The types of hybrid orbitals of nitrogen in N | vedclass

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(B) To determine the hybridization $(H)$,we use the formula: $H = \frac{1}{2}(V + M - C + A)$Where $V$ = number of valence electrons of the central at

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$sp, sp^2, sp^3$

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(B) To determine the hybridization $(H)$,we use the formula: $H = \frac{1}{2}(V + M - C + A)$Where $V$ = number of valence electrons of the central atom,$M$ = number of monovalent atoms,$C$ = cationic charge,and $A$ = anionic charge.$1$. For $NO_2^{+}$:$H = \frac{1}{2}(5 + 0 - 1 + 0) = 2$,which corresponds to $sp$ hybridization.$2$. For $NO_3^{-}$:$H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$,which corresponds to $sp^2$ hybridization.$3$. For $NH_4^{+}$:$H = \frac{1}{2}(5 + 4 - 1 + 0) = 4$,which corresponds to $sp^3$ hybridization.Thus,the hybridization types are $sp, sp^2, sp^3$ respectively.

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 types of hybrid orbitals of nitrogen in $NO_2^{+}$,$NO_3^{-}$ and $NH_4^{+}$ respectively are

Similar Questions

Identify the compound which is polar and has the central atom in $sp^2$ hybridization.

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) $sp^3d^2$

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