Question

The percentage of s-character of the hybrid orbital of $C$ involved in the bond formation of $>C=C<$ is:

• A. $25$%
• B. $50$%
• C. $33.33$%
• D. $100$%

Answer 1

Answer: (C)

$33.33$%

Here the carbon atoms are $s{p}^2$ hybridised.

Percentage of s-character = $=\frac{no.ofs-orbitals}{totalno.oforbitals}\times 100$

$=\frac{1}{3}\times 100$ $=33\cdot 33$%

Option C is correct.