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(D) To determine the $s-$character,we look at the hybridization of the nitrogen atom in each molecule:$1$. In $NH_3$,$N$ is $sp^3$ hybridized ($25\% \

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$N-H$ bond in $HN \equiv NH$

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(D) To determine the $s-$character,we look at the hybridization of the nitrogen atom in each molecule:$1$. In $NH_3$,$N$ is $sp^3$ hybridized ($25\% \ s-$character).$2$. In $NH_4^+$,$N$ is $sp^3$ hybridized ($25\% \ s-$character).$3$. In $H_2N-NH_2$ (hydrazine),$N$ is $sp^3$ hybridized ($25\% \ s-$character).$4$. In $HN \equiv NH$ (diazene/diimide),the nitrogen atoms are $sp$ hybridized.For $sp$ hybridization,the $s-$character is $\frac{1}{2} 	imes 100 = 50\%$.Since $50\% > 25\%$,the $N-H$ bond in $HN \equiv NH$ has the highest $s-$character.

The highest amount of $s-$character is observed in

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