JEE Main · 2022 · 26 Jul 2022 · Shift 2 · d and f Block Elements

To determine the strongest oxidizing agent, we examine the oxidation states and electronic configurations:

• $\mathrm{MnF}{}_4$: Mn is in +4 state, electronic configuration $[\mathrm{Ar}]3d^3$
• $\mathrm{MnF}{}_3$: Mn is in +3 state, electronic configuration $[\mathrm{Ar}]3d^4$
• $\mathrm{MnF}{}_2$: Mn is in +2 state, electronic configuration $[\mathrm{Ar}]3d^5$

$\mathrm{MnF}{}_3$ is the strongest oxidizing agent because Mn³⁺ can easily be reduced to the stable Mn²⁺ state (which has the half-filled $d^5$ configuration).

For $\mathrm{MnF}{}_3$ with 4 unpaired electrons ($d^4$):
$\mu =\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.89\approx$ 5 B.M.