### 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

### Related Formula textIE propto frac1textStability of electronic configuration$\text{IE} \propto \frac{1}{\text{Stability of electronic configuration}}$ ### Core Logic Let's examine the configurations: - For Mn^2+$Mn^{2+}$, the electronic configuration is [Ar]3d^5$[Ar]3d^5$, which features a highly stable, symmetric half-filled d-subshell. - For Fe^2+$Fe^{2+}$, the configuration is [Ar]3d^6$[Ar]3d^6$. Because of the extra exchange energy and stability of the half-filled 3d^5$3d^5$ state, it is harder to remove an electron from Mn^2+$Mn^{2+}$ than from Fe^2+$Fe^{2+}$. Therefore, the ionisation enthalpy of Mn^2+$Mn^{2+}$ is greater than that of Fe^2+$Fe^{2+}$: textIE(Mn^2+) > textIE(Fe^2+)$\text{IE}(Mn^{2+}) > \text{IE}(Fe^{2+})$ Hence, the expression Mn^2+ < Fe^2+$Mn^{2+} < Fe^{2+}$ is incorrect. ### Pattern Recognition Whenever you see manganese (Mn$Mn$) in the +2$+2$ oxidation state, remember its exceptionally stable d^5$d^5$ config. This creates anomalous spikes in successive ionisation energies compared to neighboring iron (Fe$Fe$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements

### Core Logic Let's map the electronic configurations and count the unpaired d-orbital electrons for each option: * For pair (1): V^2+ implies [Ar] 3d^3 4s^0 implies 3 text unpaired electrons$V^{2+} \implies [Ar] 3d^3 4s^0 \implies 3 \text{ unpaired electrons}$ Co^2+ implies [Ar] 3d^7 4s^0 implies t_2g^5 e_g^2 implies 3 text unpaired electrons$Co^{2+} \implies [Ar] 3d^7 4s^0 \implies t_{2g}^5 e_g^2 \implies 3 \text{ unpaired electrons}$ Both ions contain exactly 3 unpaired electrons. * For other ions: Ti^2+ implies [Ar] 3d^2 implies 2 text unpaired e-, quad Fe^3+ implies [Ar] 3d^5 implies 5 text unpaired e-$Ti^{2+} \implies [Ar] 3d^2 \implies 2 \text{ unpaired e-}, \quad Fe^{3+} \implies [Ar] 3d^5 \implies 5 \text{ unpaired e-}$ Cr^2+ implies [Ar] 3d^4 implies 4 text unpaired e-, quad Ti^3+ implies [Ar] 3d^1 implies 1 text unpaired e-$Cr^{2+} \implies [Ar] 3d^4 \implies 4 \text{ unpaired e-}, \quad Ti^{3+} \implies [Ar] 3d^1 \implies 1 \text{ unpaired e-}$ Mn^2+ implies [Ar] 3d^5 implies 5 text unpaired e-$Mn^{2+} \implies [Ar] 3d^5 \implies 5 \text{ unpaired e-}$ ### Pattern Recognition D-orbital counts follow a predictable symmetry: a 3d^n$3d^n$ system contains the same number of unpaired electrons as a 3d^10-n$3d^{10-n}$ system under high-spin conditions. This explains why 3d^3$3d^3$ (V^2+$V^{2+}$) and 3d^7$3d^7$ (Co^2+$Co^{2+}$) match perfectly with 3 unpaired electrons each. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements