MnO_4^2-, in acidic medium, disproportionates to :

Solution & Explanation

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

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D And F Block Elements

More d and f Block Elements Previous-Year Questions — Page 3

Q27 jee_main_2025_04_april_evening Ionisation Enthalpy Trends
The incorrect relationship in the following pairs in relation to ionisation enthalpies is :
  • A. Mn^+ < Cr^+
  • B. Mn^+ < Mn^2+
  • C. Fe^2+ < Fe^3+
  • D. Mn^2+ < Fe^2+

Solution

### Related Formula textIE propto frac1textStability of electronic configuration ### Core Logic Let's examine the configurations: - For Mn^2+, the electronic configuration is [Ar]3d^5, which features a highly stable, symmetric half-filled d-subshell. - For Fe^2+, the configuration is [Ar]3d^6. Because of the extra exchange energy and stability of the half-filled 3d^5 state, it is harder to remove an electron from Mn^2+ than from Fe^2+. Therefore, the ionisation enthalpy of Mn^2+ is greater than that of Fe^2+: textIE(Mn^2+) > textIE(Fe^2+) Hence, the expression Mn^2+ < Fe^2+ is incorrect. ### Pattern Recognition Whenever you see manganese (Mn) in the +2 oxidation state, remember its exceptionally stable d^5 config. This creates anomalous spikes in successive ionisation energies compared to neighboring iron (Fe). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements
Q44 jee_main_2025_04_april_morning Magnetic Properties
Pair of transition metal ions having the same number of unpaired electrons is:
  • A. V^2+, Co^2+
  • B. Ti^2+, Co^2+
  • C. Fe^3+, Cr^2+
  • D. Ti^3+, Mn^2+

Solution

### 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 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- 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- ### Pattern Recognition D-orbital counts follow a predictable symmetry: a 3d^n system contains the same number of unpaired electrons as a 3d^10-n system under high-spin conditions. This explains why 3d^3 (V^2+) and 3d^7 (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
Q47 jee_main_2025_04_april_morning Chemical Properties of KMnO4
KMnO_4 acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y is ______.
Numerical Answer. Answer: 10 to 10

Solution

### Core Logic Let's resolve both components step by step: 1. **Finding X:** In an acidic medium, the permanganate ion (KMnO_4, where Mn is in the +7 state) is reduced to the divalent manganese cation (Mn^2+, state +2): X = 7 - 2 = 5 2. **Finding Y:** During qualitative salt analysis, the acetate ion reacts with neutral ferric chloride to produce a characteristic blood-red coordination solution. Boiling this solution throws down a **brown-red precipitate** of basic ferric acetate, [Fe(OH)_2(CH_3COO)]. In this complex, Iron retains its +3 oxidation state: Fe^3+ implies [Ar] 3d^5 4s^0 implies textNumber of d-electrons (Y) = 5 Summing the values yields: X + Y = 5 + 5 = 10 ### Pattern Recognition This problem elegantly links standard redox transitions with qualitative inorganic salt tests. Remember that throughout the basic ferric acetate precipitation test, Iron remains steadily in its ferric +3 (d^5) core configuration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements Inorganic Qualitative Analysis
Q35 jee_main_2025_24_jan_evening Magnetic Properties of Transition Metals
Match List-I with List-II.
List-I (Transition metal ion)List-II (Spin only magnetic moment (B.M.))
(A) mathrmTi^3+(I) 3.87
(B) mathrmV^2+(II) 0.00
(C) mathrmNi^2+(III) 1.73
(D) mathrmSc^3+(IV) 2.84
Choose the correct answer from the options given below :
  • A. \text{(A)-(III), (B)-(I), (C)-(II), (D)-(IV)}
  • B. \text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}
  • C. \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
  • D. \text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}

Solution

### Related Formula mu = sqrtn(n+2) text B.M. where n represents the number of unpaired electrons. ### Core Logic Let's calculate the number of unpaired d-electrons (n) and the resulting spin-only magnetic moment for each transition metal ion: * (A) mathrmTi^3+: Electronic configuration = [Ar] 3d^1 ightarrow n = 1 mu = sqrt1(1+2) = sqrt3 approx 1.73text B.M. ightarrow text(III) * (B) mathrmV^2+: Electronic configuration = [Ar] 3d^3 ightarrow n = 3 mu = sqrt3(3+2) = sqrt15 approx 3.87text B.M. ightarrow text(I) * (C) mathrmNi^2+: Electronic configuration = [Ar] 3d^8. The 3d subshell has 3 paired orbitals and 2 unpaired orbitals ightarrow n = 2 mu = sqrt2(2+2) = sqrt8 approx 2.84text B.M. ightarrow text(IV) * (D) mathrmSc^3+: Electronic configuration = [Ar] 3d^0 ightarrow n = 0 mu = 0.00text B.M. ightarrow text(II) Matching these values yields the sequence: (A)-(III), (B)-(I), (C)-(IV), (D)-(II). ### Pattern Recognition Shortcut: The digit before the decimal point in a spin-only magnetic moment matches the number of unpaired electrons (n). For example, a value of 3.87text B.M. means there are exactly 3 unpaired electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements
Q31 jee_main_2025_24_jan_morning Lanthanoids Oxidation States
Which of the following ions is the strongest oxidizing agent? [Atomic Number of Ce=58, Eu=63, Tb=65, Lu=71]
  • A. Lu^3+
  • B. Eu^2+
  • C. Tb^4+
  • D. Ce^3+

Solution

### Core Logic The most common and chemically robust oxidation state for lanthanoid elements is +3. Consequently, ions existing in unstable +4 oxidation states exhibit a pronounced thermodynamic driving force to capture electrons and revert to the +3 form. Among the options, Tb^4+ acts as a potent oxidizing agent due to this stability drive. ### Pattern Recognition Ln^4+ forms naturally act as electron grabbers to sink back into the thermodynamic sweet spot of +3 states. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d-and f-Block Elements

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