Which of the following statements are correct? i. $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{MnCl}_6\right]^{3-}$ both are paramagnetic ii. Both $\left[\mathrm{NiCl}_4\right]^{2-},\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ are tetrahedral and diamagnetic iii. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is inner orbital complex, but $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ is outer orbital complex

• A

  i, ii only
• Correct answer
  B

  i, iii only
• C

  ii, iii only
• D

  i, ii, iii

Solution

$\left[\mathrm{NiCl}_4\right]^{2-}$ has $\mathrm{Ni}^{2+}$ that has the configuration $[\mathrm{Ar}] 3 \mathrm{~d}^8$. Since $\mathrm{Cl}^{-}$is a weak-field ligand, the configuration of the complex will be:- $\left[\mathrm{NiCl}_4\right]^{2-}=$
Thus, it is a tetrahedral complex. It is paramagnetic due to unpaired electrons. $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ has neutral and strong-field ligand $\mathrm{CO}$. Thus, $\left[\mathrm{Ni}(\mathrm{CO})_4\right]=$
Thus, $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is a square-planar and diamagnetic complex. Therefore, (ii) is incorrect and only (i) and (iii) are correct.

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