Oxidation states of transition elements

The stability of the +2 state in aqueous solution is judged from the standard electrode potentials $E^\circ(M^{2+}/M)$: the more negative the value, the more stable the $M^{2+}$ ion.
$E^\circ(M^{2+}/M)$ values: Mn = −1.18 V, Cr = −0.91 V, Fe = −0.44 V, Co = −0.28 V.
$Mn^{2+}$ ($3d^5$, half-filled) is the most stable. $Cr^{2+}$ ($3d^4$), though it has a fairly negative $E^\circ(M^{2+}/M)$, is a strong reducing agent and readily changes to $Cr^{3+}$ ($t_{2g}^3$), so its +2 state is less stable than that of $Fe^{2+}$.
$Co^{2+}$ has the least negative value among these and is placed last.
The order accepted in the answer key is Mn > Fe > Cr > Co.

Transition elements show variable oxidation states because both the ns and the (n − 1)d electrons can take part in bonding; the maximum oxidation state equals the total number of these electrons (up to manganese).
$3d^24s^2$ (Ti): 4 such electrons, maximum oxidation state +4.
$3d^34s^2$ (V): 5 such electrons, maximum +5.
$3d^54s^1$ (Cr): 6 such electrons, maximum +6.
$3d^54s^2$ (Mn): 7 such electrons, maximum +7; Mn shows all the states from +2 to +7.
Hence the element with configuration $3d^54s^2$ exhibits the largest number of oxidation states.