Identify the correct option.

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(A) $1$. In $NO_3^-$,the bond angle is $120^\circ$. In $NH_3$,it is $107^\circ$. In $NH_2^-$,it is $104.5^\circ$. Thus,the order $NO_3^- > NH_3 > NH_2

Vedclass · Accepted Answer

$NO_3^- > NH_3 > NH_2^-$ (order of bond angle)

Vedclass · Answer

(A) $1$. In $NO_3^-$,the bond angle is $120^\circ$. In $NH_3$,it is $107^\circ$. In $NH_2^-$,it is $104.5^\circ$. Thus,the order $NO_3^- > NH_3 > NH_2^-$ is correct.$2$. $(CH_3)_3B$ has $3$ bond pairs and $0$ lone pairs on $B$,making it $sp^2$ hybridized and planar. This is also correct.$3$. In $NH_4^+$,$N$ is $sp^3$ hybridized. Thus,in $NH_4Cl$,$N$ is $sp^3$ hybridized. This is also correct.$4$. In $S_8$,each $S$ atom forms $2$ bonds and has $2$ lone pairs. Total lone pairs = $8 	imes 2 = 16$. Total non-bonded electrons = $16 	imes 2 = 32$. Thus,option $D$ is incorrect.Given the nature of the question,all options $A, B, C$ are factually correct. However,in standard competitive contexts,if one must choose,$A$ is a fundamental comparison of bond angles.

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