NEET UG 2011 — Chemistry Inorganic Chemistry

2011mcqeasy

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. no. $\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$ )

Official previous-year question

Held on 30 Apr 2011 · Verified 9 Jul 2026.

Options

A
Fe $>\mathrm{Mn}>\mathrm{Co}>\mathrm{Cr}$
B
Co $>\mathrm{Mn}>$ Fe $>\mathrm{Cr}$
C
Cr $>$ Mn $>$ Co $>$ Fe
D
Mn $>$ Fe $>$ Cr $>$ Co

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Solution

This can be understood on the basis of $E^{\circ}$ values for $M^{2+} / M$ . $\begin{array}{lrrcr}E \% V & \mathrm{Cr} & \mathrm{Mn} & \mathrm{Fe} & \text { Co } \\ M^{2+} / M & -0.90 & -1.18 & -0.44 & -0.28\end{array}$ $E^{\circ}$ value for $\mathrm{Mn}$ is more negative than expected from general trend due to extra stability of half-filled $\mathrm{Mn}^{2+}$ ion. Thus the correct order should be, $\mathrm{Mn}>\mathrm{Cr}>\mathrm{Fe}>\mathrm{CO}$ an examination of $E^{\circ}$ values for redox couple $M^{3+} / M^{2+}$ shows that $\mathrm{Cr}^{2+}$ is strong reducing $\operatorname{agent}\left(E_{M^{3+}}^{\circ} / M^{2+}=0.41 \mathrm{~V}\right)$ and liberates $\mathrm{H}_2$ from dilute acids. $2 \mathrm{Cr}^{2+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow 2 \mathrm{Cr}^{3+}(a q)+\mathrm{H}_2 \uparrow(g)$ $\therefore$ The correct order is $\mathrm{Mn}>\mathrm{Fe}>\mathrm{Cr}>\mathrm{Co}$ .

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NEET UG 2011 — Chemistry Inorganic Chemistry

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NEET UG 2011 — Chemistry Inorganic Chemistry

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NEET UG Inorganic Chemistry in other years

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Other Chemistry topics for NEET UG

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