Question

For the four successive transition elements $(Cr,Mn,Fe\text{and}Co)$ , the stability of $+2$ oxidation state in gaseous state will be there in which of the following order ?
(Atomic no. $Cr=24,Mn=25,Fe=26,Co=27$)

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• A. $Fe>Mn>Co>Cr$
• B. $Co>Mn>Fe>Cr$
• C. $Mn>Fe>Cr>Co$
• D. $Cr>Mn>Co>Fe$

Answer 1

The correct option is C $Mn>Fe>Cr>Co$

$Cr\to 3d^{5}4s^1\stackrel{C{r}^{2+}}{⟶}3d^4$
$Mn\to 3d^{5}4s^2\stackrel{M{n}^{2+}}{⟶}3d^5\left(\text{More stable; Half filled configuration}\right)$
$Fe\to 3d^{6}4s^2\stackrel{F{e}^{2+}}{⟶}3d^6$
$Co\to 3d^{7}4s^2\stackrel{C{o}^{2+}}{⟶}3d^7$

Again stability of oxidation potential can be predicted from standard reduction potential .
Higher the reduction potential greater the tendency of $M^{+2}$ to get reduced and lower will be the oxidation state stability.

$E_{\left(M^{+2}/M\right)}^0$

$CrMnFeCo$
$-0.90-1.18-0.44-0.28$
$Mn$ has lowest reduction potential, so more stable

So order should be

$Mn>Cr>Fe>Co$

But again $C{r}^{+2}$ is strongest reducing agent and is more stable in its +3 oxidation state.

$C{r}^{+2}\to C{r}^{+3}\left(3d^3\right)\left(t_{2g}^{3}e{g}^0\right)$

So order will be

$Mn>Fe>Cr>Co$

correct optoion will be (a)