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The correct answer is:
$\begin{array}{llll}4 & 1 & 2 & 3\end{array}$
\begin{array}{lllll}
\hline S.No. & Molecule & Steric no. & \begin{array}{l}
Hybridi- \\
sation
\end{array} & Shape \\
\hline (A) & \mathrm{NO}_3^{-} & \frac{5+1}{2}=3 & \begin{array}{l}
\mathrm{sp}^2 \\
(4)
\end{array} & \begin{array}{l}
Trigonal \\
planar
\end{array} \\
\hline (B) & \mathrm{NH}_2^{-} & \frac{5+2+1}{2}=4 & \mathrm{sp}^3 & Bent/angular \\
\hline (C) & \mathrm{SCN}^{-} & \frac{3+1}{2}=2 & \begin{array}{l}
\mathrm{sp} \\
(2)
\end{array} & Linear \\
\hline (D) & \mathrm{ICl}_2^{-} & \frac{7+3}{2}=5 & \begin{array}{l}
\mathrm{spd}^3 \mathrm{~d} \\
(3)
\end{array} & Linear \\
\hline
\end{array}
Hence, (A-4), (B-1), (C-2), (D-3).
\hline S.No. & Molecule & Steric no. & \begin{array}{l}
Hybridi- \\
sation
\end{array} & Shape \\
\hline (A) & \mathrm{NO}_3^{-} & \frac{5+1}{2}=3 & \begin{array}{l}
\mathrm{sp}^2 \\
(4)
\end{array} & \begin{array}{l}
Trigonal \\
planar
\end{array} \\
\hline (B) & \mathrm{NH}_2^{-} & \frac{5+2+1}{2}=4 & \mathrm{sp}^3 & Bent/angular \\
\hline (C) & \mathrm{SCN}^{-} & \frac{3+1}{2}=2 & \begin{array}{l}
\mathrm{sp} \\
(2)
\end{array} & Linear \\
\hline (D) & \mathrm{ICl}_2^{-} & \frac{7+3}{2}=5 & \begin{array}{l}
\mathrm{spd}^3 \mathrm{~d} \\
(3)
\end{array} & Linear \\
\hline
\end{array}
Hence, (A-4), (B-1), (C-2), (D-3).
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