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A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :
  • a)
    C2A3
  • b)
    C3A2
  • c)
    C3A4
  • d)
    C4A3
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A compound is formed by cation C and anion A. The anions form hexagona...
Anions(A) are in hcp, so number of anions (A) = 6
Cations(C) are in 75% O.V., so number of cations (C)

= 18/4
= 9/2
So formula of compound will be 

C9A12 ⇒ C3A4
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A compound is formed by cation C and anion A. The anions form hexagona...
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A compound is formed by cation C and anion A. The anions form hexagona...
Explanation:


To determine the formula of the compound, we need to analyze the arrangement of the cations and anions in the crystal lattice.

1. Given Information:

- Anions form a hexagonal close-packed (hcp) lattice.
- Cations occupy 75% of octahedral voids.

2. Hexagonal Close-Packed (HCP) Lattice:

In a hcp lattice, the anions are arranged in a close-packed manner with a hexagonal arrangement in one plane and the third layer directly above the first layer. Each anion is surrounded by six neighboring anions in a hexagonal arrangement.

3. Octahedral Voids:

Octahedral voids are the spaces between the anions in a hcp lattice. These voids are located above and below each hexagonal close-packed layer, forming a repeating pattern.

4. Cations in Octahedral Voids:

According to the given information, the cations occupy 75% of the octahedral voids. This means that for every four octahedral voids, three are occupied by cations.

5. Determining the Formula:

Based on the arrangement of cations and anions, we can determine the formula of the compound.

- Each cation occupies 75% of the octahedral voids, so the ratio of cations to octahedral voids is 3:4.
- Each octahedral void is surrounded by six anions.
- In the formula, the ratio of cations to anions should be such that the charges balance out.

Considering the information above, the formula of the compound is C3A4. This means that there are three cations (C) for every four anions (A), ensuring charge neutrality in the compound.

Therefore, the correct answer is option 'C' (C3A4).
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Read the passage given below and answer the following questions:The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of up to some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since d-electron bonding levels are involved, the cations-exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group.In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage.Assertion (

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A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :a)C2A3b)C3A2c)C3A4d)C4A3Correct answer is option 'C'. Can you explain this answer?
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