K₂Cr₂O₇ + H₂O₂ + H₂SO₄ $\underset{conditions}{\overset{ethercold}{\to }}$ compound ‘X’ X is a chromium compound, what is the oxidation state of chromium in compound ‘X’.

The correct answer is +6.

Step-by-step Explanation:

1. Identify the reaction and Compound 'X':
When potassium dichromate ($K_2{\mathrm{Cr}}_2{O}_7$) reacts with hydrogen peroxide ($H_2{O}_2$) in an acidic medium (such as $H_2{\mathrm{SO}}_4$) in the presence of ether under cold conditions, a deep blue colored chromium compound 'X' is formed.
This blue compound 'X' is chromium pentoxide (also known as chromium oxide peroxide), which has the molecular formula ${\mathrm{CrO}}_5$. The ether layer helps to stabilize this otherwise unstable blue-colored compound.

2. Analyze the chemical structure of ${\mathrm{CrO}}_5$:
Chromium pentoxide (${\mathrm{CrO}}_5$) has a unique butterfly-like structure. In this molecule:
- There is one oxo oxygen atom that is double-bonded directly to the chromium atom ($\mathrm{Cr}=O$). This oxygen atom has a standard oxidation state of -2.
- There are four peroxide oxygen atoms forming two peroxide linkages ($-O-O-$) bonded to the chromium atom. Each oxygen atom involved in a peroxide linkage has an oxidation state of -1.

3. Calculate the oxidation state of chromium (Cr) in ${\mathrm{CrO}}_5$:
Let the oxidation state of chromium be represented as $x$.
Since ${\mathrm{CrO}}_5$ is a neutral molecule, the sum of the oxidation states of all the atoms in it must be equal to 0.

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

Simplifying the equation:

$x-2-4=0$

$x-6=0$

$x=+6$

Therefore, the oxidation state of chromium in compound 'X' (${\mathrm{CrO}}_5$) is +6.