The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, [V(CN)6]4-,…

🧠 Configure Each as Low-Spin Octahedral

| Ion | d-count | LS config | Unpaired | |---|---|---|---| | V²⁺ | d³ | $\mathrm{t_{2g}^3}$ (LS = HS) | 3 | | Cr²⁺ | d⁴ | $\mathrm{t_{2g}^4}$ | 2 | | Ru³⁺ | d⁵ | $\mathrm{t_{2g}^5}$ | 1 | | Fe²⁺ | d⁶ | $\mathrm{t_{2g}^6}$ | 0 |

Strictly chemistry-defensible order of μ: $\mathrm{V^{2+}}\,(3) > \mathrm{Cr^{2+}}\,(2) > \mathrm{Ru^{3+}}\,(1) > \mathrm{Fe^{2+}}\,(0)$.

🗺️ Why LS Doesn't Help d³

For $\mathrm{d^3}$ in an octahedral field, the three electrons each occupy a separate $t_{2g}$ orbital with parallel spins — regardless of whether the field is strong or weak (HS = LS configuration). So V²⁺ in a strong-field complex still has 3 unpaired.

For $\mathrm{d^4}$ LS, the fourth electron pairs in $t_{2g}$ rather than entering $e_g$, dropping unpaired from 4 to 2.

⚡ Spin-only μ from n

$\mu = \sqrt{n(n+2)}$ BM:

| n | μ (BM) | |---|---| | 0 | 0 | | 1 | 1.73 | | 2 | 2.83 | | 3 | 3.87 | | 4 | 4.90 |

⚠️ Track d-count, Not Just OS

V²⁺ vs Cr²⁺ both look like "M²⁺", but their d-counts differ by 1, and that flips the LS unpaired count from 3 to 2.

$\boxed{\text{Answer: (1) } \mathrm{V^{2+}} > \mathrm{Cr^{2+}} > \mathrm{Ru^{3+}} > \mathrm{Fe^{2+}}}$