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

0 a n d + 7

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 2 a n d + 7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0 a n d + 6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 1 a n d + 7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0 a n d + 7$
$A l + M n O_{4}^{-} \to A l (O H)_{4}^{-} + M n O_{2}$
-The oxidation number of an element in its free (uncombined) state is zero.
So the oxidation state of $A l$ is $= 0$ at reactant side.
-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 $- 2$ . Therefore the oxidation state of $M n$ in ${M n O_{4}}^{-}$ is
$x + 4 (- 2) = - 1$
$∴ x = + 7$

Suggest Corrections

Similar questions

A l + M n O_{4}^{-} \to A l (O H)_{4}^{-} + M n O_{2}

A l

M n

A l + M n O_{4}^{-} \to A l {(O H)}_{4}^{-} + M n O_{2}

A l + M n O_{4}^{-} \to A l (O H)_{4}^{-} + M n O_{2}

X

X

Y

M n

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

4

+ 6

+ 7

A l + N O_{2}^{-} \to A l (O H)_{4}^{-} + N H_{3}

Join BYJU'S Learning Program