MnO 4-ions can be reduced in strong alkaline medium to giveA- Mn 2-B- MnO 2C- MnO 42D- MnO 3-

The correct option is C MnO^2 _4MnO^ _4 → MnO^2 _4 (Maganate) ...

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Byju's Answer

Standard XII

Chemistry

Werner's Coordination Theory

MnO 4-ions ca...

Question

$M n O_{4}^{-}$ ions can be reduced in strong alkaline medium to give

M n^{2 +}

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M n O_{2}

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M n O_{4}^{2 -}

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M n O_{3}^{-}

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Solution

The correct option is C $M n O_{4}^{2 -}$
$M n O_{4}^{-} \to M n O_{4}^{2 -} (M a g a n a t e)$

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Similar questions

For titrating certain volume of a reducing substance by

1 M

K M n O_{4}

, it is found that

20

mL was used in acidic medium,

33.4

mL was used in alkaline medium and

100

mL was used in neutral medium.

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

is reduced to

M n^{2 +}

in acidic Medium.

To what oxidation state is

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

reduced to in alkaline medium?

Q. In dilute alkaline solution,

M n O_{4}^{-}

changes to:

100

mL of

1 M K M n O_{4}

oxidised

100

mL of

H_{2} O_{2}

in acidic medium (when

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

is reduced to

M n^{2 +}

). Volume of same

K M n O_{4}

required to oxidise

100

mL of

H_{2} O_{2}

in neutral medium (when

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

is reduced to

M n O_{2}

) will be:

H_{2} O_{2}

reduces

M n O_{4}^{-}

ion to

Q. The total number of

H_{2} O

molecules present in the balanced chemical reaction is :

M n O_{4}^{⊖} + {F e}^{2 +} ⟶ {M n}^{2 +} + {F e}^{3 +}

(in alkaline medium)

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MnO−4 ions can be reduced in strong alkaline medium to give

The correct option is C MnO2−4MnO−4→MnO2−4(Maganate)

$M n O_{4}^{-}$ ions can be reduced in strong alkaline medium to give

The correct option is C $M n O_{4}^{2 -}$
$M n O_{4}^{-} \to M n O_{4}^{2 -} (M a g a n a t e)$