The hybridisations of the atomic orbitals of | vedclass

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(A) To determine the hybridisation,we use the formula: $	ext{Steric Number} = (	ext{Number of sigma bonds}) + (	ext{Number of lone pairs on the cen

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$sp^{2}, sp$ and $sp^{3}$

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(A) To determine the hybridisation,we use the formula: $	ext{Steric Number} = (	ext{Number of sigma bonds}) + (	ext{Number of lone pairs on the central atom})$.$1$. For $NO_{2}^{-}$: The central nitrogen atom is bonded to two oxygen atoms (two sigma bonds) and has one lone pair. Steric number = $2 + 1 = 3$,which corresponds to $sp^{2}$ hybridisation.$2$. For $NO_{2}^{+}$: The central nitrogen atom is bonded to two oxygen atoms (two sigma bonds) and has no lone pairs. Steric number = $2 + 0 = 2$,which corresponds to $sp$ hybridisation.$3$. For $NH_{4}^{+}$: The central nitrogen atom is bonded to four hydrogen atoms (four sigma bonds) and has no lone pairs. Steric number = $4 + 0 = 4$,which corresponds to $sp^{3}$ hybridisation.Thus,the hybridisations are $sp^{2}, sp$ and $sp^{3}$ 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

The hybridisations of the atomic orbitals of nitrogen in $NO_{2}^{-}$,$NO_{2}^{+}$ and $NH_{4}^{+}$ respectively are.

Similar Questions

Number of unhybridised orbitals in vinyl acetylene is

What is the hybridization of the $Cl$ atom in $ClO^{-}_2$?

Identify the correct set from the following table:
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

The trigonal bipyramidal geometry results from the hybridization:

Explain the geometry of $SF_{6}$.

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