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 2

Q39 jee_main_2025_07_april_morning Properties of Oxides
The first transition series metal 'M' has the highest enthalpy of atomisation in its series. One of its aquated ions (mathbfM^n+) exists in green colour. The nature of the oxide formed by the above M ion is:
  • A. textneutral
  • B. textacidic
  • C. textbasic
  • D. textamphoteric

Solution

### Core Logic 1. In the 3mathrmd transition series, **Vanadium (mathrmV)** has the highest enthalpy of atomisation (515 text kJ mol^-1). 2. One of its aquated ions, mathrmV^3+mathrm(aq) [specifically [mathrmV(H_2O)_6]^3+], has a characteristic **green colour**. 3. The corresponding oxide for this state is mathrmV_2mathrmO_3 (Vanadium(III) oxide). 4. Metal oxides in lower oxidation states (+2, +3) are typically **basic** in nature, while intermediate states like mathrmV_2mathrmO_4 are amphoteric, and high states like mathrmV_2mathrmO_5 are acidic. Therefore, mathrmV_2mathrmO_3 is purely a basic oxide. ### Pattern Recognition Vanadium (V) stands out with high atomisation enthalpy and characteristic oxidation states. Lower oxides of transition metals are always basic, higher oxidation state oxides are acidic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements
Q36 jee_main_2025_08_april_evening Magnetic Properties
The correct decreasing order of spin-only magnetic moment values (BM) of textCu^+, \text{Cu}^{2+}, \text{Cr}^{2+}, and \text{Cr}^{3+}$ ions is:
  • A. textCu^+ > textCu^2+ > textCr^3+ > textCr^2+
  • B. textCu^2+ > textCu^+ > textCr^2+ > textCr^3+
  • C. textCr^2+ > textCr^3+ > textCu^2+ > textCu^+
  • D. textCr^3+ > textCr^2+ > textCu^+ > textCu^2+

Solution

### Related Formula Spin-only magnetic moment equation: mu = sqrtn(n+2) quad textBM where n is the exact count of unpaired d-shell electrons. ### Execution Let us compute the unpaired electron distribution for each transition metal ion: 1. **textCu^+**: Electronic configuration is [textAr]3d^10. All electrons are paired up. n = 0 implies mu = 0 text BM 2. **textCu^2+**: Electronic configuration is [textAr]3d^9. Has one unpaired hole. n = 1 implies mu = sqrt1(1+2) = sqrt3 approx 1.73 text BM 3. **textCr^3+**: Electronic configuration is [textAr]3d^3. Has three unpaired parallel spins. n = 3 implies mu = sqrt3(3+2) = sqrt15 approx 3.87 text BM 4. **textCr^2+**: Electronic configuration is [textAr]3d^4. Has four unpaired spins. n = 4 implies mu = sqrt4(4+2) = sqrt24 approx 4.90 text BM Arranging these values in decreasing structural order: mu(textCr^2+) > mu(textCr^3+) > mu(textCu^2+) > mu(textCu^+) ### Pattern Recognition The value of the spin-only magnetic moment scales monotonically with the number of unpaired electrons (n). More unpaired electrons directly translate to a higher magnetic moment, bypassing any tedious square-root calculations during testing. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements
Q44 jee_main_2025_28_jan_morning Oxidizing Properties of KMnO4 and K2Cr2O7
Which of the following oxidation reactions are carried out by both mathrmK_2mathrmCr_2mathrmO_7 and mathrmKMnO_4 in acidic medium? A. mathrmI^- rightarrow mathrmI_2 B. mathrmS^2- rightarrow mathrmS C. mathrmFe^2+ rightarrow mathrmFe^3+ D. mathrmI^- rightarrow mathrmIO_3^- E. mathrmS_2mathrmO_3^2- rightarrow mathrmSO_4^2- Choose the correct answer from the options given below:
  • A. textB, C and D only
  • B. textA, D and E only
  • C. textA, B and C only
  • D. textC, D and E only

