1. Determine oxidation state of the metal in each complex

For $[MCl_4]^{2-}$:

• Each $Cl^-$ contributes $-1$
• Let oxidation state of metal be $x$

$x + 4(-1) = -2$ $x - 4 = -2$ $x = +2$

So all metals are in the $+2$ oxidation state.

2. Write the electronic configuration of each $M^{2+}$ ion

• $Mn$ $(Z=25)$: $Mn = [Ar]3d^5 4s^2 \\ Mn^{2+} = [Ar]3d^5$

• $Co$ $(Z=27)$: $Co = [Ar]3d^7 4s^2 \\ Co^{2+} = [Ar]3d^7$

• $Ni$ $(Z=28)$: $Ni = [Ar]3d^8 4s^2 \\ Ni^{2+} = [Ar]3d^8$

• $Zn$ $(Z=30)$: $Zn = [Ar]3d^{10} 4s^2 \\ Zn^{2+} = [Ar]3d^{10}$

3. Nature of the complexes

$Cl^-$ is a weak field ligand, and these are tetrahedral complexes. Tetrahedral complexes with weak ligands are generally high spin.

So we count unpaired electrons directly for tetrahedral $d^n$ configurations.

4. Number of unpaired electrons in each complex

(i) $[MnCl_4]^{2-}$

Metal ion: $Mn^{2+} = 3d^5$

For tetrahedral high-spin $d^5$:

• Number of unpaired electrons, $n = 5$

So, $\mu = \sqrt{n(n+2)} = \sqrt{5(7)} = \sqrt{35}$

(ii) $[CoCl_4]^{2-}$

Metal ion: $Co^{2+} = 3d^7$

For tetrahedral high-spin $d^7$:

• Number of unpaired electrons, $n = 3$

So, $\mu = \sqrt{3(5)} = \sqrt{15}$

(iii) $[NiCl_4]^{2-}$

Metal ion: $Ni^{2+} = 3d^8$

For tetrahedral high-spin $d^8$:

• Number of unpaired electrons, $n = 2$

So, $\mu = \sqrt{2(4)} = \sqrt{8}$

(iv) $[ZnCl_4]^{2-}$

Metal ion: $Zn^{2+} = 3d^{10}$

For $d^{10}$:

• Number of unpaired electrons, $n = 0$

So, $\mu = 0$

5. Compare magnetic moments

Since spin-only magnetic moment increases with number of unpaired electrons:

$[MnCl_4]^{2-} > [CoCl_4]^{2-} > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$

6. Match with the given options

This corresponds to Option D.

7. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.