Question

Which of the following order regarding the size of hybrid orbital of carbon is correct?

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

Answer 1

Answer: (B)

$sp<s{p}^2<s{p}^3$

$S{P}^3$ $S{P}^2$ $SP$
%$S=25$ %$S=33$ %$S=50$

As the $s$ character increases, electronegativity increases and the bond length decreases as more overlapping takes place.

Thus the correct order is $sp<s{p}^2<s{p}^3$

Hence, the correct option is $B$