JEE Main 2026 — d and f Block Elements Question with Solution

The highest manganese fluoride is $Mn{F}_4$, where the oxidation state of Mn is $+4$.

The highest manganese oxide is $M{n}_2{O}_7$, where the oxidation state of Mn is $+7$.

The difference in oxidation states is $|x|=|7-4|=3$.

The number of unpaired electrons in the given ions are:
$S{c}^{3+}$: $[Ar]3d^0\Rightarrow 0$ unpaired electrons
$Z{n}^{2+}$: $[Ar]3d^{10}\Rightarrow 0$ unpaired electrons
$V^{2+}$: $[Ar]3d^3\Rightarrow 3$ unpaired electrons
$F{e}^{2+}$: $[Ar]3d^6\Rightarrow 4$ unpaired electrons
$C{o}^{2+}$: $[Ar]3d^7\Rightarrow 3$ unpaired electrons

Thus, $V^{2+}$ (C) and $C{o}^{2+}$ (E) have $3$ unpaired electrons.

Answer: C and E Only