Q.In a metallic oxide, oxide ions are arranged in cubic close packing. One sixth of the tetrahedral voids are occupied by cations P and one third of octahedral voids are occupied by the cation Q. Deduce the formula of the compound.

For an fcc lattice, $r = \dfrac{\sqrt{2}}{4}a$; with $a = 495$ pm, $r \approx 175.0$ pm.

In a face-centred cubic (fcc) unit cell, atoms touch each other along the face diagonal. The relation between the edge length $a$ and atomic radius $r$ is:

$4r = \sqrt{2}\,a \Rightarrow r = \dfrac{\sqrt{2}}{4}a$ …

FCC unit cell: $8\times\frac18 + 6\times\frac12 = 4$ particles.

In an FCC unit cell, particles sit at the 8 corners and the centres of the 6 faces.

Contribution of corner atoms $= 8 \times \dfrac{1}{8} = 1$

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AgBr is the classic example of an ionic solid that shows both Schottky and Frenkel defects.

A Schottky defect occurs when a cation and an anion of similar size leave their lattice sites together, creating a pair of vacancies (seen in NaCl, KCl, CsCl). A Frenkel defect occurs when a smaller ion (usually the cation) leaves its normal site and squeezes into an interstitial site, leaving a vacancy behind (seen in AgCl, AgBr, AgI, ZnS) — here there is a large difference between cation and anion size.

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In a body-centred cubic lattice, atoms touch along the body diagonal, giving $4r=\sqrt3\,a$, i.e. $r=(\sqrt3/4)a$.

In a bcc unit cell, the corner atoms and the body-centre atom touch along the body diagonal of the cube. The body diagonal has length $\sqrt3\,a$ (where $a$ is the edge length), and along this diagonal there are 4 atomic radii in contact (corner atom + centre atom + centre …

Counting each corner atom as 1/8 (shared by 8 cells) and each face-centred atom as 1/2 (shared by 2 cells) gives $4$ atoms per f.c.c. unit cell.

Counting contributions in an f.c.c. lattice

An f.c.c. (cubic close-packed) unit cell has atoms at the 8 corners of the cube AND at the centre of each of the 6 faces.

• Corner atoms: each of the 8 corners is shared simultaneously among 8 adjoining unit cells, so each corner atom contributes only $\frac{1}{8}$ to this cell: $8 \times \frac{1}{8} = 1$ atom.
• Face-centred atoms: each of the 6 face centres is shared between exactly 2 adjoining unit cells (the cell on each side of that face), so each contributes $\frac{1}{2}$: $6 \times \frac{1}{2} = 3$ atoms. …

The coordination number of a body-centred cubic lattice is 8.

In a body-centred cubic (bcc) unit cell, the atom at the body centre is in direct contact with the 8 atoms located at the corners of the cube (each corner atom is equidistant from the centre atom, along the body diagonal). Correspondingly, each corner atom also touches 8 …

The stem gives N as the number of UNIT CELLS. Each ccp/fcc unit cell contains 8 tetrahedral voids, so the total number in N unit cells is 8N. Answer: (d).

A bcc unit cell has 8 corner atoms (1/8 share each) and 1 body-centre atom (full share), giving 2 effective atoms per cell.

Counting the atoms

In a body-centred cubic (bcc) lattice, atoms sit at the 8 corners of the cube and one additional atom sits at the centre of the cube.

• Each corner atom is shared among 8 adjacent unit cells, so it contributes $\dfrac{1}{8}$ atom to this cell: $8 \times \dfrac{1}{8} = 1$ atom.
• The body-centre atom lies wholly inside this one unit cell, so it contributes a full atom: $1$ atom.

$\text{Total} = 1 + 1 = 2 \text{ atoms per unit cell}$

Why the other options are wrong …

A space lattice is the abstract geometric skeleton of a crystal — an ordered, repeating array of points, each standing for one constituent particle, all with identical surroundings.

Defining a space lattice

In a crystalline solid, the constituent particles (atoms, ions, or molecules) are arranged in a highly ordered, repetitive pattern extending in all three dimensions. If each particle in this arrangement is represented by a single point, the resulting three-dimensional array of points is called a space lattice (or crystal lattice).

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A body-centred cubic unit cell contains 2 atoms.

In a bcc lattice, atoms sit at the 8 corners and 1 at the body centre. Each corner atom is shared among 8 neighbouring unit cells, so it contributes 1/8 to a given cell; the body-centre atom belongs entirely …

Face-centred A (3 per cell) and corner B (1 per cell) give the ratio $A_3B$.

In an FCC lattice, atoms sit at the 8 corners and the centres of the 6 faces of the cube. Corner atoms are shared among 8 unit cells, so each corner contributes $\frac{1}{8}$ atom per cell; face-centred atoms are shared between 2 unit cells, so each contributes $\frac{1}{2}$ atom per cell.

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Counting shared corner and face contributions in an FCC unit cell gives 4 lattice points/atoms per cell.

In a face-centred cubic (FCC) lattice, particles occupy the 8 corners and the centres of all 6 faces of the cube.

• Each of the 8 corner particles is shared among 8 adjacent unit cells, so each contributes $\dfrac{1}{8}$ to this cell: $8 \times \dfrac{1}{8} = 1$ …