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 :

### Related Formula 3mathrmMnO_4^2- + 4mathrmH^+ rightarrow 2mathrmMnO_4^- + mathrmMnO_2 + 2mathrmH_2mathrmO$$3\mathrm{MnO}_4^{2-} + 4\mathrm{H}^+ \rightarrow 2\mathrm{MnO}_4^- + \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$$ ### Core Logic Manganate ion (mathrmMnO_4^2-$\mathrm{MnO}_4^{2-}$), where Mn is in +6 oxidation state, is unstable in acidic medium and undergoes disproportionation. It oxidizes to Permanganate (mathrmMnO_4^-$\mathrm{MnO}_4^-$, +7 state) and reduces to Manganese dioxide (mathrmMnO_2$\mathrm{MnO}_2$, +4 state). ### Pattern Recognition Manganate (green, +6) disproportionates in acid to Permanganate (purple, +7) and MnO_2$MnO_2$ (brown/black precipitate, +4). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements

### Core Logic The first reaction is the classical **Chromyl Chloride Test**: 4mathrmNaCl + mathrmK_2Cr_2O_7 + 6mathrmH_2SO_4 rightarrow 2mathrmCrO_2Cl_2 (textA) + 2mathrmKHSO_4 + 4mathrmNaHSO_4 + 3mathrmH_2O$4\mathrm{NaCl} + \mathrm{K_2Cr_2O_7} + 6\mathrm{H_2SO_4} \rightarrow 2\mathrm{CrO_2Cl_2} (\text{A}) + 2\mathrm{KHSO_4} + 4\mathrm{NaHSO_4} + 3\mathrm{H_2O}$ Product A is Chromyl chloride (mathrmCrO_2Cl_2$\mathrm{CrO_2Cl_2}$), a red-orange gas. When Chromyl chloride gas is passed into NaOH solution, it forms a yellow solution of sodium chromate (B): mathrmCrO_2Cl_2 (textA) + 4mathrmNaOH rightarrow mathrmNa_2CrO_4 (textB) + 2mathrmNaCl + 2mathrmH_2O$\mathrm{CrO_2Cl_2} (\text{A}) + 4\mathrm{NaOH} \rightarrow \mathrm{Na_2CrO_4} (\text{B}) + 2\mathrm{NaCl} + 2\mathrm{H_2O}$ Acidifying the sodium chromate solution with H_2SO_4$H_2SO_4$ and adding H_2O_2$H_2O_2$ yields a deep blue solution of Chromium(VI) peroxide, CrO_5$CrO_5$ (C): mathrmNa_2CrO_4 (textB) + mathrmH_2SO_4 + 2mathrmH_2O_2 rightarrow mathrmCrO_5 (textC) + mathrmNa_2SO_4 + 3mathrmH_2O$\mathrm{Na_2CrO_4} (\text{B}) + \mathrm{H_2SO_4} + 2\mathrm{H_2O_2} \rightarrow \mathrm{CrO_5} (\text{C}) + \mathrm{Na_2SO_4} + 3\mathrm{H_2O}$ Structure of CrO_5$CrO_5$:

- It has a butterfly structure. - Number of peroxy units (O_2^2-$O_2^{2-}$), X = 2$X = 2$. - Total number of oxygen atoms, Y = 5$Y = 5$. - Oxidation state of Cr, Z = +6$Z = +6$. Sum: X + Y + Z = 2 + 5 + 6 = 13$X + Y + Z = 2 + 5 + 6 = 13$. ### Step 1: Final Calculation X + Y + Z = 13$X + Y + Z = 13$ ### Pattern Recognition Chromyl chloride test rightarrow$\rightarrow$ CrO_2Cl_2$CrO_2Cl_2$ (red gas). Absorbed in NaOH rightarrow$\rightarrow$ Na_2CrO_4$Na_2CrO_4$ (yellow). Tested with H_2O_2/H^+$H_2O_2/H^+$ rightarrow$\rightarrow$ CrO_5$CrO_5$ (butterfly structure, blue, two peroxy links, Cr in +6). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

### 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