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Interstitial compounds and alloys: NEET MCQs with answers

89 questions on this topic of d- and f-Block Elements (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Nichrome is a widely used heating element alloy. What are its principal metallic components?

  1. Cu, Ni, and Zn
  2. Ni, Cr, and Fe
  3. Fe, W, and Co
  4. Ni, Al, and Ti
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Answer: (B) Ni, Cr, and Fe

Nichrome is an alloy primarily composed of nickel (about 80%) and chromium (about 20%), frequently along with small amounts of iron.

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2. Given below are two statements: Statement I: Interstitial compounds are neither typically ionic nor covalent. Statement II: The melting points of interstitial compounds are typically higher than those of the pure parent metals. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
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Answer: (A) Both Statement I and Statement II are true.

According to NCERT, interstitial compounds do not exhibit standard ionic or covalent bonds; they retain metallic-type bonding with enhanced metal-nonmetal interactions. Because of these strong localized bonds, their melting points are higher than those of the pure transition metals.

hard

3. A non-stoichiometric interstitial hydride of scandium has the formula ScH₁.₅. Given the atomic masses Sc = 45 g mol⁻¹ and H = 1.0 g mol⁻¹, what is the percentage by mass of scandium in this compound?

  1. 95.3%
  2. 93.4%
  3. 98.2%
  4. 96.8%
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Answer: (D) 96.8%

Molar mass of ScH₁.₅ = 45 + (1.5 × 1.0) = 46.5 g mol⁻¹. The percentage by mass of Sc = (45 / 46.5) × 100% ≈ 96.77% ≈ 96.8%.

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4. Transition metals readily form alloys with each other because:

  1. They are all non-metals
  2. They do not form metallic bonds
  3. Their atomic radii are similar (within about 15 per cent)
  4. They have very different sizes
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Answer: (C) Their atomic radii are similar (within about 15 per cent)

Atoms of similar size can replace each other in a crystal lattice. Transition metals have similar radii, so they easily form alloys, which are often hard with high melting points.

NCERT Class 12

easy

5. Bronze is an important alloy widely used since ancient times. What are its primary metallic constituents?

  1. Copper and Tin
  2. Copper and Zinc
  3. Copper and Nickel
  4. Zinc and Tin
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Answer: (A) Copper and Tin

Bronze is an alloy composed primarily of copper and tin (Cu and Sn). In contrast, brass is an alloy of copper and zinc.

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6. Bell metal is a historic alloy composed of copper and tin. If a bell metal artifact consists of Cu and Sn in an 4:1 molar ratio, what is the approximate percentage by mass of tin in the artifact? (Atomic mass: Cu = 63.5 g mol⁻¹, Sn = 118.7 g mol⁻¹)

  1. 31.8%
  2. 21.8%
  3. 42.5%
  4. 15.2%
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Answer: (A) 31.8%

Mass of 4 moles of Cu = 4 × 63.5 = 254.0 g. Mass of 1 mole of Sn = 118.7 g. Total mass = 254.0 + 118.7 = 372.7 g. Percentage by mass of Sn = (118.7 / 372.7) × 100 ≈ 31.85% ≈ 31.8%.

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7. In a hexagonal close-packed (hcp) interstitial carbide of a transition metal M, carbon atoms occupy one-sixth of the total tetrahedral voids. What is the stoichiometric formula of the resulting interstitial carbide?

  1. MC₀.₂₅
  2. MC₀.₅
  3. MC₀.₃₃
  4. MC
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Answer: (C) MC₀.₃₃

In an hcp lattice of n metal atoms, there are 2n tetrahedral voids. If one-sixth of these tetrahedral voids are filled, the number of carbon atoms is (1/6) × 2n = n/3. Therefore, the ratio of M to C is n : n/3, which corresponds to MC₁/₃ or MC₀.₃₃.

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8. In a cubic close-packed (ccp) interstitial carbide of a transition metal M, the metal atoms M form the lattice points, and carbon atoms occupy 50% of the octahedral voids while all tetrahedral voids remain empty. What is the empirical formula of this interstitial carbide?

  1. MC₂
  2. M₄C
  3. MC
  4. M₂C
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Answer: (D) M₂C

In a ccp lattice, for n metal atoms, there are n octahedral voids. Since carbon occupies 50% of the octahedral voids, the number of C atoms = 0.5 × n. The atomic ratio M : C = n : 0.5n = 2 : 1. Hence, the empirical formula is M₂C.

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9. Which of the following statements is INCORRECT regarding interstitial compounds of transition metals?

  1. They retain metallic electrical conductivity.
  2. They possess melting points higher than those of pure transition metals.
  3. They are chemically more reactive than their parent metals.
  4. Some of their borides approach diamond in hardness.
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Answer: (C) They are chemically more reactive than their parent metals.

Interstitial compounds are chemically inert and show resistance to chemical attack (e.g., acids) compared to the parent pure transition metals. The remaining properties—high melting points, retaining metallic conductivity, and high hardness—are characteristic features described in NCERT.

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10. Which of the following physical or chemical characteristics is NOT typically exhibited by interstitial compounds of transition metals?

  1. They are chemically extremely reactive
  2. They retain metallic conductivity
  3. They possess high melting points, higher than those of pure metals
  4. They are very hard; some borides approach diamond in hardness
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Answer: (A) They are chemically extremely reactive

Interstitial compounds of transition metals are chemically inert, not reactive. They retain metallic conductivity, have melting points higher than the parent metals, and exhibit exceptional hardness.

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11. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Interstitial carbide TiC approaches diamond in hardness. Reason (R): Introduction of carbon into octahedral interstitial sites creates directional, highly rigid metal-carbon bonding that prevents lattice planes from sliding. In light of the above statements, choose the correct answer from the options given below:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

TiC is exceptionally hard and approaches diamond in hardness. This extreme hardness arises because the interstitial carbon atoms form strong, directional metal-carbon interactions within the interstitial cavities, effectively pinning dislocations and preventing slip between crystal planes.

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12. Interstitial compounds are usually non-stoichiometric. This means they:

  1. Do not have fixed whole-number ratios of their elements
  2. Contain no metal
  3. Are always ionic
  4. Have exact formulae like TiC₂ only
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Answer: (A) Do not have fixed whole-number ratios of their elements

Interstitial compounds such as TiC or VH₀.₅₆ are typically non-stoichiometric; neither typically ionic nor covalent.

NCERT Class 12

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