### 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 Let's look at the enthalpy of atomisation values for the given 3d transition metals: - **Chromium (mathrmCr$\mathrm{Cr}$)**: 397 text kJ mol^-1$397 \text{ kJ mol}^{-1}$ - **Iron (mathrmFe$\mathrm{Fe}$)**: 416 text kJ mol^-1$416 \text{ kJ mol}^{-1}$ - **Cobalt (mathrmCo$\mathrm{Co}$)**: 425 text kJ mol^-1$425 \text{ kJ mol}^{-1}$ - **Nickel (mathrmNi$\mathrm{Ni}$)**: 430 text kJ mol^-1$430 \text{ kJ mol}^{-1}$ Among the choices, **Chromium (mathrmCr$\mathrm{Cr}$)** has the lowest enthalpy of atomisation (397 text kJ mol^-1$397 \text{ kJ mol}^{-1}$), 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$\mathrm{Cr}$ is: mathrmCr = [mathrmAr] 3mathrmd^5 4mathrms^1$\mathrm{Cr} = [\mathrm{Ar}] 3\mathrm{d}^5 4\mathrm{s}^1$ Total valence electrons = 5 + 1 = 6$= 5 + 1 = 6$. ### Pattern Recognition In transition metals, manganese (Mn$Mn$) has the absolute lowest enthalpy of atomisation in the 3d series because of its completely half-filled d^5$d^5$ and completely filled s^2$s^2$ stability. Since Mn$Mn$ is not in the list, Chromium ("Cr"$"Cr"$) is next, having 6$6$ valence electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements

### 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