The correct order regarding electronegativity | vedclass

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(B) The electronegativity $(EN)$ of hybrid orbitals is directly proportional to the percentage of $s$-character in the hybrid orbital.For $sp$ hybridi

Vedclass · Accepted Answer

$sp > sp^2 > sp^3$

Vedclass · Answer

(B) The electronegativity $(EN)$ of hybrid orbitals is directly proportional to the percentage of $s$-character in the hybrid orbital.For $sp$ hybridization,the $s$-character is $50\%$.For $sp^2$ hybridization,the $s$-character is $33.3\%$.For $sp^3$ hybridization,the $s$-character is $25\%$.Therefore,the order of electronegativity is $sp (50\%) > sp^2 (33.3\%) > sp^3 (25\%)$.

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

The correct order regarding electronegativity of hybrid orbitals of carbon is

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Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

The hybridization in methane,ethene and ethyne respectively is

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