Which represents correct comparison of the stability of ions?

M n O_{4}^{-} < M n^{2 +}

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C r^{2 +} < C r^{3 +}

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C r O_{4}^{2 -} < C r^{3 +}

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All of these

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Solution

The correct option is C All of these
In higher states these compounds have a high tendency to get electron and reduce, this is the reason they are a good oxidising agent.

So, order is - $C r O_{4}^{2 -} < C r^{3 +}$
$M n O_{4}^{-} < M n^{2 +}$

Also, $C r$ is more stable in the $+ 3$ oxidation state rather than in the $+ 2$ oxidation state.

Hence, the correct option is $D$

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Similar questions

F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}

F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}

V^{2 +} < C r^{2 +} < M n^{2} < F e^{2 +}

N i^{2 +} < C o^{2 +} < F e^{2 +} < M n^{2 +}

C o^{3 +} < F e^{3 +} < C r^{3 +} < S c^{3 +}

S c < T i < C r < M n

E_{C l_{2} / C l^{-}}^{\circ} = 1.36 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{\circ} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = 1.51 V

C r O_{4}^{2 -} + 8 H^{+} + 6 e^{-} \to C r^{3 +} + 4 H_{2} O

I^{-} + 6 O H^{-} \to I O_{3}^{-} + 3 H_{2} O + 6 e^{-}

M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}

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