A first-row transition metal dipositive ion forms an octahedral aqua complex $\mathbf{P}$, which is paramagnetic. The addition of excess ammonia to $\mathbf{P}$ gives $\mathbf{Q}$, which also is paramagnetic. Adding excess NaCN to $\mathbf{P}$ gives a diamagnetic complex $\mathbf{R}$. The complexes $\mathbf{P}, \mathbf{Q}$, and $\mathbf{R}$, respectively, are

• Correct answer
  A

  $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$, and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$
• B

  $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{2+}$, and $\left[\mathrm{Ti}(\mathrm{CN})_6\right]^{4-}$
• C

  $\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{V}\left(\mathrm{NH}_3\right)_6\right]^{2+}$, and $\left[\mathrm{V}(\mathrm{CN})_6\right]^{4-}$
• D

  $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_6\right]^{2+}$, and $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$

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