Question

Which of the following order is correct regarding the electronegativity of the hybrid orbitals of a carbon atom?

• 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: (A)

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

Higher is the %$s$ character, higher is the electronegativity of the $C$ atom because more $s$ character of hybrid orbitals means that electrons are held close to the nucleus of the atom. Due to this strong attraction between nucleus and electrons, the electronegativity of the carbon atom increases.

%$s$ in $sp$=$50$%

%$s$ in $s{p}^2$=$33$%

%$s$ in $s{p}^3$=$25$%

$sp$ hybrid orbital has highest %$s$ character, thus it makes carbon atom more electronegative.

Hence, option $A$ is correct.