JEE Chemistry 2026 Question with Solution

Given: The reactions are:

$$NaCl+K_{2}Cr_{2}O_{7}+H_{2}SO_{4} \rightarrow A +...$$ $$A + NaOH \rightarrow B +...$$ $$B + H_{2}SO_{4}+H_{2}O_{2} \rightarrow C +...$$

Find: The value of $X+Y+Z$ for product $C$, where $X$ is the number of $O_{2}^{2-}$ units, $Y$ is the total number of oxygen atoms, and $Z$ is the oxidation state of $Cr$.

Identify the compounds:

1. In the chromyl chloride test,

$$A = CrO_2Cl_2$$

2. Chromyl chloride reacts with $NaOH$ to form sodium chromate,

$$B = Na_2CrO_4$$

3. Chromate in acidic medium with hydrogen peroxide forms chromium peroxide,

$$C = CrO_5$$

Analyze the structure of $CrO_5$:

• It contains one terminal oxygen and two peroxo groups.
• Therefore, the number of $O_{2}^{2-}$ units is

$$X = 2$$

• Total number of oxygen atoms in $CrO_5$ is

$$Y = 5$$

Oxidation state of chromium: In $CrO_5$, one oxygen has oxidation state $-2$ and four oxygens in the two peroxo groups have oxidation state $-1$ each.

$$x + 1(-2) + 4(-1) = 0$$ $$x - 2 - 4 = 0$$ $$x = +6$$

So,

$$Z = 6$$

Calculate:

$$X+Y+Z = 2+5+6 = 13$$

Therefore, the value of $X+Y+Z$ is $13$.

Note: The solution's lists the correct answer as $18$, but the extracted solution working gives $13$. Using the solution working, the defensible answer is $13$.

Given: Product $C$ is formed when chromate reacts with $H_2SO_4$ and $H_2O_2$.

Find: $X+Y+Z$.

The key idea is that $C$ is chromium peroxide,

$$CrO_5$$

which has the well-known butterfly structure.

This structure has:

• one normal oxo oxygen,
• two peroxide linkages, each contributing one $O_2^{2-}$ unit.

So,

$$X = 2$$

Since the formula is $CrO_5$,

$$Y = 5$$

For oxidation state of chromium, peroxide oxygen is counted as $-1$, not $-2$. Hence,

$$Z + (-2) + 4(-1) = 0$$ $$Z - 2 - 4 = 0$$ $$Z = 6$$

Thus,

$$X+Y+Z = 2+5+6 = 13$$

Therefore, the correct numerical value from the shown chemistry is $13$.