Arrange the following manganese species in increasing order of the nu…

The correct answer is Option (2) → (B) < (D) < (A) < (C)

Manganese (Z = 25) is [Ar] 3d⁵ 4s², with seven electrons outside the argon core. The 4s electrons are lost first, so in the oxidation state +x the number of 3d electrons left is 7 − x. The higher the oxidation state, the fewer the 3d electrons.

(B) \( \ce{MnO4^-} \) — first, the fewest. x + 4(−2) = −1, so Mn is +7: 3d⁰. This is why permanganate has no unpaired electron and is diamagnetic.

(D) \( \ce{MnO4^2-} \) — second. x + 4(−2) = −2, so Mn is +6: 3d¹. The green manganate is paramagnetic because of this one unpaired electron.

(A) \( \ce{MnO2} \) — third. x + 2(−2) = 0, so Mn is +4: 3d³.

(C) \( \ce{Mn^2+} \) — fourth, the most. Mn is +2: only the two 4s electrons are lost, leaving 3d⁵, the half-filled d subshell.

NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “The manganate and permanganate ions are tetrahedral; the π- bonding takes place by overlap of p orbitals of oxygen with d orbitals of manganese. The green manganate is paramagnetic because of one unpaired electron but the permanganate is diamagnetic due to the absence of unpaired electron.”

4.3.6 Trends in the M³⁺/M²⁺ Standard Electrode Potentials – “…The comparatively high value for Mn shows that Mn²⁺(d⁵) is particularly stable, whereas comparatively low value for Fe shows the extra stability of Fe³⁺ (d⁵).”

Example 4.9 – “…For example, manganese (VI) becomes unstable relative to manganese(VII) and manganese (IV) in acidic solution.

3Mn^VIO₄^2– + 4 H⁺ → 2 Mn^VIIO₄^– + Mn^IVO₂ + 2H₂O”