Solution

### Core Logic In an acidic medium, both mathrmK_2mathrmCr_2mathrmO_7 and mathrmKMnO_4 act as strong oxidizing agents and carry out the following transformations: - **A:** Oxidize iodide to iodine: mathrmI^- rightarrow mathrmI_2 - **B:** Oxidize sulfide to elemental sulfur: mathrmS^2- rightarrow mathrmS - **C:** Oxidize ferrous ions to ferric ions: mathrmFe^2+ rightarrow mathrmFe^3+ For reactions D and E: - Iodide is oxidized to iodate (mathrmIO_3^-) by mathrmKMnO_4 primarily in a neutral or faintly alkaline medium, not acidic. - Thiosulfate (mathrmS_2mathrmO_3^2-) undergoes a disproportionation/precipitation reaction in acid rather than clean oxidation to sulfate by dichromate. Thus, statements A, B, and C are valid for both under acidic conditions. ### Pattern Recognition Sees: Shared oxidation products in an acidic environment. Shortcut: Remember that mathrmI^- rightarrow mathrmIO_3^- is the signature reaction for alkaline permanganate, allowing quick elimination of choices containing D. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements
Q29 jee_main_2025_03_april_morning Magnetic Properties of Transition Metal Ions
The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are: A. Cr^2+ B. Fe^2+ C. Fe^3+ D. Co^2+ E. Mn^3+ Choose the correct answer from the options given below:
  • A. A, C and E only
  • B. A, D and E only
  • C. B and E only
  • D. A, B and E only

Solution

### Related Formula The spin-only magnetic moment (mu) is given by: mu = sqrtn(n+2)text B.M. ### Core Logic Given mu = 4.9text B.M., we can solve for the number of unpaired electrons (n): 4.9 = sqrtn(n+2) implies 24.01 = n^2 + 2n implies n = 4 ### Step 1: Electron Configuration Audit Let us compute the number of unpaired electrons (n) for each ion: * **A.** _24textCr^2+: [textAr] 3d^4 implies n = 4 * B. _{26}\text{Fe}^{2+}: [\text{Ar}] 3d^{6} implies n = 4 * C. $_26textFe^3+: [textAr] 3d^5 implies n = 5 * D. _27textCo^2+: [textAr] 3d^7 implies n = 3 * E. _{25}\text{Mn}^{3+}: [\text{Ar}] 3d^{4} implies n = 4$ ### Step 2: Selection Thus, ions A, B, and E possess exactly 4 unpaired electrons and give a magnetic moment of 4.9\text{ B.M.} ### Pattern Recognition Shortcut: The value of the magnetic moment always starts with the integer equal to the number of unpaired electrons (4.x implies n = 4). Instantly filter configurations with d^4 or d^6$ profiles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: d- and f-Block Elements
Q50 jee_main_2025_03_april_morning Potassium Dichromate - Preparation and Structure
Consider the following reactions: A+NaCl+H_2SO_4 ightarrow CrO_2Cl_2+textSide Products textCrO*2textCl*2(textVapour) + NaOH ightarrow B + NaCl + H_2O B+H^+ ightarrow C+H_2O The number of terminal 'O' present in the compound 'C' is
Numerical Answer. Answer: 6 to 6

Solution

### Core Logic Let us identify the sequential chemical components via the chromyl chloride test pathway: 1. Reactant **A** represents a dichromate salt like K_2Cr_2O_7. Heating it with metal chloride and concentrated acid generates deep red chromyl chloride vapors (CrO_2Cl_2). 2. Passing these vapors into sodium hydroxide dissolves them, producing yellow sodium chromate compound **B** (Na_2CrO_4). 3. Acidifying the chromate solution dimerizes it into orange sodium dichromate compound **C** (Na_2Cr_2O_7). ### Step 1: Structural Analysis of Dichromate The dichromate ion (Cr_2O_7^2-) consists of two tetrahedral chromium units sharing a bridging oxygen atom (textCr-textO-textCr). Each chromium atom retains 3 localized terminal oxygen units. Thus, the total count of terminal oxygen atoms in the structure is 2 times 3 = 6.
Dichromate structural topology breakdown diagram for Q50
Dichromate structural topology breakdown diagram for Q50
### Pattern Recognition Shortcut: Chromyl chloride path loops directly from dichromate back to dichromate via chromate intermediate salts. Total oxygen atoms in textCr_2textO_7^2- is 7, out of which 1 is bridging, leaving exactly 6 terminal ones. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements

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