Which of the two ions from the list given below, have the geometry that is explained by the same hybridisation of orbitals, $\mathrm{NO}_2^{-}, \mathrm{NO}_3^{-}, \mathrm{NH}_2^{-}, \mathrm{NH}_4^{+}, \mathrm{SCN}^{-}$?

• A

  $\mathrm{NH}_4^{+}$and $\mathrm{NO}_3^{-}$
• B

  $\mathrm{SCN}^{-}$and $\mathrm{NH}_2^{-}$
• C

  $\mathrm{NO}_2^{-}$and $\mathrm{NH}_2^{-}$
• Correct answer
  D

  $\mathrm{NO}_2^{-}$and $\mathrm{NO}_3^{-}$

Solution

$$ \begin{array}{ll} \mathrm{NO}_2^{-} & \rightarrow s p^2 \\ \mathrm{NO}_3^{-} & \rightarrow s p^2 \\ \mathrm{NH}_2^{+} & \rightarrow s p^3 \end{array} $$ $$ \begin{array}{ll} \mathrm{NH}_4^{-} & \rightarrow s p^3 \\ \mathrm{SCN}^{-} & \rightarrow s p \end{array} $$ $\therefore \quad \mathrm{NO}_2^{-}$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.

Hence, the correct answer is option: 'D'.