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In acidic medium, H $_2O_2$ changes Cr $_2O_7^{2-}$ to CrO $_5$ which has two (-O-O-) bonds. The oxidation state of Cr in CrO $_5$ is

medium

RedoxReactions

2014

chemistry

+5

+3

+6

-10

Explanation

To solve this problem, we need to determine the oxidation state of Cr in CrO $_5. \\$ Let's analyze the structure of CrO $_5: \\$

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• CrO

_5

is known to have two

(-O-O-)

bonds, which are peroxo bonds.

\\

• The structure can be visualized as

O

_2

Cr(O-O)_2

. \\

In this structure:

\\

• Each peroxo group

(-O-O-)

contributes a total charge of

-2. \\

• There are two peroxo groups, contributing a total charge of

-4. \\

• The remaining oxygen (not part of the peroxo groups) is a typical oxide, contributing

-2. \\

Let's calculate the oxidation state of Cr:

\\

• Let the oxidation state of Cr be

x. \\

• The total charge of the compound is neutral (0).

\\

Setting up the equation:

\\ x + (-2) + 2(-2) = 0 \\

Simplify the equation:

\\ x - 2 - 4 = 0 \\ x - 6 = 0 \\ x = +6 \\

Therefore, the oxidation state of Cr in

CrO

_5

is

+6. \\

This corresponds to Option 3.