Question

In acidic medium, $H_2O_2 $ changes $ Cr_2O_7^{2-} $ to $CrO_5 $ which has two (-O-O-) bonds. Oxidation state of Cr in $CrO_5 $ is

• A. 5
• B. 3
• C. 6
• D. -10

Answer 1

The Correct Option is C

To determine the oxidation state of chromium in CrO_5, we need to consider the composition and oxidation states of its constituent atoms.

Step 1: Analyze the molecular structure of CrO_5

The compound CrO_5 consists of one chromium (Cr) atom and five oxygen (O) atoms. It is known that CrO_5 includes two peroxo (–O–O–) bonds and one double-bonded oxygen to chromium.

Step 2: Assign oxidation states to the atoms

• Each oxygen in the –O–O– bond has an oxidation state of -1. As there are two such peroxo bonds, that contributes a total of four oxygen atoms with oxidation number -1.
• Each of these four oxygen atoms contributes -1 for a total oxidation state from this group of 4 \times (-1) = -4.
• The remaining oxygen (connected to Cr by a double bond) will have an oxidation state of -2.

Step 3: Calculate the oxidation state of Chromium

Assume the oxidation state of chromium is x. The sum of oxidation states in a neutral molecule should be zero.

Setting up the equation, we have:

x + 4 \times (-1) + (-2) = 0

x - 4 - 2 = 0

x = 6

Therefore, the oxidation state of chromium in CrO_5 is 6.

Conclusion: The correct answer is the oxidation state of Cr in CrO_5 is 6. Hence, the correct option is 6.