Question

Decreasing order of p-orbital character in the following is:
A) $Si{O}_2$ B) $C{O}_2$ C) Graphite

• A. A>B>C
• B. B>A>C
• C. B>C>A
• D. A>C>B

Answer 1

The correct option is D A>C>B

The more the number of p orbitals involved in hybridisation, the more is the p-character of the hybrid orbital and hence, more is the p-character of the bonds. The hybridisations are:
$Si{O}_2$ is $s{p}^3$ hybridised with 3 p-orbitals involved. $C{O}_2$ is $sp$ hybridised with 1 p-orbital involved. Graphite is $s{p}^2$ hybridised with 2 p-orbitals involved. Hence, p character is maximum in $Si{O}_2$, then graphite, and the least in $C{O}_2$.