The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is…

🧠 Difference (HS Unpaired) − (LS Unpaired) = 2

Walk each candidate ion's HS vs LS unpaired count for octahedral:

| Ion | d-count | HS unpaired | LS unpaired | Diff | |---|---|---|---|---| | Fe²⁺ | $d^6$ | 4 ($t_{2g}^4 e_g^2$) | 0 ($t_{2g}^6$) | 4 | | Co²⁺ | $d^7$ | 3 ($t_{2g}^5 e_g^2$) | 1 ($t_{2g}^6 e_g^1$) | 2 ✓ | | Ni²⁺ | $d^8$ | 2 ($t_{2g}^6 e_g^2$) | 2 (same — d⁸ has no LS option) | 0 | | Mn²⁺ | $d^5$ | 5 ($t_{2g}^3 e_g^2$) | 1 ($t_{2g}^5$) | 4 |

Only Co²⁺ gives a difference of 2.

🗺️ The d⁴ and d⁷ Pattern

For diff = 2, the metal needs $d^4$ or $d^7$:

• d⁴ HS: $t_{2g}^3 e_g^1$ (4 unpaired); LS: $t_{2g}^4$ (2 unpaired). Diff = 2.
• d⁷ HS: $t_{2g}^5 e_g^2$ (3 unpaired); LS: $t_{2g}^6 e_g^1$ (1 unpaired). Diff = 2.

Among the options, Co²⁺ (d⁷) is the only fit. Cr²⁺ and Mn³⁺ would also satisfy d⁴ but neither is on the list.

⚡ Why d⁵ and d⁶ Have Diff = 4

For d⁵: HS = 5 unpaired, LS = 1 — diff = 4. For d⁶: HS = 4, LS = 0 — diff = 4.

The "gap" between HS and LS unpaired counts is:

• 0 for d¹, d², d³, d⁸, d⁹, d¹⁰ (no HS/LS distinction).
• 2 for d⁴ and d⁷.
• 4 for d⁵ and d⁶.

⚠️ Read the Question Carefully

"Difference is two" excludes the most populous d-counts (d⁵ Mn²⁺, d⁶ Fe²⁺) — both have diff 4. Only d⁴ or d⁷ fits.

$\boxed{\text{Answer: (2) Co²⁺}}$