σPx, 2Py, 2Pz and four are anti-bonding molecules orbitals i.e. σ*2s,σ*2Px, π*2Py, π*2Pz. thus each nitrogen atom contributes five electrons to N2 molecule from its valence shell.  

Oxygen molecules (O2):

In oxygen molecule each oxygen atom contributes for atomic orbitals i.e.  2S²,2P_x1, 2P_y1, 2P_z2. Thus eight molecular orbitals are formed. Four of them are bonding molecular orbitals i.e σ2S, σ2Px, π2Py, π2Pz and four are anti-bonding molecular orbitals i.e σ*2s, σ*2px, π*2py, π*2pz. Thus each oxygen atom contributes six electrons to O2 molecule fromits valence shell.

Example No.5 Ammonia (NH₃)

In NH₃ nitrogen is a central atom, which contains four electron pairs. Thus the electron pair geometry should be tetrahedral. There are three bond pair s and one lone pair. The lone pair occupies more space than bond pair. So the N—H bond pairs are pushed closer and the bond angle decrease to 107.5̊. Thus the shape of the NH₃ is pyramidal in which nitrogen is in the centre and the fourth vertex is occupied by a lone pair.

Methane (CH₄) 

In methane carbon is SP3 hybrid i.e one 2S and three 2P orbitals of carbon mix together to form four SP³ hybrid orbitals. The ratio of combination is 1:3. These four sp³ hybrid orbitals are directed to the four corners of a regular tetrahedron. The angle between any two-sp³ hybrid orbitals is 109.5°. Now each sp³ hybrid orbital each containing one electron overlaps axially (head to head) with the four s- orbitals of four hydrogen atoms each containing one electron, forming four sigma bonds (single covalent bonds).The C-H bond length is 1.09A°.

This theory explains quite successfully the formation of covalent i.e how atoms share electrons. According to this theory: A covalent bond is formed by the overlapping of orbitals of two atoms. Overlapping means that the two orbitals share a common region in which the shared pair of electrons is present. In covalent bond formation only those orbitals take part that have only one electron and are half filled. The orbitals containing a pair of electrons are not capable of combination and can’t take part in bond formation. The spin of electron in a covalent bond have to be opposite. The overlapping of orbitals occure in two different ways due to which two types of bonds are formed.
Sigma (σ) bond
Pi (π) bond

1. It predicts either a bond is feasible. For example the formation of He molecule is completely ruled out as its bond order is zero.
2. Bond order is correctly predicted by molecular orbital theory. This is helpful as it givens the following information.
3. Stability of the molecule i.e. a molecule is stable, if n_b>n_a and vice versa. (Where n_a is number of antibonding electrons)
4. Bond dissociation energy depends upon bond order. Greater the bond order greater is the bond dissociation energy.
5. Bond length is inversely proportional to bond r. her higher is the bond order, smaller is the bond length.
6. Molecular orbital theory explained presence of unpaired electrons in molecules which makes it paramagnetic as discussed in oxygen molecule.

The valence bond theory and VSEPR theory fail to show unpaired electrons in O2 molecule but these unpaired electrons can be shown by M.O theory Therefore, in this respect M.O theory is superior to VSEPR theory and valence bond theory.