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The correct order regarding the electronegativity of hybrid orbitals of carbon is :-
A
$sp \gt sp^2 \lt sp^3$
B
$sp \gt sp^2 \gt sp^3$
C
$sp \lt sp^2 \gt sp^3$
D
$sp \lt sp^2 \lt sp^3$
Detailed Solution
The electronegativity of a hybrid orbital depends on its percentage of s-character.
An s orbital is closer to the nucleus than a p orbital, so electrons in an orbital with more s-character are held more strongly by the nucleus.
s-character: $sp$ = 50 percent, $sp^2$ = 33.3 percent, $sp^3$ = 25 percent.
Greater the s-character, greater is the electronegativity of the hybrid orbital.
So the order of electronegativity is $sp \gt sp^2 \gt sp^3$.
This explains why the hydrogen of ethyne (sp carbon) is more acidic than that of ethene or ethane.
An s orbital is closer to the nucleus than a p orbital, so electrons in an orbital with more s-character are held more strongly by the nucleus.
s-character: $sp$ = 50 percent, $sp^2$ = 33.3 percent, $sp^3$ = 25 percent.
Greater the s-character, greater is the electronegativity of the hybrid orbital.
So the order of electronegativity is $sp \gt sp^2 \gt sp^3$.
This explains why the hydrogen of ethyne (sp carbon) is more acidic than that of ethene or ethane.