Exams › JEE Advanced › Chemistry

In a cubic close-packed (ccp) arrangement of oxygen atoms within a mineral's unit cell, aluminium ions occupy a fraction m of the octahedral voids, while magnesium ions occupy a fraction n of the tetrahedral voids. What are the values of m and n?

1. 1/2, 1/8
2. 1, 1/4
3. 1/2, 1/4
4. 1/4, 1/8

Correct answer: 1/2, 1/8

Solution

In a ccp arrangement, the number of octahedral voids equals the number of atoms, and the number of tetrahedral voids is twice that. Given the fractions m = 1/2 for aluminum and n = 1/8 for magnesium, this matches the occupancy of these voids.

Related JEE Advanced Chemistry questions

• Identify the accurate statement regarding lattice energy:
• An element forms a crystal with a face-centered cubic unit cell and an edge length of 200 pm. Calculate the density of the element if 200 g of it contains 24 × 10²³ atoms.
• Copper exhibits a face-centered cubic (fcc) arrangement. If the atomic radius of copper is 128 pm, what is the length of the unit cell edge?
• Sodium chloride forms cubic crystals when crystallized from water, but octahedral crystals when crystallized from a urea solution. How does this observation relate to the law of constancy of interfacial angles?
• If the edge length of a unit cell is 200 pm (200 × 10⁻¹² m), the unit cell's volume is calculated as (200 × 10⁻¹² m)³ = 8 × 10⁻³⁰ m³. For a face-centered cubic (fcc) structure, the number of atoms per unit cell (n) is 4. The mass of the unit cell is determined as n × A = 4 × 200 / 24 × 10²³ = 33.3 × 10⁻²³ g. Using the formula density = mass of unit cell / volume of unit cell, what is the density of the unit cell in g cm⁻³?
• In a face-centered cubic (fcc) lattice, the face diagonal equals four times the atomic radius. If the face diagonal of a copper crystal is given as 512 pm, what is the atomic radius?

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →