Question

The order of electronegativity of carbon in $sp,s{p}^2$ and $s{p}^3$ hybridized states follows:

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

Answer 1

The correct option is A $sp>s{p}^2>s{p}^3$

% S character $sp=50$%, $s{p}^2=33.33$%, $s{p}^3=25$%

More the % S character more the electronegativity [other factor being equal]