CBSE Class 12 · 2011 · All India · d and f Block ElementsShort Answer

(i) Both are d${}^4$, but they move in opposite directions toward stability:
Cr${}^{2+}$ (d${}^4$) loses an electron to become Cr${}^{3+}$ (d${}^3$, stable half-filled t${}_{2g}^3$) — so Cr${}^{2+}$ is reducing.
Mn${}^{3+}$ (d${}^4$) gains an electron to become Mn${}^{2+}$ (d${}^5$, exactly half-filled) — so Mn${}^{3+}$ is oxidising.
(ii) Early members have too few d electrons, late members have too many paired d electrons; in the middle (e.g. Mn) the maximum number of unpaired d electrons plus the ns pair are all available for bonding, so the middle shows the widest range of states (+2 … +7).