JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Coordination Compounds

A spin-only moment above 3.0 BM needs at least three unpaired electrons, since three give $\sqrt{3(3+2)}=3.87$ BM while two give only $2.83$ BM. In high-spin octahedral fields, ${\mathrm{V}}^{3+}$ ($3d^2$), ${\mathrm{Ti}}^{2+}$ ($3d^2$) and ${\mathrm{Ni}}^{2+}$ ($3d^8$) each have two unpaired electrons at $2.83$ BM and are excluded, whereas ${\mathrm{Co}}^{2+}$ ($3d^7$) has three unpaired electrons ($3.87$ BM) and ${\mathrm{Fe}}^{2+}$ ($3d^6$) has four ($4.90$ BM), both clearing the limit. Adding the unpaired electrons of the two qualifying ions, $3+4$, gives a sum of 7.