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Identify correct statements from below: A. The chromate ion is square planar. B. Dichromates are generally prepared from chromates. C. The green manganate ion is diamagnetic. D. Dark green coloured K_2MnO_4 disproportionates in a neutral or acidic medium to give permanganate. E. With increasing oxidation number of transition metal, ionic character of the oxides decreases. Choose the correct answer from the options given below:

Solution & Explanation

### Step 1: Statement A Analysis CrO_4^2- (chromate ion) is tetrahedral, not square planar. Statement A is incorrect. ### Step 2: Statement B Analysis 2Na_2CrO_4 + 2H^+ rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O. Dichromates are indeed prepared from chromates. Statement B is correct. ### Step 3: Statement C Analysis The green manganate ion (MnO_4^2-) has manganese in the +6 oxidation state (3d^1). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect. ### Step 4: Statement D Analysis Dark green coloured K_2MnO_4 undergoes disproportionation in neutral or acidic media to yield permanganate (MnO_4^-) and manganese dioxide (MnO_2). Statement D is correct. ### Step 5: Statement E Analysis Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct. ### Final Conclusion The correct statements are B, D, and E. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

Reference Study Guides

The D And F Block Elements

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

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
Q86 jee_main_2024_31_jan_evening Chromyl Chloride Test
In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+) ________
Numerical Answer. Answer: 6 to 6

Solution

### Related Formula K_2Cr_2O_7(s) + 4KCl(s) + 6H_2SO_4(conc.) rightarrow 2CrO_2Cl_2(g) + 6KHSO_4 + 3H_2O ### Core Logic This reaction represents the Chromyl Chloride test used to detect the presence of chloride ions. When potassium dichromate is heated with a metal chloride in concentrated sulfuric acid, red vapors of chromyl chloride (CrO_2Cl_2) are evolved. ### Step 1: Oxidation State Calculation In chromyl chloride (CrO_2Cl_2): Let the oxidation state of Chromium be x. Oxygen is typically -2 and Chlorine is -1. x + 2(-2) + 2(-1) = 0 x - 4 - 2 = 0 x = +6 Thus, the oxidation state of Chromium in the product is 6. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements Class 11 Chemistry: Practical Chemistry

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