Consider |x| is the difference in oxidation states of Mn in highest...

The highest manganese fluoride is $MnF_4$, where the oxidation state of Mn is $+4$.

The highest manganese oxide is $Mn_2O_7$, 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:
$Sc^{3+}$: $[Ar] 3d^0 \Rightarrow 0$ unpaired electrons
$Zn^{2+}$: $[Ar] 3d^{10} \Rightarrow 0$ unpaired electrons
$V^{2+}$: $[Ar] 3d^3 \Rightarrow 3$ unpaired electrons
$Fe^{2+}$: $[Ar] 3d^6 \Rightarrow 4$ unpaired electrons
$Co^{2+}$: $[Ar] 3d^7 \Rightarrow 3$ unpaired electrons

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

Answer: C and E Only