In SO_2,NO_2^- and N_3^- the hybridizations a | vedclass

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(A) To determine the hybridization of the central atom,we use the formula: $	ext{Steric Number} = (	ext{Number of sigma bonds}) + (	ext{Number of l

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$sp^2, sp^2$ and $sp$

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(A) To determine the hybridization of the central atom,we use the formula: $	ext{Steric Number} = (	ext{Number of sigma bonds}) + (	ext{Number of lone pairs})$.$1$. For $SO_2$: The central atom $S$ has $2$ sigma bonds and $1$ lone pair. $	ext{Steric Number} = 2 + 1 = 3$,which corresponds to $sp^2$ hybridization.$2$. For $NO_2^-$: The central atom $N$ has $2$ sigma bonds and $1$ lone pair. $	ext{Steric Number} = 2 + 1 = 3$,which corresponds to $sp^2$ hybridization.$3$. For $N_3^-$: The central atom $N$ has $2$ sigma bonds and $0$ lone pairs. $	ext{Steric Number} = 2 + 0 = 2$,which corresponds to $sp$ hybridization.Therefore,the hybridizations are $sp^2, sp^2$ and $sp$ respectively.

Molecule	Hybridisation of central atom	Shape
$(a)$ $PCl_5$	$dsp^3$	square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$	$sp^3$	tetrahedral
$(c)$ $SF_6$	$sp^3d^2$	octahedral
$(d)$ $IF_3$	$dsp^3$	pyramidal

In $SO_2$,$NO_2^-$ and $N_3^-$ the hybridizations at the central atom are respectively $:$

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Identify the correct set from the following table:
Molecule Hybridisation of central atom Shape
$(a)$ $PCl_5$ $dsp^3$ square pyramidal
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$(d)$ $IF_3$ $dsp^3$ pyramidal

$sp^3$ hybridisation is found in

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