The spin-only magnetic moment value of $$M^{n+}$$ ion formed among Ni, Zn, Mn and Cu that has the least enthalpy of atomisation is _________. (in nearest integer)

Here n is equal to the number of diamagnetic complexes among $$K_2[NiCl_4]$$, $$[Zn(H_2O)_6]Cl_2$$, $$K_3[Mn(CN)_6]$$ and $$[Cu(PPh_3)_3I]$$

The four complexes given are:
  (i) $$K_2[NiCl_4]$$   (ii) $$[Zn(H_2O)_6]Cl_2$$   (iii) $$K_3[Mn(CN)_6]$$   (iv) $$[Cu(PPh_3)_3I]$$

Step 1 : Identify which of them are diamagnetic.

Case 1: $$[NiCl_4]^{2-}$$ (in $$K_2[NiCl_4]$$)
Ni oxidation state: $$x+4(-1) = -2 \Rightarrow x = +2$$, so $$Ni^{2+}$$ is $$3d^8$$. In a tetrahedral $$Cl^-$$ field (weak field) the ion is high spin with two unpaired electrons ⇒ paramagnetic.

Case 2: $$[Zn(H_2O)_6]^{2+}$$ (in $$[Zn(H_2O)_6]Cl_2$$)
$$Zn^{2+}$$ is $$3d^{10}$$, all electrons paired ⇒ diamagnetic.

Case 3: $$[Mn(CN)_6]^{3-}$$ (in $$K_3[Mn(CN)_6]$$)
$$x+6(-1)=-3 \Rightarrow x = +3$$, so $$Mn^{3+}$$ is $$3d^4$$. $$CN^-$$ is a strong‐field ligand giving a low-spin configuration $$t_{2g}^4e_g^0$$ with two unpaired electrons ⇒ paramagnetic.

Case 4: $$[Cu(PPh_3)_3I]$$ (neutral complex)
Oxidation state: $$Cu + (-1) = 0 \Rightarrow Cu = +1$$, so $$Cu^{+}$$ is $$3d^{10}$$ ⇒ diamagnetic.

Thus the diamagnetic complexes are Case 2 and Case 4. Number of diamagnetic complexes $$= 2$$, hence $$n = 2$$.

Step 2 : Among the elements Ni, Zn, Mn and Cu, compare enthalpies of atomisation. For 3d metals: $$\Delta H_{atom}$$ rises to a maximum near the middle of the series (Cr, Mn, Fe, Co) and then falls, being minimum for Zn.
Approximate values (kJ mol$$^{-1}$$): $$Zn \approx 120$$, $$Cu \approx 330$$, $$Mn \approx 280$$, $$Ni \approx 420$$. Therefore the element with the least enthalpy of atomisation is $$Zn$$.

So the required ion is $$Zn^{n+} = Zn^{2+}$$.

Step 3 : Calculate its spin-only magnetic moment. For an ion with $$n_u$$ unpaired electrons,
$$\mu_{spin} = \sqrt{n_u(n_u+2)}\; \text{BM}$$.

For $$Zn^{2+}$$, electronic configuration $$3d^{10}$$ ⇒ $$n_u = 0$$.
Hence $$\mu_{spin} = \sqrt{0(0+2)} = 0\; \text{BM}$$.

The spin-only magnetic moment, to the nearest integer, is 0.