Question

The hybridization of carbon in diamond, graphite and acetylene is respectively:

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

Answer 1

The correct option is C $s{p}^3,s{p}^2,sp$

In diamond, each carbon combine with 4 other carbon atoms with four sigma bonds. Hence, the hybridization of carbon is $s{p}^3$.

In graphite, each carbon combines with 3 other carbon atoms with three sigma bonds. so $s{p}^2$ hybridization.

In acetylene, one carbon combines with another carbon atom with three bonds (1 sigma and 2 pi bonds). so $s{p}^2$ hybridization.

The hybridisation of carbon atom in diamond, graphite and acetylene are respectively $s{p}^3,s{p}^2,sp$

Hence, option C is correct.