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Which of the two ions from the list given below, have the geometry that is explained by the same hybridisation of orbitals, $\mathrm{NO}_2^{-}, \mathrm{NO}_3^{-}, \mathrm{NH}_2^{-}, \mathrm{NH}_4^{+}, \mathrm{SCN}^{-}$?
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Verified Answer
The correct answer is:
$\mathrm{NO}_2^{-}$and $\mathrm{NO}_3^{-}$
$$
\begin{array}{ll}
\mathrm{NO}_2^{-} & \rightarrow s p^2 \\
\mathrm{NO}_3^{-} & \rightarrow s p^2 \\
\mathrm{NH}_2^{+} & \rightarrow s p^3
\end{array}
$$
$$
\begin{array}{ll}
\mathrm{NH}_4^{-} & \rightarrow s p^3 \\
\mathrm{SCN}^{-} & \rightarrow s p
\end{array}
$$
$\therefore \quad \mathrm{NO}_2^{-}$and $\mathrm{NO}_3^{-}$both have the same hybridisation.
$\therefore\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ has minimum number of unpaired electrons and thus, minimum paramagnetic behaviour.
\begin{array}{ll}
\mathrm{NO}_2^{-} & \rightarrow s p^2 \\
\mathrm{NO}_3^{-} & \rightarrow s p^2 \\
\mathrm{NH}_2^{+} & \rightarrow s p^3
\end{array}
$$
$$
\begin{array}{ll}
\mathrm{NH}_4^{-} & \rightarrow s p^3 \\
\mathrm{SCN}^{-} & \rightarrow s p
\end{array}
$$
$\therefore \quad \mathrm{NO}_2^{-}$and $\mathrm{NO}_3^{-}$both have the same hybridisation.
$\therefore\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ has minimum number of unpaired electrons and thus, minimum paramagnetic behaviour.
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