(B) In $NO_{2}^{+}$,the nitrogen atom is $sp$-hybridized. The $s$-character in $sp$-hybridization is $\frac{1}{2} \times 100 = 50 \%$.In $NO_{3}^{-}$,

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(B) In $NO_{2}^{+}$,the nitrogen atom is $sp$-hybridized. The $s$-character in $sp$-hybridization is $\frac{1}{2} \times 100 = 50 \%$.In $NO_{3}^{-}$,the nitrogen atom is $sp^{2}$-hybridized. The $s$-character in $sp^{2}$-hybridization is $\frac{1}{3} \times 100 = 33.3 \%$.Therefore,the percentages are $50 \%$ and $33.3 \%$ respectively.

Class 11 Chemistry Chemical Bonding and Molecular Structure Hybridisation

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The percentage of $s$-character in the hybrid orbitals of nitrogen in $NO_{2}^{+}$ and $NO_{3}^{-}$ respectively are:

KCET 2020 All KCET

A
$33.3 \%, 50 \%$
B
$50 \%, 33.3 \%$
C
$25 \%, 50 \%$
D
$50 \%, 25 \%$

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Chemical Bonding and Molecular Structure

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Hybridisation

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The percentage of $s$-character in the hybrid orbitals of nitrogen in $NO_{2}^{+}$ and $NO_{3}^{-}$ respectively are:

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Why is the bond angle in $CH_4$ not $90^{\circ}$,even though the outermost $p$-orbitals of carbon are at $90^{\circ}$ to each other?

Suggest reasons why the $B-F$ bond lengths in $BF_{3}$ $(130 \ pm)$ and $BF_{4}^{-}$ $(143 \ pm)$ differ.

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