JEE Main 2026 — Coordination Compounds Question with Solution

For low spin octahedral complexes, electrons fill the $t_{2g}$ orbitals first before pairing or moving to $e_g$ orbitals. The configuration depends on the number of d-electrons.
For $C{r}^{3+}$: Electronic configuration is $3d^3$. In an octahedral field, the configuration is $t_{2g}^3{e}_g^0$. Number of unpaired electrons ($n$) = 3.
For $M{n}^{3+}$: Electronic configuration is $3d^4$. For low spin, the configuration is $t_{2g}^4{e}_g^0$. Number of unpaired electrons ($n$) = 2.
For $F{e}^{3+}$: Electronic configuration is $3d^5$. For low spin, the configuration is $t_{2g}^5{e}_g^0$. Number of unpaired electrons ($n$) = 1.
For $C{o}^{3+}$: Electronic configuration is $3d^6$. For low spin, the configuration is $t_{2g}^6{e}_g^0$. Number of unpaired electrons ($n$) = 0.
Comparing the number of unpaired electrons:
$C{r}^{3+}(3)>M{n}^{3+}(2)>F{e}^{3+}(1)>C{o}^{3+}(0)$.
The correct decreasing order is $C{r}^{3+}>M{n}^{3+}>F{e}^{3+}>C{o}^{3+}$.