Chapter Exam

Unit 11: d- and f-Block Elements — Online MCQ Test

Q: The calculated spin only magnetic moment of Cr 2+ ion is:

Answer: A. 4.90 BM — Cr atom is [Ar] 3d5 4s1. Cr 2+ is [Ar] 3d4. It has 4 unpaired electrons. Magnetic moment = sqrt(n(n+2)) = sqrt(4(6)) = sqrt(24) ≈ 4.90 BM.

Q: Calculated spin-only magnetic moment of Cr2+ ion is:

Answer: A. 4.90 BM — Cr2+ has a d4 configuration. The number of unpaired electrons (n) = 4. Magnetic moment = sqrt(n(n+2)) = sqrt(4(6)) = sqrt(24) ≈ 4.90 BM.

Q: Given below are two statements: Statement I: Ferromagnetism is considered as an extreme form of paramagnetism. Statement II: The number of unpaired electrons in a Cr²⁺ ion (Z=24) is the same as that of a Nd³⁺ ion (Z=60).

Answer: C. Statement I is true but Statement II is false — Cr²⁺: [Ar] 3d⁴ (4 unpaired). Nd³⁺: [Xe] 4f³ (3 unpaired). Statement II is false.

Q: The electronic configurations of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are:

Answer: C. [Xe]4f7 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f9 6s2 — Eu (63) is [Xe]4f7 6s2 (half-filled f-subshell is stable), Gd (64) is [Xe]4f7 5d1 6s2 (extra stability of half-filled f-subshell), and Tb (65) is [Xe]4f9 6s2.

Q: Match items of Column I with the items of Column II and assign the correct code: Column I Column II (a) Cyanide process (i) Ultrapure Ge (b) Froth floatation process (ii) Dressing of Zns (c) Electrolytic reduction (iii) Extraction of Al (d) Zone refining (iv) Extraction of Au (v) Purification of Ni

Answer: D. (4) (iv) (ii) (iii) (i) — Cyanide process is used for Au extraction (iv), Froth floatation for ZnS (ii), Electrolytic reduction for Al (iii), and Zone refining for high-purity Ge (i).

Q: The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of

Answer: A. d4 to d5 configuration — Mn2+ has a particularly stable half-filled d5 configuration. Reduction of Mn3+ to Mn2+ gives d4 to d5, which is highly favorable, making E° more positive.

Q: The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of

Answer: A. d4 to d5 configuration — Mn2+ has a particularly stable half-filled d5 configuration. The reduction Mn3+ → Mn2+ changes the configuration from d4 to d5, making the E° value more positive.

Q: ‘Spin only’ magnetic moment is same for which of the following ions? A. Ti3+ B. Cr2+ C. Mn2+ D. Fe2+ E. Sc3+ Choose the most appropriate answer from the options given below:

Answer: B. A and D only — Ti3+ is d1 and Fe2+ is d6; both have four unpaired electrons? No. The intended pair with same spin-only moment is Ti3+ (d1) and Sc3+ (d0) not same. However among the options, A and D both correspond to different unpaired counts. Re-evaluating: Cr2+ is d4 (4 unpaired), Mn2+ is d5 (5 unpaired), Fe2+ is d6 (4 unpaired), Ti3+ is d1 (1 unpaired), Sc3+ is d0. Thus B and D both have 4 unpaired electrons and same spin-only magnetic moment. The correct option is B and D only.

Q: Which one is malachite from the following?

Answer: A. CuCO3.Cu(OH)2 — Malachite is a copper carbonate hydroxide mineral with the chemical formula Cu2CO3(OH)2, commonly written as CuCO3.Cu(OH)2.

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Unit 11: d- and f-Block Elements — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 11: d- and f-Block Elements, with answers and explanations. These are sample questions; the exam has its own separate question set.

The calculated spin only magnetic moment of Cr 2+ ion is:

A. 4.90 BM ✓
B. 5.92 BM
C. 2.84 BM
D. 3.87 BM

Answer: A. 4.90 BM
Cr atom is [Ar] 3d5 4s1. Cr 2+ is [Ar] 3d4. It has 4 unpaired electrons. Magnetic moment = sqrt(n(n+2)) = sqrt(4(6)) = sqrt(24) ≈ 4.90 BM.

Calculated spin-only magnetic moment of Cr2+ ion is:

A. 4.90 BM ✓
B. 5.92 BM
C. 2.84 BM
D. 3.87 BM

Answer: A. 4.90 BM
Cr2+ has a d4 configuration. The number of unpaired electrons (n) = 4. Magnetic moment = sqrt(n(n+2)) = sqrt(4(6)) = sqrt(24) ≈ 4.90 BM.

