The oxidation state of Cr in ${\mathrm{CrO}}_5$is

To determine the oxidation state of Cr in the compound CrO₅ (Chromium pentoxide), we can analyze the structure and oxidation states of its constituent elements as follows:

Step-by-Step Calculation

1. Understand the Oxidation States of Oxygen: In CrO₅, there are five oxygen atoms. Generally, oxygen has an oxidation state of -2. However, in peroxides, oxygen has an oxidation state of -1. The structure of CrO₅ includes peroxo (O-O) bonds, which affect the oxidation state calculations.
2. Analyze the Structure of CrO₅: The compound consists of one chromium atom and five oxygen atoms. The structure includes:
   • Two peroxo (O-O) linkages, accounting for four oxygen atoms, each with an oxidation state of -1.
   • One oxygen atom in the oxide form (double-bonded to chromium) with an oxidation state of -2.

3. Set Up the Equation: Let the oxidation state of Cr be x. Since CrO₅ is a neutral compound, the sum of all oxidation states must equal zero. The equation can be set up as:

   x + (4 × -1) + (-2) = 0

4. Calculate the Contribution from Oxygen:

   • Four oxygen atoms in peroxo groups: 4 × -1 = -4
   • One oxygen atom in the oxide form: -2

   Total contribution from oxygen = -4 - 2 = -6.

5. Solve for the Oxidation State of Cr: Substitute the total oxygen contribution into the equation:

   x - 6 = 0

   Solving for x gives:

   x = +6

Final Answer

The oxidation state of Cr in CrO₅ is +6.

Why is the Oxidation State of Cr in CrO₅ +6?

Chromium (Cr) achieves its maximum oxidation state of +6 in CrO₅ by bonding with oxygen atoms in both peroxo and oxide forms. This aligns with chromium's ability to reach a high oxidation state due to its electronic configuration and bonding capacity.

The calculation demonstrates that the oxidation state of Cr is consistent with its chemical properties and the structure of CrO₅.