JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Coordination Compounds

Distributing the $d$ electrons over the ${\mathrm{t}}_{2g}$ and ${\mathrm{e}}_{\mathrm{g}}$ sets of an octahedral field: ${\mathrm{d}}^3$ is ${\mathrm{t}}_{2g}^3$ with 3 unpaired electrons; ${\mathrm{d}}^4$ low spin is ${\mathrm{t}}_{2g}^4$ with 2 unpaired; ${\mathrm{d}}^5$ high spin is ${\mathrm{t}}_{2g}^3{\mathrm{e}}_{\mathrm{g}}^2$ with 5 unpaired; ${\mathrm{d}}^6$ high spin is ${\mathrm{t}}_{2g}^4{\mathrm{e}}_{\mathrm{g}}^2$ with 4 unpaired; and ${\mathrm{d}}^7$ low spin is ${\mathrm{t}}_{2g}^6{\mathrm{e}}_{\mathrm{g}}^1$ with 1 unpaired. Summing gives $3+2+5+4+1=15$ unpaired electrons.