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Common Oxidation States

When kMnO4 ...

Question

When $k M n O_{4}$ acts as an oxidising agent and ultimately forms $M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3},$ and $M n^{2 +}$ , determine the number of electrons transferred in each case.

A

1, 3, 4, 5

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B

1, 2, 3, 5

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C

1, 3, 5, 4

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D

1, 2, 4, 5

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Solution

The correct option is A $1, 3, 4, 5$
$K M n O_{4} 1 e^{-} - - \to M n O_{4}^{2 -}$
$K M n O_{4} 3 e^{-} - - \to M n O_{2}$
$K M n O_{4} 4 e^{-} - - \to M n_{2} O_{3}$
$K M n O_{4} 5 e^{-} - - \to M n^{2 +}$

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Similar questions

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 per molecule of

K M n O_{4}

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:

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

K M n O_{4}

acts as an oxidizing 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 are:

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