Q.Are the outermost electrons of carbons in diamond localized or delocalized?

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:

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

Given $a = 495$ pm:

$r = \dfrac{1.4142}{4} \times 495 = 0.35355 \times 495 \approx 175.0\ pm$

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 $= 8 \times \dfrac{1}{8} = 1$

Contribution of face-centred atoms $= 6 \times \dfrac{1}{2} = 3$

Total particles per unit cell $= 1 + 3 = 4$.

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.

In AgBr, the Ag$^+$ ion is small enough relative to Br$^-$ to migrate to an interstitial position (giving a Frenkel defect), and AgBr is also cited in NCERT-level chemistry as the standard example that exhibits both Schottky and Frenkel defects simultaneously, unlike NaCl (Schottky only) or ZnS (Frenkel only, due to large size difference) or CsCl (Schottky only, similar-sized ions).

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 atom + corner atom, i.e. $r + 2r + r = 4r$):

$4r = \sqrt3\,a \quad\Rightarrow\quad r = \dfrac{\sqrt3}{4}a$

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.

Total $= 1 + 3 = 4$ atoms per unit cell. This is why f.c.c. is also called cubic close packing (ccp) with a packing efficiency of 74%, and why metals like $Cu$, $Ag$, $Au$, $Al$, and $Ni$ (which crystallise in this structure) use $Z = 4$ directly in the density formula $\rho = \dfrac{ZM}{N_A a^3}$.

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 body-centre atoms of the surrounding cells. Hence the coordination number of bcc is 8 (compare: simple cubic = 6, fcc/hcp = 12).

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) 1 would be true only for a simple cubic cell (8 corners $\times$ 1/8 = 1, no body-centre atom).

• (c) 3 does not correspond to any standard cubic cell — it isn't obtained by any correct fractional count.

• (d) 4 is the count for a face-centred cubic (fcc) cell (8 corners $\times$ 1/8 + 6 face atoms $\times$ 1/2 = 1 + 3 = 4), not bcc.

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).

The defining property of a space lattice is that every lattice point has exactly the same environment — the same number, kind, and arrangement of neighbouring points at the same distances and orientations — as every other lattice point. Joining the nearest lattice points by straight lines reveals the geometry of the lattice, and the smallest repeating unit of this pattern (which, when repeated in 3-D, generates the entire lattice) is called the unit cell.

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 to that one cell. Total contribution:

$Z = 8 \times \tfrac{1}{8} + 1 \times 1 = 1 + 1 = 2$

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.

Number of B ions (at the 8 corners) per unit cell $= 8 \times \frac{1}{8} = 1$.

Number of A ions (at the 6 face centres) per unit cell $= 6 \times \frac{1}{2} = 3$.

So the ratio A : B $= 3 : 1$, giving the formula $A_3B$.

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: $8 \times \dfrac{1}{8} = 1$
• Each of the 6 face-centred particles is shared between 2 adjacent unit cells, so each contributes $\dfrac{1}{2}$: $6 \times \dfrac{1}{2} = 3$

Total number of particles per FCC unit cell $= 1 + 3 = 4$