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 Type:
Enter a numerical value Answer: 10 to 10 +4 marks

Solution & Explanation

### 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

Reference Study Guides

The D And F Block Elements

More The d- and f-Block Elements Previous-Year Questions — Page 7

Q71 jee_main_2024_30_january_evening Preparation and Properties of KMnO4
Alkaline oxidative fusion of MnO_2 gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:
  • A. textMn_2textO_7 text and MnO_4^-
  • B. textMnO_4^2- text and MnO_4^-
  • C. textMn_2textO_3 text and MnO_4^2-
  • D. textMnO_4^2- text and Mn_2textO_7

Solution

### Core Logic Step 1: Alkaline oxidative fusion of MnO_2 (pyrolusite ore) with KOH in the presence of O_2 (or an oxidizing agent like KNO_3) yields the green-colored manganate ion (MnO_4^2-). 2mathrmMnO_2 + 4mathrmOH^- + mathrmO_2 rightarrow 2mathrmMnO_4^2- + 2mathrmH_2mathrmO So, A is mathrmMnO_4^2-. Step 2: Electrolytic oxidation of the manganate ion (MnO_4^2-) in an alkaline medium converts it to the purple-colored permanganate ion (MnO_4^-). mathrmMnO_4^2- rightarrow mathrmMnO_4^- + mathrme^- So, B is mathrmMnO_4^-. ### Pattern Recognition Industrial preparation sequence of KMnO_4: MnO_2 xrightarrowtextfusion, KOH, O_2 MnO_4^2- text (green) xrightarrowtextelectrolytic oxidation MnO_4^- text (purple). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements
Q78 jee_main_2024_30_january_evening Compounds of Transition Elements
A and B formed in the following reactions are: mathrmCrO_2mathrmCl_2 + 4mathrmNaOH rightarrow mathrmA + 2mathrmNaCl + 2mathrmH_2mathrmO mathrmA + 2mathrmHCl + 2mathrmH_2mathrmO_2 rightarrow mathrmB + 3mathrmH_2mathrmO
  • A. textA = Na_2textCrO_4text, B = CrO_5
  • B. textA = Na_2textCr_2textO_4text, B = CrO_4
  • C. textA = Na_2textCr_2textO_7text, B = CrO_3
  • D. textA = Na_2textCr_2textO_7text, B = CrO_5

Solution

### Core Logic Step 1: Chromyl chloride (CrO_2Cl_2) reacts with an alkali like NaOH to give a yellow solution of sodium chromate (Na_2CrO_4). CrO_2Cl_2 + 4NaOH rightarrow Na_2CrO_4 (A) + 2NaCl + 2H_2O Step 2: Sodium chromate (Na_2CrO_4) reacts with hydrogen peroxide (H_2O_2) in an acidic medium (HCl) to yield the deep blue colored chromium pentoxide (CrO_5, also known as chromium(VI) oxide peroxide). Na_2CrO_4 + 2H_2O_2 + 2HCl rightarrow CrO_5 (B) + 2NaCl + 3H_2O Note: NaCl formation implies the overall balanced reaction uses the acid for neutralization/salt formation. ### Pattern Recognition Chromyl chloride test intermediate: Yellow solution = Na_2CrO_4. Reaction of chromate with H_2O_2 in acid = Blue peroxide CrO_5 (butterfly structure). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements Class 11 Chemistry: Redox Reactions
Q66 jee_main_2024_30_jan_morning Lanthanoids
Diamagnetic Lanthanoid ions are:
  • A. Nd^3+text and Eu^3+
  • B. La^3+text and Ce^4+
  • C. Nd^3+text and Ce^4+
  • D. Lu^3+text and Eu^3+

Solution

### Core Logic An ion is diamagnetic if all its electrons are paired (i.e., zero unpaired electrons). Let's write the electronic configuration for the elements in question. ### Step 1: Checking configurations Cerium (Ce, Z=58): [Xe] 4f^1 5d^1 6s^2 rightarrow Ce^4+: [Xe] 4f^0 (0 unpaired electrons rightarrow Diamagnetic) Lanthanum (La, Z=57): [Xe] 4f^0 5d^1 6s^2 rightarrow La^3+: [Xe] 4f^0 (0 unpaired electrons rightarrow Diamagnetic) ### Pattern Recognition Ions with an empty f-subshell (f^0, e.g., La^3+, Ce^4+) or a completely filled f-subshell (f^14, e.g., Lu^3+, Yb^2+) are invariably diamagnetic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements
Q76 jee_main_2024_30_jan_morning Transition Elements
Match List-I with List-II.
List-I (Species)List-II (Electronic distribution)
(A) Cr^+2(I) 3d^8
(B) Mn^+(II) 3d^54s^1
(C) Ni^+2(III) 3d^4
(D) V^+(IV) 3d^34s^1
Choose the correct answer from the options given below:
  • A. text(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  • B. text(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  • C. text(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  • D. text(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Solution

### Core Logic Let's determine the electronic configuration for each species by first writing the neutral atom's configuration, and then removing electrons starting from the outermost 4s orbital. (A) Cr (Z=24): [Ar] 3d^5 4s^1 rightarrow Cr^2+: [Ar] 3d^4 (B) Mn (Z=25): [Ar] 3d^5 4s^2 rightarrow Mn^+: [Ar] 3d^5 4s^1 (C) Ni (Z=28): [Ar] 3d^8 4s^2 rightarrow Ni^2+: [Ar] 3d^8 (D) V (Z=23): [Ar] 3d^3 4s^2 rightarrow V^+: [Ar] 3d^3 4s^1 ### Step 1: Match execution A rightarrow III B rightarrow II C rightarrow I D rightarrow IV ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements
Q78 jee_main_2024_31_jan_evening Properties of Transition Metal Oxides
Choose the correct statements from the following A. Mn_2O_7 is an oil at room temperature B. V_2O_4 reacts with acid to give VO_2^2+ C. CrO is a basic oxide D. V_2O_5 does not react with acid Choose the correct answer from the options given below:
  • A. text(1) A, B and D only
  • B. text(2) A and C only
  • C. text(3) A, B and C only
  • D. text(4) B and C only

Solution

### Core Logic (A) Mn_2O_7 is a covalent oxide and exists as a green oil at room temperature. (Correct) (B) V_2O_4 dissolves in acids to give VO^2+ (vanadyl) salts, not VO_2^2+. (Incorrect) (C) CrO has chromium in the +2 oxidation state. Lower oxidation state metal oxides are typically basic in nature. (Correct) (D) V_2O_5 is an amphoteric oxide; it reacts with both acids as well as bases. (Incorrect) ### Step 1: Final Selection Only statements A and C are correct, which corresponds to option (2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

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