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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Solution

\(Cr\to 3d^{5}4s^{1}\overset{Cr ^{2+}}{\longrightarrow}3d^{4}\)
\(Mn \to 3d^{5}4s^{2}\overset{Mn ^{2+}}{\longrightarrow}3d^{5}\left ( \text{More stable; Half filled configuration}\right )\)
\(Fe \to 3d^{6}4s^{2}\overset{Fe ^{2+}}{\longrightarrow}3d^{6}\)
\(Co \to 3d^{7}4s^{2}\overset{Co ^{2+}}{\longrightarrow}3d^{7}\)

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.

\(E^{0}_{\left ( M^{+2} /M\right )}\)

\(
CrMnFe~Co\)
\(-0.90~-1.18~-0.44~-0.28\)
\(Mn\) has lowest reduction potential, so more stable

So order should be

\(Mn>Cr>Fe>Co\)

But again \(Cr^{+2}\) is strongest reducing agent and is more stable in its +3 oxidation state.

\(Cr^{+2}\to Cr^{+3}\left ( 3d^{3} \right )\left ( t^{3}_{2g} eg^{0}\right )\)

So order will be

\(Mn>Fe>Cr>Co\)

correct optoion will be (a)

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