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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 Type:
Enter a numerical value Answer: 6 to 6 +4 marks

Solution & Explanation

### 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

Reference Study Guides

D And F Block Elements

More d- and f-Block Elements Previous-Year Questions

Q36 2025 Enthalpy of Atomisation
The number of valence electrons present in the metal among mathrmCr, mathrmCo, mathrmFe and mathrmNi which has the lowest enthalpy of atomisation is:
  • A. 8
  • B. 9
  • C. 6
  • D. 10

Solution

### Core Logic Let's look at the enthalpy of atomisation values for the given 3d transition metals: - **Chromium (mathrmCr)**: 397 text kJ mol^-1 - **Iron (mathrmFe)**: 416 text kJ mol^-1 - **Cobalt (mathrmCo)**: 425 text kJ mol^-1 - **Nickel (mathrmNi)**: 430 text kJ mol^-1 Among the choices, **Chromium (mathrmCr)** has the lowest enthalpy of atomisation (397 text kJ mol^-1), due to a highly stable half-filled d-subshell configuration which leads to weaker metallic bonding relative to the other metals listed. The valence electronic configuration of mathrmCr is: mathrmCr = [mathrmAr] 3mathrmd^5 4mathrms^1 Total valence electrons = 5 + 1 = 6. ### Pattern Recognition In transition metals, manganese (Mn) has the absolute lowest enthalpy of atomisation in the 3d series because of its completely half-filled d^5 and completely filled s^2 stability. Since Mn is not in the list, Chromium ("Cr") is next, having 6 valence electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements
Q39 2025 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 2025 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
Q29 2025 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

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