Which orbital is more electronegative among s | vedclass

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(A) The electronegativity order is $sp > sp^2 > sp^3$.This is because the electronegativity of an orbital is directly proportional to its $s$-characte

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$sp$

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(A) The electronegativity order is $sp > sp^2 > sp^3$.This is because the electronegativity of an orbital is directly proportional to its $s$-character.						Hybridization			$s$-character (%)							$sp$			$50\%$							$sp^2$			$33.3\%$							$sp^3$			$25\%$			Since the $sp$ hybrid orbital has the highest $s$-character $(50\%)$,it is the most electronegative.

Which orbital is more electronegative among $sp$,$sp^2$,and $sp^3$? Why?

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