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

Reference Study Guides

D And F Block Elements

More d and f Block Elements Previous-Year Questions — Page 8

Q86 jee_main_2024_31_jan_evening Chromyl Chloride Test
In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+) ________
Numerical Answer. Answer: 6 to 6

Solution

### Related Formula K_2Cr_2O_7(s) + 4KCl(s) + 6H_2SO_4(conc.) rightarrow 2CrO_2Cl_2(g) + 6KHSO_4 + 3H_2O ### Core Logic This reaction represents the Chromyl Chloride test used to detect the presence of chloride ions. When potassium dichromate is heated with a metal chloride in concentrated sulfuric acid, red vapors of chromyl chloride (CrO_2Cl_2) are evolved. ### Step 1: Oxidation State Calculation In chromyl chloride (CrO_2Cl_2): Let the oxidation state of Chromium be x. Oxygen is typically -2 and Chlorine is -1. x + 2(-2) + 2(-1) = 0 x - 4 - 2 = 0 x = +6 Thus, the oxidation state of Chromium in the product is 6. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements Class 11 Chemistry: Practical Chemistry
Q71 jee_main_2024_31_jan_morning Potassium Dichromate and Permanganate
Identify correct statements from below: A. The chromate ion is square planar. B. Dichromates are generally prepared from chromates. C. The green manganate ion is diamagnetic. D. Dark green coloured K_2MnO_4 disproportionates in a neutral or acidic medium to give permanganate. E. With increasing oxidation number of transition metal, ionic character of the oxides decreases. Choose the correct answer from the options given below:
  • A. textB, C, D only
  • B. textA, D, E only
  • C. textA, B, C only
  • D. textB, D, E only

Solution

### Step 1: Statement A Analysis CrO_4^2- (chromate ion) is tetrahedral, not square planar. Statement A is incorrect. ### Step 2: Statement B Analysis 2Na_2CrO_4 + 2H^+ rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O. Dichromates are indeed prepared from chromates. Statement B is correct. ### Step 3: Statement C Analysis The green manganate ion (MnO_4^2-) has manganese in the +6 oxidation state (3d^1). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect. ### Step 4: Statement D Analysis Dark green coloured K_2MnO_4 undergoes disproportionation in neutral or acidic media to yield permanganate (MnO_4^-) and manganese dioxide (MnO_2). Statement D is correct. ### Step 5: Statement E Analysis Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct. ### Final Conclusion The correct statements are B, D, and E. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

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