The hybridizations of atomic orbitals of nitr | vedclass

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(C) To determine the hybridization of the central nitrogen atom,we use the formula: $	ext{Number of hybrid orbitals} = \frac{1}{2} [V + M - C + A]$,w

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$sp$,$sp^2$ and $sp^3$

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(C) To determine the hybridization of the central nitrogen atom,we use the formula: $	ext{Number of hybrid orbitals} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1)$ For $NO_2^+$: $V = 5$,$M = 0$,$C = 1$,$A = 0$. $	ext{Hybrid orbitals} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. Hence,$sp$ hybridization.$2)$ For $NO_3^-$: $V = 5$,$M = 0$,$C = 0$,$A = 1$. $	ext{Hybrid orbitals} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Hence,$sp^2$ hybridization.$3)$ For $NH_4^+$: $V = 5$,$M = 4$,$C = 1$,$A = 0$. $	ext{Hybrid orbitals} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$. Hence,$sp^3$ hybridization.Therefore,the hybridizations are $sp$,$sp^2$,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 hybridizations of atomic orbitals of nitrogen in $NO_2^+$,$NO_3^-$ and $NH_4^+$ respectively are

Similar Questions

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

What is the decreasing order of carbon-carbon single bond length in the following compounds?
$(I) CH_2 = CH - C \equiv CH$
$(II) CH \equiv C - C \equiv CH$
$(III) CH_3 - CH = CH_2$
$(IV) CH_2 = CH - CH = CH_2$

What is the type of hybridization in $SO_3$?

What is the type of hybridization in $IF_5$?

Which type of hybridization occurs in diborane $(B_2H_6)$?

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