Consider the following reactions: NaCl + K_2Cr_2O_7 + H_2SO_4 rightarrow A + KHSO_4 + NaHSO_4 + H_2O A + NaOH rightarrow B + NaCl + H_2O B + H_2SO_4 + H_2O_2 rightarrow C + Na_2SO_4 + H_2O In the product 'C', 'X' is the number of O_2^2- units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of X + Y + Z is ____.

Numerical Answer Type:
Enter a numerical value Answer: 13 to 13 +4 marks

Solution & Explanation

### Core Logic The first reaction is the classical **Chromyl Chloride Test**: 4mathrmNaCl + mathrmK_2Cr_2O_7 + 6mathrmH_2SO_4 rightarrow 2mathrmCrO_2Cl_2 (textA) + 2mathrmKHSO_4 + 4mathrmNaHSO_4 + 3mathrmH_2O Product A is Chromyl chloride (mathrmCrO_2Cl_2), a red-orange gas. When Chromyl chloride gas is passed into NaOH solution, it forms a yellow solution of sodium chromate (B): mathrmCrO_2Cl_2 (textA) + 4mathrmNaOH rightarrow mathrmNa_2CrO_4 (textB) + 2mathrmNaCl + 2mathrmH_2O Acidifying the sodium chromate solution with H_2SO_4 and adding H_2O_2 yields a deep blue solution of Chromium(VI) peroxide, CrO_5 (C): mathrmNa_2CrO_4 (textB) + mathrmH_2SO_4 + 2mathrmH_2O_2 rightarrow mathrmCrO_5 (textC) + mathrmNa_2SO_4 + 3mathrmH_2O Structure of CrO_5:
Chromyl Chloride Test diagram for Q75 - JEE Main 2026 Morning
Chromyl Chloride Test diagram for Q75 - JEE Main 2026 Morning
- It has a butterfly structure. - Number of peroxy units (O_2^2-), X = 2. - Total number of oxygen atoms, Y = 5. - Oxidation state of Cr, Z = +6. Sum: X + Y + Z = 2 + 5 + 6 = 13. ### Step 1: Final Calculation X + Y + Z = 13 ### Pattern Recognition Chromyl chloride test rightarrow CrO_2Cl_2 (red gas). Absorbed in NaOH rightarrow Na_2CrO_4 (yellow). Tested with H_2O_2/H^+ rightarrow CrO_5 (butterfly structure, blue, two peroxy links, Cr in +6). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

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