$A l + M n O_{4}^{-} \to A l (O H)_{4}^{-} + M n O_{2}$
For this reaction the oxidation state of $A l$ and $M n$ in the product are respectively:

+ 3 a n d + 4

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+ 4 a n d + 4

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+ 3 a n d + 7

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+ 2 a n d + 4

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Solution

The correct option is B $+ 3 a n d + 4$
The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. The oxidation number of oxygen in a compound is usually $- 2$ . The oxidation state of hydrogen in a compound is usually $+ 1$ .
The oxidation state of $A l$ in ${A l (O H)_{4}}^{-}$
$x + 4 (+ 1 - 2) = - 1$
$∴ x = + 3$
The oxidation state of $M n$ in $M n O_{2}$
$y + 2 (- 2) = 0$
$∴ y = + 4$

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