The correct order of H-N-H bond angles of amm | vedclass

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(B) According to $VSEPR$ theory,the bond angle depends on the number of lone pairs on the central nitrogen atom.As the number of lone pairs increases,

Vedclass · Accepted Answer

$NH_4^{+} > NH_3 > NH_2^{-}$

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(B) According to $VSEPR$ theory,the bond angle depends on the number of lone pairs on the central nitrogen atom.As the number of lone pairs increases,the repulsion between lone pair-bond pair increases,which compresses the bond angle.- $NH_4^{+}$: $0$ lone pairs,$4$ bond pairs,bond angle $\approx 109^{\circ} 28'$.- $NH_3$: $1$ lone pair,$3$ bond pairs,bond angle $\approx 107^{\circ}$.- $NH_2^{-}$: $2$ lone pairs,$2$ bond pairs,bond angle $\approx 105^{\circ}$.Therefore,the correct order of bond angles is $NH_4^{+} > NH_3 > NH_2^{-}$.

List-$I$ (molecules/ions)	List-$II$ (No. of lone pairs of $e^-$ on central atom)
$A$. $IF_7$	$I$. Three
$B$. $ICl_4^-$	$II$. One
$C$. $XeF_6$	$III$. Two
$D$. $XeF_2$	$IV$. Zero

The correct order of $H-N-H$ bond angles of ammonia,ammonium ion,and amide ion is:

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List-$I$ (molecules/ions) List-$II$ (No. of lone pairs of $e^-$ on central atom)

$A$. $IF_7$ $I$. Three

$B$. $ICl_4^-$ $II$. One

$C$. $XeF_6$ $III$. Two

$D$. $XeF_2$ $IV$. Zero

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