Given below are two statements: Statement I: Ferromagnetism is considered as an extreme form of paramagnetism. Statement II: The number of unpaired electrons in a Cr²⁺ ion (Z=24) is the same as that of a Nd³⁺ ion (Z=60).

A. Both Statement I and Statement II are true
B. Both Statement I and Statement II are false
C. Statement I is true but Statement II is false ✓
D. Statement I is false but Statement II is true

Answer: C. Statement I is true but Statement II is false
Cr²⁺: [Ar] 3d⁴ (4 unpaired). Nd³⁺: [Xe] 4f³ (3 unpaired). Statement II is false.

The electronic configurations of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are:

A. [Xe]4f6 5d1 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f9 6s2
B. [Xe]4f6 5d1 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f8 5d1 6s2
C. [Xe]4f7 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f9 6s2 ✓
D. [Xe]4f7 6s2, [Xe]4f8 6s2 and [Xe]4f8 5d1 6s2

Answer: C. [Xe]4f7 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f9 6s2
Eu (63) is [Xe]4f7 6s2 (half-filled f-subshell is stable), Gd (64) is [Xe]4f7 5d1 6s2 (extra stability of half-filled f-subshell), and Tb (65) is [Xe]4f9 6s2.

Match items of Column I with the items of Column II and assign the correct code: Column I Column II (a) Cyanide process (i) Ultrapure Ge (b) Froth floatation process (ii) Dressing of Zns (c) Electrolytic reduction (iii) Extraction of Al (d) Zone refining (iv) Extraction of Au (v) Purification of Ni

A. (1) (ii) (iii) (i) (v)
B. (2) (i) (ii) (iii) (iv)
C. (3) (iii) (iv) (v) (i)
D. (4) (iv) (ii) (iii) (i) ✓

Answer: D. (4) (iv) (ii) (iii) (i)
Cyanide process is used for Au extraction (iv), Froth floatation for ZnS (ii), Electrolytic reduction for Al (iii), and Zone refining for high-purity Ge (i).

The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of

A. d4 to d5 configuration ✓
B. d3 to d5 configuration
C. d5 to d4 configuration
D. d5 to d2 configuration

Answer: A. d4 to d5 configuration
Mn2+ has a particularly stable half-filled d5 configuration. Reduction of Mn3+ to Mn2+ gives d4 to d5, which is highly favorable, making E° more positive.

The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of

A. d4 to d5 configuration ✓
B. d3 to d5 configuration
C. d5 to d4 configuration
D. d5 to d2 configuration

Answer: A. d4 to d5 configuration
Mn2+ has a particularly stable half-filled d5 configuration. The reduction Mn3+ → Mn2+ changes the configuration from d4 to d5, making the E° value more positive.

‘Spin only’ magnetic moment is same for which of the following ions? A. Ti3+ B. Cr2+ C. Mn2+ D. Fe2+ E. Sc3+ Choose the most appropriate answer from the options given below:

A. B and C only
B. A and D only ✓
C. B and D only
D. A and E only

Answer: B. A and D only
Ti3+ is d1 and Fe2+ is d6; both have four unpaired electrons? No. The intended pair with same spin-only moment is Ti3+ (d1) and Sc3+ (d0) not same. However among the options, A and D both correspond to different unpaired counts. Re-evaluating: Cr2+ is d4 (4 unpaired), Mn2+ is d5 (5 unpaired), Fe2+ is d6 (4 unpaired), Ti3+ is d1 (1 unpaired), Sc3+ is d0. Thus B and D both have 4 unpaired electrons and same spin-only magnetic moment. The correct option is B and D only.

Which one is malachite from the following?

A. CuCO3.Cu(OH)2 ✓
B. CuFeS2
C. Cu(OH)2
D. Fe3O4

Answer: A. CuCO3.Cu(OH)2
Malachite is a copper carbonate hydroxide mineral with the chemical formula Cu2CO3(OH)2, commonly written as CuCO3.Cu(OH)2.

Identify the correct statement from the following:

A. Vapour phase refining is carried out for Nickel by Van Arkel method.
B. Pig iron can be moulded into a variety of shapes. ✓
C. Wrought iron is impure iron with 4% carbon.
D. Blister copper has blistered appearance due to evolution of CO₂.

Answer: B. Pig iron can be moulded into a variety of shapes.
Cast iron (pig iron) can be moulded into a variety of shapes because it is brittle but can be cast. Van Arkel method is used for Zirconium/Titanium (not Nickel — Nickel is refined by Mond's process). Wrought iron is the purest form of iron with very little carbon (<0.1%). Blister copper's blistered appearance is due to evolution of SO₂, not CO₂.

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