### Related Formula In transition metal groups: - Stability of higher oxidation states increases down the group: textStability: mathrmCr(VI) < mathrmMo(VI) < mathrmW(VI)$\text{Stability: } \mathrm{Cr(VI)} < \mathrm{Mo(VI)} < \mathrm{W(VI)}$ - Oxidizing power is inversely proportional to the stability of the high oxidation state. ### Core Logic Statement I Analysis: - Since mathrmCr(VI)$\mathrm{Cr(VI)}$ is less stable than mathrmMo(VI)$\mathrm{Mo(VI)}$, chromium is easily reduced from +6$+6$ to +3$+3$, making mathrmCrO_3$\mathrm{CrO}_3$ a much stronger oxidizing agent than mathrmMoO_3$\mathrm{MoO}_3$. Statement I is True. ### Step 1: Analyze Statement II - Statement II asserts that mathrmCr(VI)$\mathrm{Cr(VI)}$ is more stable than mathrmMo(VI)$\mathrm{Mo(VI)}$. As we go down a transition metal group, the higher oxidation states become increasingly stable due to better shielding of the core electrons and relativistic effects. Hence, mathrmMo(VI)$\mathrm{Mo(VI)}$ is more stable than mathrmCr(VI)$\mathrm{Cr(VI)}$. Statement II is False. ### Step 2: Conclusion Therefore, Statement I is True but Statement II is False, matching Option (2). ### Pattern Recognition For d-block elements, higher oxidation states are more stable down the group (e.g., mathrmMo(VI)$\mathrm{Mo(VI)}$ and mathrmW(VI)$\mathrm{W(VI)}$ are very stable and non-oxidizing, whereas mathrmCr(VI)$\mathrm{Cr(VI)}$ is unstable and strongly oxidizing). This is the exact opposite of p-block elements where the inert pair effect makes lower oxidation states more stable down the group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

### Related Formula Spin-only magnetic moment (mu$\mu$) is given by: mu = sqrtn(n+2)mathrm~BM$\mu = \sqrt{n(n+2)}\mathrm{~BM}$ where n$n$ is the number of unpaired d-electrons. ### Core Logic Enthalpies of atomization of the given 3d transition elements (in mathrmkJ/mol$\mathrm{kJ/mol}$): - Scandium (mathrmSc$\mathrm{Sc}$): 326$326$ - Manganese (mathrmMn$\mathrm{Mn}$): 281$281$ - Cobalt (mathrmCo$\mathrm{Co}$): 425$425$ - Copper (mathrmCu$\mathrm{Cu}$): 339$339$ Thus, Cobalt (mathrmCo$\mathrm{Co}$) has the highest enthalpy of atomization. ### Step 1: Determine unpaired electrons in mathrmCo^2+$\mathrm{Co}^{2+}$ Electronic configuration of Cobalt (Z=27$Z=27$): mathrmCo: [mathrmAr] 3d^7 4s^2$\mathrm{Co}: [\mathrm{Ar}] 3d^7 4s^2$ For divalent Cobalt ion (mathrmCo^2+$\mathrm{Co}^{2+}$): mathrmCo^2+: [mathrmAr] 3d^7$\mathrm{Co}^{2+}: [\mathrm{Ar}] 3d^7$ In the d-subshell (five orbitals): - Three orbitals are paired, and three are unpaired (n=3$n=3$). ### Step 2: Calculate spin-only magnetic moment mu = sqrt3(3+2) = sqrt15 approx 3.87mathrm~BM$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\mathrm{~BM}$ Rounding to the nearest integer gives 4$4$. ### Pattern Recognition Enthalpy of atomization generally peaks near the middle of transition series due to maximum metallic bonding. However, mathrmMn$\mathrm{Mn}$ (3d^5 4s^2$3d^5 4s^2$) is an anomaly with an exceptionally low value (281mathrm~kJ/mol$281\mathrm{~kJ/mol}$) due to its highly stable half-filled d^5$d^5$ subshell configuration which reduces electron delocalization in metallic bonding. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The 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 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