The correct order regarding the electronegativity of hybrid orbitals of carbon is:
The correct order regarding the electronegativity of hybrid orbitals of carbon is:
$s p < s p^2 < s p^3$
$s p > s p^2 < s p^3$
$s p > s p^2 > s p^3$
$s p < s p^2 > s p^3$
Solution
In $s p, s p^2$ and $s p^3$ hybrid orbitals, the $s$-orbital character is $50 \%, 33.3 \%$ and $25 \%$, respectively, and because of the higher $s$-orbital character, electronegativity, increases.
Related Theory
The electrons of $s p^3$ hybridized atom are farther from the nucleus than those in $s p^2$ hybridized species. Therefore, $s p^2$ hybrid species are more stable than $s p^3$ hybrid species. This is because the stability is greater when the electrons are close to the nucleus.