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Valency and Oxidation State

When an oxidi...

Question

When an oxidising agent reacts, it consumes electrons. In the following changes, which one of the oxidising agents has consumed maximum number of electrons per mole?

A

C r O_{4}^{2 -} ⟶ {C r}^{3 +}

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B

M n O_{2} ⟶ {M n}^{3 +}

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C

M n O_{4}^{2 -} ⟶ {M n}^{2 +}

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D

M n O_{4}^{-} ⟶ {M n}^{2 +}

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Solution

The correct option is D $M n O_{4}^{-} ⟶ {M n}^{2 +}$
Oxidation number of $M n$ in $M n O_{4}$ is $+ 7$
Therefore, conversion of ${M n}^{2 +}$ from ${M n}^{+ 7}$ will have to accept $5$ electrons, which is highest.

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Similar questions

K M n O_{4}

acts as an oxidising agent, it can form

M n O_{4}^{2 -}, M n O_{2}, {M n}_{2} O_{3}

{M n}^{2 +}

, the numbers of electrons transferred in each case respectively, are :

K M n O_{4}

act as an oxidising agent and ultimately form

M n O_{4}^{2 -}, M n O_{2}, {M n}_{2} O_{3}

{M n}^{2 +}

, then the numbers of electrons transferred in each case, respectively, are :

Q. Number of electrons transfer in each case when

K M n O_{4}

acts as an oxidising agent to given:

M n O_{2}, {M n}^{2 +}, M n {(O H)}_{3}

M n O_{4}^{2 -}

are respectively:

K M n O_{4}

acts as an oxidising agent and ultimately forms

[M n O_{4}]^{2 -}, M n O_{2}, M n_{2} O_{3}, M n^{2 +}

then the number of electrons transferred in each case respectively is

K M n O_{4}

acts as an oxidising agent and ultimately forms

[M n O_{4}]^{2 -}, M n O_{2}, M n_{2} O_{3}, M n^{2 +}

then the number of electrons transferred in each case respectively is

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