The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, [V(CN)$$_6$$]$$^{4-}$$, [Fe(CN)$$_6$$]$$^{4-}$$, [Ru(NH$$_3$$)$$_6$$]$$^{3+}$$ and [Cr(NH$$_3$$)$$_6$$]$$^{2+}$$, is:

First, we recall the spin-only magnetic moment formula for any metal ion:

$$\mu_{\text{spin}}=\sqrt{n\,(n+2)}\ \text{Bohr magneton (BM)}$$

where $$n$$ is the number of unpaired electrons.

All the complexes given are stated to be low-spin octahedral, which means that all electrons prefer to pair in the lower-energy $$t_{2g}$$ set before entering the higher-energy $$e_g$$ set. We now examine each complex one by one.

(1) [V(CN)6]$$^{4-}$$

The charge balance gives the oxidation state of vanadium:

$$x + 6(-1) = -4 \;\; \Longrightarrow \;\; x = +2$$

So the ion inside is $$V^{2+}$$. Neutral vanadium is $$[Ar]\,3d^3\,4s^2$$; removing two 4s electrons leaves

$$V^{2+} : [Ar]\,3d^3$$

This is a $$d^3$$ case. For low spin $$d^3$$, the electrons occupy $$t_{2g}^3 e_g^0$$, one in each of the three $$t_{2g}$$ orbitals, so

$$n = 3$$

Hence

$$\mu_{V^{2+}} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}$$

(2) [Fe(CN)6]$$^{4-}$$

Again equating charges:

$$x + 6(-1) = -4 \;\; \Longrightarrow \;\; x = +2$$

So we have $$Fe^{2+}$$. Neutral iron is $$[Ar]\,3d^6\,4s^2$$; removing two 4s electrons gives

$$Fe^{2+} : [Ar]\,3d^6$$

This is a $$d^6$$ system. With the strong-field ligand CN$$^-$$ and low spin, the filling is $$t_{2g}^6 e_g^0$$, all electrons paired, thus

$$n = 0$$

Therefore

$$\mu_{Fe^{2+}} = \sqrt{0(0+2)} = 0\ \text{BM}$$

(3) [Ru(NH3)6]$$^{3+}$$

Here NH3 is a neutral ligand, so

$$x + 6(0) = +3 \;\; \Longrightarrow \;\; x = +3$$

Thus the ion is $$Ru^{3+}$$. Neutral ruthenium is $$[Kr]\,4d^7\,5s^1$$; removing three electrons (first the one 5s and then two 4d) yields

$$Ru^{3+} : [Kr]\,4d^5$$

This is a $$d^5$$ case in the 4d series. Because 4d orbitals experience a larger crystal field splitting, the complex is low spin: electrons fill $$t_{2g}^5 e_g^0$$, leaving one unpaired electron, so

$$n = 1$$

Hence

$$\mu_{Ru^{3+}} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \text{BM}$$

(4) [Cr(NH3)6]$$^{2+}$$

Again, NH3 is neutral, so

$$x + 6(0) = +2 \;\; \Longrightarrow \;\; x = +2$$

Thus the ion is $$Cr^{2+}$$. Neutral chromium is $$[Ar]\,3d^5\,4s^1$$; removing two electrons (4s then 3d) produces

$$Cr^{2+} : [Ar]\,3d^4$$

This is a $$d^4$$ configuration. Under the supplied “low-spin” condition, the electrons occupy $$t_{2g}^4 e_g^0$$. The first three electrons go singly into the three $$t_{2g}$$ orbitals; the fourth pairs up in one of them. Thus we have

$$n = 2$$

and therefore

$$\mu_{Cr^{2+}} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\ \text{BM}$$

Ordering the magnetic moments

We have obtained

$$\mu_{V^{2+}} \approx 3.87\ \text{BM}$$

$$\mu_{Cr^{2+}} \approx 2.83\ \text{BM}$$

$$\mu_{Ru^{3+}} \approx 1.73\ \text{BM}$$

$$\mu_{Fe^{2+}} = 0\ \text{BM}$$

Thus the descending order is

$$V^{2+} \; > \; Cr^{2+} \; > \; Ru^{3+} \; > \; Fe^{2+}$$

This sequence matches Option A.

Hence, the correct answer is Option A.