Question

Q-6 (a) Identify the hybridization of carbon atoms numbered 1-6 in the structure below: (3 marks) CH 5 CH2 C 3 H3C 2 4 C CH2 (b) Construct the molecular orbital energy level diagram for second period small interaction heteronuclear diatomic molecule Li2 and calculate its bond order. (2 marks

Answer 1

6a)

To check the hybridisation of the C atoms, count the number of atoms the Carbon is attached to.

If it is attached to 4 other atoms , then the hybridisation is sp³. (to remember :1s + 3p = 4 total atoms bonded to Carbon)

Similarly, it is sp² , then it will be attached to 3 other atoms, and for sp, Carbon will be attached to 2 more atoms.

Alternatively, if a carbon has a triple bond then it will be sp hybridised, if it has a double bond, it is sp2 hybridised, and if it has only single bonds, then it will be sp3 hybridised.

However, if an allene is given : CH₂=C=CH₂ , then the two terminal carbons are sp2 hybridised (as expected) while the central carbon is sp hybridised.

In the given question :

C1 = sp3

C2 = sp3,

C3 = sp2

C4 = sp

C5 = sp

C6 = sp2

(b)

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Bond order = (No. of bonding electrons - No. of antibonding electrons)/2

For calculating this, we can either consider all the electrons, or we can consider only the valence MOs. They both will give same result.

If we consider all electrons : No. of bonding electrons = 4 (from σ1s and 2s) , No. of antibonding electrons = 2 (from σ*1s)

So, bond order = (4-2)/2 = 1

If we consider only valence MO (here, it is σ2s and σ*2s) then bonding electrons = 2(σ2s) and antibonding electrons = 0, which gives bond order = (2-0)/2 = 1