Match List-I with List-II.

List-I (Transition metal ion)	List-II (Spin only magnetic moment (B.M.))
(A) mathrmTi^3+$\mathrm{Ti}^{3+}$	(I) 3.87$3.87$
(B) mathrmV^2+$\mathrm{V}^{2+}$	(II) 0.00$0.00$
(C) mathrmNi^2+$\mathrm{Ni}^{2+}$	(III) 1.73$1.73$
(D) mathrmSc^3+$\mathrm{Sc}^{3+}$	(IV) 2.84$2.84$

Choose the correct answer from the options given below :

### Core Logic 1. In the 3mathrmd$3\mathrm{d}$ transition series, **Vanadium (mathrmV$\mathrm{V}$)** has the highest enthalpy of atomisation (515 text kJ mol^-1$515 \text{ kJ mol}^{-1}$). 2. One of its aquated ions, mathrmV^3+mathrm(aq)$\mathrm{V}^{3+}\mathrm{(aq)}$ [specifically [mathrmV(H_2O)_6]^3+$[\mathrm{V(H_2O)_6}]^{3+}$], has a characteristic **green colour**. 3. The corresponding oxide for this state is mathrmV_2mathrmO_3$\mathrm{V}_2\mathrm{O}_3$ (Vanadium(III) oxide). 4. Metal oxides in lower oxidation states (+2, +3$+2, +3$) are typically **basic** in nature, while intermediate states like mathrmV_2mathrmO_4$\mathrm{V}_2\mathrm{O}_4$ are amphoteric, and high states like mathrmV_2mathrmO_5$\mathrm{V}_2\mathrm{O}_5$ are acidic. Therefore, mathrmV_2mathrmO_3$\mathrm{V}_2\mathrm{O}_3$ is purely a basic oxide. ### Pattern Recognition Vanadium (V$V$) 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

### Related Formula Spin-only magnetic moment equation: mu = sqrtn(n+2) quad textBM$$\mu = \sqrt{n(n+2)} \quad \text{BM}$$ where n$n$ is the exact count of unpaired d-shell electrons. ### Execution Let us compute the unpaired electron distribution for each transition metal ion: 1. **textCu^+$\text{Cu}^+$**: Electronic configuration is [textAr]3d^10$[\text{Ar}]3d^{10}$. All electrons are paired up. n = 0 implies mu = 0 text BM$$n = 0 \implies \mu = 0 \text{ BM}$$ 2. **textCu^2+$\text{Cu}^{2+}$**: Electronic configuration is [textAr]3d^9$[\text{Ar}]3d^9$. Has one unpaired hole. n = 1 implies mu = sqrt1(1+2) = sqrt3 approx 1.73 text BM$$n = 1 \implies \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$$ 3. **textCr^3+$\text{Cr}^{3+}$**: Electronic configuration is [textAr]3d^3$[\text{Ar}]3d^3$. Has three unpaired parallel spins. n = 3 implies mu = sqrt3(3+2) = sqrt15 approx 3.87 text BM$$n = 3 \implies \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$$ 4. **textCr^2+$\text{Cr}^{2+}$**: Electronic configuration is [textAr]3d^4$[\text{Ar}]3d^4$. Has four unpaired spins. n = 4 implies mu = sqrt4(4+2) = sqrt24 approx 4.90 text BM$$n = 4 \implies \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$$ Arranging these values in decreasing structural order: mu(textCr^2+) > mu(textCr^3+) > mu(textCu^2+) > mu(textCu^+)$$\mu(\text{Cr}^{2+}) > \mu(\text{Cr}^{3+}) > \mu(\text{Cu}^{2+}) > \mu(\text{Cu}^+)$$ ### Pattern Recognition The value of the spin-only magnetic moment scales monotonically with the number of unpaired electrons (n$n$). 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

### Core Logic In an acidic medium, both mathrmK_2mathrmCr_2mathrmO_7$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and mathrmKMnO_4$\mathrm{KMnO}_4$ act as strong oxidizing agents and carry out the following transformations: - **A:** Oxidize iodide to iodine: mathrmI^- rightarrow mathrmI_2$\mathrm{I}^- \rightarrow \mathrm{I}_2$ - **B:** Oxidize sulfide to elemental sulfur: mathrmS^2- rightarrow mathrmS$\mathrm{S}^{2-} \rightarrow \mathrm{S}$ - **C:** Oxidize ferrous ions to ferric ions: mathrmFe^2+ rightarrow mathrmFe^3+$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}$ For reactions D and E: - Iodide is oxidized to iodate (mathrmIO_3^-$\mathrm{IO}_3^-$) by mathrmKMnO_4$\mathrm{KMnO}_4$ primarily in a neutral or faintly alkaline medium, not acidic. - Thiosulfate (mathrmS_2mathrmO_3^2-$\mathrm{S}_2\mathrm{O}_3^{2-}$) 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^-$\mathrm{I}^- \rightarrow \mathrm{IO}_3^-$ 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

### Related Formula The spin-only magnetic moment (mu$\mu$) is given by: mu = sqrtn(n+2)text B.M. $$ mu = \sqrt{n(n+2)}\text{ B.M.} $$ ### Core Logic Given mu = 4.9text B.M.$\mu = 4.9\text{ B.M.}$, we can solve for the number of unpaired electrons (n$n$): 4.9 = sqrtn(n+2) implies 24.01 = n^2 + 2n implies n = 4$$ 4.9 = \sqrt{n(n+2)} implies 24.01 = n^2 + 2n implies n = 4$$ ### Step 1: Electron Configuration Audit Let us compute the number of unpaired electrons (n$n$) for each ion: * **A.** _24textCr^2+: [textAr] 3d^4 implies n = 4$_{24}\text{Cr}^{2+}: [\text{Ar}] 3d^{4} implies n = 4$ * B. $ * B. $_{26}\text{Fe}^{2+}: [\text{Ar}] 3d^{6} implies n = 4 * C. $_26textFe^3+: [textAr] 3d^5 implies n = 5$$ * C. $_{26}\text{Fe}^{3+}: [\text{Ar}] 3d^{5} implies n = 5$$ * D. _27textCo^2+: [textAr] 3d^7 implies n = 3$_{27}\text{Co}^{2+}: [\text{Ar}] 3d^{7} implies n = 3$ * E. $ * 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 $ ### Step 2: Selection Thus, ions A, B, and E possess exactly 4 unpaired electrons and give a magnetic moment of $4.9\text{ B.M.} ### Pattern Recognition Shortcut: The value of the magnetic moment always starts with the integer equal to the number of unpaired electrons ($ ### 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 $). Instantly filter configurations with $d^4 or $ or $d^6$ profiles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: d- and f-Block Elements

### 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$K_2Cr_2O_7$. Heating it with metal chloride and concentrated acid generates deep red chromyl chloride vapors (CrO_2Cl_2$CrO_2Cl_2$). 2. Passing these vapors into sodium hydroxide dissolves them, producing yellow sodium chromate compound **B** (Na_2CrO_4$Na_2CrO_4$). 3. Acidifying the chromate solution dimerizes it into orange sodium dichromate compound **C** (Na_2Cr_2O_7$Na_2Cr_2O_7$). ### Step 1: Structural Analysis of Dichromate The dichromate ion (Cr_2O_7^2-$Cr_2O_7^{2-}$) consists of two tetrahedral chromium units sharing a bridging oxygen atom (textCr-textO-textCr$\text{Cr}-\text{O}-\text{Cr}$). 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$2 \times 3 = 6$.

### 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-$\text{Cr}_2\text{O}_7^{2-}$ 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