Question

The hybridisation of carbon in diamond, graphite and acetylene are respectively:

• 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

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

Hybridization of carbon in:

Diamond: $s{p}^3$. In diamond, each carbon combine with 4 other carbon atoms with four sigma bonds. So, $s{p}^3$ hybridization.

Graphite: $s{p}^2$. In graphite, each carbon combine with 3 other carbon atoms with three sigma bonds. So, $s{p}^2$ hybridization.

Acetylene ($C_2{H}_2$) : $sp$. In acetylene one carbon combine with another carbon atom with three bonds (1 sigma and 2 pi bonds) and with one 1 hydrogen atom. So, $sp$ hybridization.

Hence option B is correct.