1. Find $n$ = number of diamagnetic complexes

We examine each complex:

(i) $\mathrm{K_2[NiCl_4]}$

• Complex ion: $[\mathrm{NiCl_4}]^{2-}$
• Oxidation state of Ni: $x+4(-1)=-2 \Rightarrow x=+2$
• So $\mathrm{Ni^{2+}}: 3d^8$
• $\mathrm{Cl^-}$ is a weak-field ligand, tetrahedral complex forms.
• Tetrahedral $d^8$ has 2 unpaired electrons.
• Hence paramagnetic.

(ii) $[\mathrm{Zn(H_2O)_6}]\mathrm{Cl_2}$

• Complex cation: $[\mathrm{Zn(H_2O)_6}]^{2+}$
• So $\mathrm{Zn^{2+}}: 3d^{10}$
• All electrons paired.
• Hence diamagnetic.

(iii) $\mathrm{K_3[Mn(CN)_6]}$

• Complex ion: $[\mathrm{Mn(CN)_6}]^{3-}$
• Oxidation state of Mn: $x+6(-1)=-3 \Rightarrow x=+3$
• So $\mathrm{Mn^{3+}}: 3d^4$
• $\mathrm{CN^-}$ is strong-field, octahedral low-spin complex.
• Configuration: $t_{2g}^4 e_g^0$
• This has 2 unpaired electrons.
• Hence paramagnetic.

(iv) $[\mathrm{Cu(PPh_3)_3I}]$

• $\mathrm{PPh_3}$ is neutral, $\mathrm{I^-}=-1$
• For neutral complex: $x+0+(-1)=0 \Rightarrow x=+1$
• So $\mathrm{Cu^+}: 3d^{10}$
• All electrons paired.
• Hence diamagnetic.

Therefore, diamagnetic complexes are:

• $[\mathrm{Zn(H_2O)_6}]\mathrm{Cl_2}$
• $[\mathrm{Cu(PPh_3)_3I}]$

So, $n=2$

2. Among Ni, Zn, Mn, Cu, identify the metal with least enthalpy of atomisation

We need the ion $\mathrm{M^{n+}} = \mathrm{M^{2+}}$ since $n=2$.

Now compare atomisation enthalpies of Ni, Zn, Mn, Cu.

• Transition metals generally have high atomisation enthalpy due to strong metallic bonding.
• Zn has comparatively least enthalpy of atomisation because of completely filled $3d^{10}4s^2$ configuration, leading to weaker metallic bonding.

Hence, $M = \mathrm{Zn}$ and required ion is $\mathrm{Zn^{2+}}$

3. Find spin-only magnetic moment of $\mathrm{Zn^{2+}}$

Electronic configuration: $\mathrm{Zn}: [Ar]3d^{10}4s^2$ $\mathrm{Zn^{2+}}: [Ar]3d^{10}$

Number of unpaired electrons: $n_u=0$

Spin-only magnetic moment: $\mu = \sqrt{n_u(n_u+2)}\;\text{BM}$ $\mu = \sqrt{0(0+2)}=0\;\text{BM}$

Nearest integer: $\boxed{0}$

4. Comparison with stored correct answer

Stored correct answer = $0$

Our derived answer = $0$

So the stored answer is correct.