Question

The hybridisation of $C$ in diamond, graphite and ethyne is in the order

• A. $s{p}^3,sp,s{p}^2$
• B. $s{p}^3,s{p}^2,sp$
• C. $sp,s{p}^2,s{p}^3$
• D. $s{p}^2,s{p}^3,sp$

Answer 1

Answer: (B)

(i) As diamond has tetrahedral structure in which all the C-atoms are bonded by $4\sigma$ -bonds with 4 neighbouring C-atoms. Thus, it has $s{p}^3$ hybridisation.

(ii) Graphite forms an hexagonal sheet like structure in which each C-atom is bonded by $3\sigma$ -bonds with $3-C$ atoms in the same plane, 4th electron of C-atom is free to move thus, show $s{p}^2$ -hybridisation.

(iii) Ethyne has $CH=CH$ structure, in which one C-atom is bonded with other C-atom through only one $\sigma$ -bond. (as $\pi$ -bonds do not take part in hybridisation)

Thus, it shows $sp$ -hybridisation.

Hence, correct order is: $s{p}^3,s{p}^2,sp$