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For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order? (At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)
A
Cr > Mn > Co > Fe
B
Mn > Fe > Cr > Co
C
Fe > Mn > Co > Cr
D
Co > Mn > Fe > Cr
Detailed Solution
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.
$E^\circ(M^{2+}/M)$ 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.