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A spinel is an important class of oxid...

A spinel is an important class of oxide consisting two types of metal ions with the oxide ions arranged in ccp layers . The normal spinel has one -eight of the tetrahedral holes occupied by one type of metal ions and one- half of the octaherdral holes occupied by another type of metal ion. Such a spine is formed by `Mg^(2+), Al^(3+) and O^(2-)` . The neutrality of the crystal is being maintained.
Type of hole occupied by `Mg^(2+)` ions is:

A

tetrahedral

B

octahedral

C

both (a) amd(b)

D

None of these

Text Solution

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The correct Answer is:
To solve the question regarding the type of hole occupied by `Mg^(2+)` ions in a spinel structure formed by `Mg^(2+)`, `Al^(3+)`, and `O^(2-)`, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Structure**: - In a spinel structure, the oxide ions (`O^(2-)`) are arranged in a cubic close-packed (CCP) or face-centered cubic (FCC) arrangement. 2. **Calculating the Number of Oxide Ions**: - In an FCC arrangement, there are 4 oxide ions per unit cell. - Contribution from corners: 8 corners × (1/8 contribution per corner) = 1 - Contribution from faces: 6 faces × (1/2 contribution per face) = 3 - Total = 1 + 3 = 4 oxide ions. 3. **Calculating Tetrahedral and Octahedral Holes**: - The number of tetrahedral holes in a CCP structure is given by \(2 \times \text{number of atoms}\). - Thus, number of tetrahedral holes = \(2 \times 4 = 8\). - The number of octahedral holes is equal to the number of atoms in the CCP structure. - Thus, number of octahedral holes = 4. 4. **Occupancy of Holes**: - According to the problem, `1/8` of the tetrahedral holes are occupied by one type of metal ion, and `1/2` of the octahedral holes are occupied by another type of metal ion. - Therefore, the number of tetrahedral holes occupied by one type of metal ion = \(1/8 \times 8 = 1\) (for `Mg^(2+)`). - The number of octahedral holes occupied = \(1/2 \times 4 = 2\) (for `Al^(3+)`). 5. **Charge Neutrality**: - The total negative charge from the oxide ions is \(4 \times (-2) = -8\). - To maintain charge neutrality: - Let’s assume `Mg^(2+)` occupies the tetrahedral holes and `Al^(3+)` occupies the octahedral holes. - Charge from `Mg^(2+)`: \(1 \times (+2) = +2\). - Charge from `Al^(3+)`: \(2 \times (+3) = +6\). - Total positive charge = \(+2 + +6 = +8\). - This balances the total negative charge of \(-8\). 6. **Conclusion**: - Since `Mg^(2+)` occupies the tetrahedral holes and `Al^(3+)` occupies the octahedral holes, the type of hole occupied by `Mg^(2+)` ions is **tetrahedral**. ### Final Answer: The type of hole occupied by `Mg^(2+)` ions is **tetrahedral**. ---
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