Question

The correct order regarding the electronegativity of hybrid orbitals of carbon is:
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

The correct order regarding the electronegativity of hybrid orbitals depends on the percentage s character. As s orbital is nearer to the nucleus so, the nuclear force of attraction increases which results in the increase of electronegativity tendency.

Complete step by step answer:
The reason behind this order of electronegativity is the amount of s-orbital character in these different hybrid orbitals. As the s-orbital character increases the electron attracting tendency increases, which means electronegativity increases.
In $s{p^3}$ hybrid orbital there are one s and three p-orbitals .So the overall percentage of s-orbital character is 25%.In $s{p^2}$ hybrid orbital there are one s and two p-orbitals .So overall the percentage of s-orbital character is 33.3%.While in in $sp$ hybrid orbital there is only one s and one p-orbital Hence the overall percentage of the s-orbital character is 50%.
So observing the trend of the s-orbital character in different hybrid orbitals we can conclude that the electron attracting tendency increases as$sp > s{p^2} > s{p^3}$. This means the correct trend of hybrid orbitals of carbon regarding electronegativity is$sp > s{p^2} > s{p^3}$.

Hence, the correct option is (C).

Note:
Thus, the correct order regarding the electronegativity of hybrid orbitals of carbon is$sp > s{p^2} > s{p^3}$. The electronegativity i.e. electron attracting tendency of the hybrid orbitals of carbon is based on the amount of s-character. The s-character in different hybrid orbitals of carbon follow the trend $sp > s{p^2} > s{p^3}$ i.e. $50\% {\rm{ }} > {\rm{ }}33.3\% {\rm{ }} > {\rm{ }}25\% .$. So, the carbon atom which is $sp$ hybridised is the most electronegative whereas the carbon atom which is $s{p^3}$hybridised is the least electronegative.