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

The correct option is C $33.33\%$

>C=C<

Here, both Carbons are $s{p}^2$ hybridized. Thus, percentage of s character = 33.33 % in $s{p}^2$ hybridization.