Question

The decreasing order of electronegativity among hybridized carbon atoms is :

• 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. $s{p}^2>s{p}^3>sp$

Answer 1

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

Electrons are closer to the nucleus in s-orbital than in p-orbital. Thus, electrons are more tightly held by nucleus in an s-orbital than in p-orbital.
As the character increases, the electronegativty increases.
The s-character in $sp$, $s{p}^2$ and $s{p}^3$ orbitals is $50\%,33\%$ and $25\%$ respectively.
Hence, the s-character and the electronegativity decreases in the order $sp>s{p}^2>s{p}^3$.