1. Find oxidation state and d-electron count of each metal ion

For the complexes:

• $[V(CN)_6]^{4-}$

  Let oxidation state of V be $x$: $x+6(-1)=-4 \Rightarrow x=+2$ So metal ion is $V^{2+}$. Vanadium: $Z=23$, electronic configuration of V is $[Ar]3d^34s^2$. Hence $V^{2+} = 3d^3$.

• $[Fe(CN)_6]^{4-}$

  Let oxidation state of Fe be $x$: $x+6(-1)=-4 \Rightarrow x=+2$ So metal ion is $Fe^{2+}$. Iron: $[Ar]3d^64s^2$. Hence $Fe^{2+}=3d^6$.

• $[Ru(NH_3)_6]^{3+}$

  Since $NH_3$ is neutral: $x=+3$ So metal ion is $Ru^{3+}$. Ruthenium: $[Kr]4d^75s^1$. Hence $Ru^{3+} = 4d^5$.

• $[Cr(NH_3)_6]^{2+}$

  Since $NH_3$ is neutral: $x=+2$ So metal ion is $Cr^{2+}$. Chromium: $[Ar]3d^54s^1$. Hence $Cr^{2+}=3d^4$.

2. Use low-spin octahedral splitting

All complexes are stated to be low spin.

For octahedral complexes, electrons fill $t_{2g}$ first if pairing energy is overcome.

• $V^{2+} : d^3$ $t_{2g}^3e_g^0$ Number of unpaired electrons, $n=3$.

• $Fe^{2+} : d^6$ low spin $t_{2g}^6e_g^0$ Number of unpaired electrons, $n=0$.

• $Ru^{3+} : d^5$ low spin $t_{2g}^5e_g^0$ Number of unpaired electrons, $n=1$.

• $Cr^{2+} : d^4$ low spin $t_{2g}^4e_g^0$ Number of unpaired electrons, $n=2$.

3. Calculate spin-only magnetic moment order

Spin-only magnetic moment is: $\mu = \sqrt{n(n+2)}\, \text{BM}$ So larger $n$ means larger magnetic moment.

Thus:

• $V^{2+}$: $n=3$
• $Cr^{2+}$: $n=2$
• $Ru^{3+}$: $n=1$
• $Fe^{2+}$: $n=0$

Therefore the order is: $V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$

4. Check options

• A: $V^{2+} > Ru^{3+} > Cr^{2+} > Fe^{2+}$ ❌
• B: $V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$ ✅
• C: $Cr^{2+} > V^{2+} > Ru^{3+} > Fe^{2+}$ ❌
• D: $Cr^{2+} > Ru^{3+} > Fe^{2+} > V^{2+}$ ❌

Hence, the correct option is B.