In the preparation of $K M n O_{4}$ , pyrolusite $(M n O_{2})$ is first converted to potassium manganate $(K_{2} M n O_{4})$ . In this conversion, the oxidation state of manganese changes from

+1 to +3

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

+2 to +4

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

+3 to +5

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

+4 to +6

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

Open in App

Solution

The correct option is D +4 to +6
Oxidation state of Mn in $(M n O_{2})$ is +4 and Oxidation state of Mn in $(K_{2} M n O_{4})$ is +6.

Suggest Corrections

Similar questions

M n O_{2}

K M n O_{4}

M n O_{2} I \to M n O_{4}^{2 -} + M n O_{4}^{-}

M n O_{2}

K M n O_{4}

M n O_{2} I ⟶ M n O_{4}^{2 -} I I ⟶ M n O_{4}^{-}

M n O_{2}

K M n O_{4}

M n O_{2} I \to M n O_{4}^{2 -} I I \to M n O_{4}^{-}

B r_{2} \to B r O_{3}^{-}

When $H_{2} S O_{3}$ is converted into $H_{2} S O_{4}$ the change in the oxidation state of sulphur is from:

Join BYJU'S Learning Program