The bond angles of NH_3,NH_4^+,and NH_2^- are | vedclass

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Question

(B) According to $VSEPR$ theory,the bond angle decreases as the number of lone pairs on the central atom increases due to increased lone pair-bond pai

Vedclass · Accepted Answer

$NH_4^+ > NH_3 > NH_2^-$

Vedclass · Answer

(B) According to $VSEPR$ theory,the bond angle decreases as the number of lone pairs on the central atom increases due to increased lone pair-bond pair repulsion.$NH_4^+$ has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109.5^\circ$.$NH_3$ has $3$ bond pairs and $1$ lone pair,which exerts repulsion on the bond pairs,reducing the bond angle to approximately $107^\circ$.$NH_2^-$ has $2$ bond pairs and $2$ lone pairs,which exert even greater repulsion,further reducing the bond angle to approximately $104.5^\circ$.Therefore,the order of bond angles is $NH_4^+ > NH_3 > NH_2^-$.

Molecule	Geometry
$(A)$ $XeF_2$	$1$. Trigonal pyramidal
$(B)$ $XeO_3$	$2$. Distorted octahedral
$(C)$ $XeF_6$	$3$. Linear
$(D)$ $XeOF_4$	$4$. Square pyramidal

The bond angles of $NH_3$,$NH_4^+$,and $NH_2^-$ are in the order:

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