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 7

Q71 jee_main_2024_30_january_evening Preparation and Properties of KMnO4
Alkaline oxidative fusion of MnO_2 gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:
  • A. textMn_2textO_7 text and MnO_4^-
  • B. textMnO_4^2- text and MnO_4^-
  • C. textMn_2textO_3 text and MnO_4^2-
  • D. textMnO_4^2- text and Mn_2textO_7

Solution

### Core Logic Step 1: Alkaline oxidative fusion of MnO_2 (pyrolusite ore) with KOH in the presence of O_2 (or an oxidizing agent like KNO_3) yields the green-colored manganate ion (MnO_4^2-). 2mathrmMnO_2 + 4mathrmOH^- + mathrmO_2 rightarrow 2mathrmMnO_4^2- + 2mathrmH_2mathrmO So, A is mathrmMnO_4^2-. Step 2: Electrolytic oxidation of the manganate ion (MnO_4^2-) in an alkaline medium converts it to the purple-colored permanganate ion (MnO_4^-). mathrmMnO_4^2- rightarrow mathrmMnO_4^- + mathrme^- So, B is mathrmMnO_4^-. ### Pattern Recognition Industrial preparation sequence of KMnO_4: MnO_2 xrightarrowtextfusion, KOH, O_2 MnO_4^2- text (green) xrightarrowtextelectrolytic oxidation MnO_4^- text (purple). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements
Q78 jee_main_2024_30_january_evening Compounds of Transition Elements
A and B formed in the following reactions are: mathrmCrO_2mathrmCl_2 + 4mathrmNaOH rightarrow mathrmA + 2mathrmNaCl + 2mathrmH_2mathrmO mathrmA + 2mathrmHCl + 2mathrmH_2mathrmO_2 rightarrow mathrmB + 3mathrmH_2mathrmO
  • A. textA = Na_2textCrO_4text, B = CrO_5
  • B. textA = Na_2textCr_2textO_4text, B = CrO_4
  • C. textA = Na_2textCr_2textO_7text, B = CrO_3
  • D. textA = Na_2textCr_2textO_7text, B = CrO_5

Solution

### Core Logic Step 1: Chromyl chloride (CrO_2Cl_2) reacts with an alkali like NaOH to give a yellow solution of sodium chromate (Na_2CrO_4). CrO_2Cl_2 + 4NaOH rightarrow Na_2CrO_4 (A) + 2NaCl + 2H_2O Step 2: Sodium chromate (Na_2CrO_4) reacts with hydrogen peroxide (H_2O_2) in an acidic medium (HCl) to yield the deep blue colored chromium pentoxide (CrO_5, also known as chromium(VI) oxide peroxide). Na_2CrO_4 + 2H_2O_2 + 2HCl rightarrow CrO_5 (B) + 2NaCl + 3H_2O Note: NaCl formation implies the overall balanced reaction uses the acid for neutralization/salt formation. ### Pattern Recognition Chromyl chloride test intermediate: Yellow solution = Na_2CrO_4. Reaction of chromate with H_2O_2 in acid = Blue peroxide CrO_5 (butterfly structure). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements Class 11 Chemistry: Redox Reactions
Q66 jee_main_2024_30_jan_morning Lanthanoids
Diamagnetic Lanthanoid ions are:
  • A. Nd^3+text and Eu^3+
  • B. La^3+text and Ce^4+
  • C. Nd^3+text and Ce^4+
  • D. Lu^3+text and Eu^3+

Solution

### Core Logic An ion is diamagnetic if all its electrons are paired (i.e., zero unpaired electrons). Let's write the electronic configuration for the elements in question. ### Step 1: Checking configurations Cerium (Ce, Z=58): [Xe] 4f^1 5d^1 6s^2 rightarrow Ce^4+: [Xe] 4f^0 (0 unpaired electrons rightarrow Diamagnetic) Lanthanum (La, Z=57): [Xe] 4f^0 5d^1 6s^2 rightarrow La^3+: [Xe] 4f^0 (0 unpaired electrons rightarrow Diamagnetic) ### Pattern Recognition Ions with an empty f-subshell (f^0, e.g., La^3+, Ce^4+) or a completely filled f-subshell (f^14, e.g., Lu^3+, Yb^2+) are invariably diamagnetic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements
Q76 jee_main_2024_30_jan_morning Transition Elements
Match List-I with List-II.
List-I (Species)List-II (Electronic distribution)
(A) Cr^+2(I) 3d^8
(B) Mn^+(II) 3d^54s^1
(C) Ni^+2(III) 3d^4
(D) V^+(IV) 3d^34s^1
Choose the correct answer from the options given below:
  • A. text(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  • B. text(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  • C. text(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  • D. text(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Solution

### Core Logic Let's determine the electronic configuration for each species by first writing the neutral atom's configuration, and then removing electrons starting from the outermost 4s orbital. (A) Cr (Z=24): [Ar] 3d^5 4s^1 rightarrow Cr^2+: [Ar] 3d^4 (B) Mn (Z=25): [Ar] 3d^5 4s^2 rightarrow Mn^+: [Ar] 3d^5 4s^1 (C) Ni (Z=28): [Ar] 3d^8 4s^2 rightarrow Ni^2+: [Ar] 3d^8 (D) V (Z=23): [Ar] 3d^3 4s^2 rightarrow V^+: [Ar] 3d^3 4s^1 ### Step 1: Match execution A rightarrow III B rightarrow II C rightarrow I D rightarrow IV ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements
Q78 jee_main_2024_31_jan_evening Properties of Transition Metal Oxides
Choose the correct statements from the following A. Mn_2O_7 is an oil at room temperature B. V_2O_4 reacts with acid to give VO_2^2+ C. CrO is a basic oxide D. V_2O_5 does not react with acid Choose the correct answer from the options given below:
  • A. text(1) A, B and D only
  • B. text(2) A and C only
  • C. text(3) A, B and C only
  • D. text(4) B and C only

Solution

### Core Logic (A) Mn_2O_7 is a covalent oxide and exists as a green oil at room temperature. (Correct) (B) V_2O_4 dissolves in acids to give VO^2+ (vanadyl) salts, not VO_2^2+. (Incorrect) (C) CrO has chromium in the +2 oxidation state. Lower oxidation state metal oxides are typically basic in nature. (Correct) (D) V_2O_5 is an amphoteric oxide; it reacts with both acids as well as bases. (Incorrect) ### Step 1: Final Selection Only statements A and C are correct, which corresponds to option (2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

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