Maharashtra State Board · Class 12 · Chemistry · Chapter 1

Solid State — Formula Sheet

Board Formulas

13 formulas

1.Contribution of Particles to a Cubic Unit Cell
$z = \frac{N _{c}}{8} + \frac{N _{f}}{2} + \frac{N _{e}}{4} + N_{b}$
$z$ : Number of particles (atoms or ions) per unit cell · $N_{c}$ : Particles at the corners · $N_{f}$ : Particles at face centres · $N_{e}$ : Particles at edge centres · $N_{b}$ : Particles at the body centre
A corner particle is shared by 8 unit cells, a face-centre particle by 2 and an edge-centre particle by 4; a body-centre particle belongs wholly to one cell. z is a pure number.
2.Particles per Unit Cell — sc, bcc, fcc★
$z_{sc} = 8 \times \frac{1}{8} = 1, z_{b cc} = 8 \times \frac{1}{8} + 1 = 2, z_{f cc} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$
These three numbers start every density and packing numerical. Coordination numbers: simple cubic 6, bcc 8, fcc (ccp) and hcp 12.
3.Radius–Edge Relation: Simple Cubic
$a = 2 r \Rightarrow r = \frac{a}{2}$
$a$ : Edge length of the unit cell (pm) · $r$ : Atomic radius (pm)
Neighbouring atoms touch along the cube edge. Nearest-neighbour distance = a.
4.Radius–Edge Relation: Body-Centred Cubic
$3 a = 4 r \Rightarrow r = \frac{3}{4} a = 0.433 a$
Atoms touch along the body diagonal, not along the edge. Nearest-neighbour distance = 2r = (√3/2)a.
5.Radius–Edge Relation: Face-Centred Cubic (ccp)
$2 a = 4 r \Rightarrow r = \frac{a}{2 2} = 0.3536 a$
Atoms touch along the face diagonal. Nearest-neighbour distance = 2r = a/√2.
6.Packing Efficiency (general)★
$Packing efficiency = \frac{z \times \frac{4}{3} π r ^{3}}{a ^{3}} \times 100%$
$z$ : Number of atoms per unit cell · $r$ : Atomic radius · $a^{3}$ : Volume of the cubic unit cell
Percentage of the unit-cell volume occupied by the spheres. Substitute the r–a relation of the lattice so that r cancels. Empty space = 100% − packing efficiency.
7.Packing Efficiency: Simple Cubic
$\frac{1 \times \frac{4}{3} π r ^{3}}{( 2 r ) ^{3}} \times 100 = \frac{π}{6} \times 100 = 52.36%$
The least efficient cubic packing: 47.64% of the space is empty.
8.Packing Efficiency: Body-Centred Cubic
$\frac{2 \times \frac{4}{3} π r ^{3}}{( \frac{4 r}{3} ) ^{3}} \times 100 = \frac{3 π}{8} \times 100 = 68%$
32% of the space in a bcc crystal is empty.
9.Packing Efficiency: Face-Centred Cubic (ccp) and hcp★
$\frac{4 \times \frac{4}{3} π r ^{3}}{( 2 2 r ) ^{3}} \times 100 = \frac{π}{3 2} \times 100 = 74%$
The most efficient packing of identical spheres; hcp has the same 74%. Only 26% of the space is empty.
10.Density of a Cubic Crystal★
$ρ = \frac{z M}{a ^{3} N _{A}}$
$ρ$ : Density (g cm⁻³) · $z$ : Atoms per unit cell (1, 2 or 4) · $M$ : Molar mass (g mol⁻¹) · $a$ : Edge length (cm); 1 pm = 10⁻¹⁰ cm · $N_{A}$ : Avogadro number = 6.022 × 10²³ mol⁻¹
z is the number of atoms per unit cell (the Maharashtra textbook writes n). With M in g mol⁻¹, a in cm and N_A = 6.022 × 10²³ mol⁻¹, ρ is in g cm⁻³. Rearrange the same equation to find z, a or M.
11.Number of Unit Cells and Atoms in x g of a Metal
$Unit cells = \frac{x}{ρ a ^{3}}, Atoms = \frac{z x}{ρ a ^{3}} = \frac{x N _{A}}{M}$
$x$ : Mass of the metal sample (g) · $ρ a^{3}$ : Mass of one unit cell (g)
ρa³ is the mass of one unit cell. The number of unit cells in a volume V of crystal is V/a³.
12.Voids in Close Packing
$N_{tetrahedral} = 2 N, N_{octahedral} = N$
For N spheres in ccp or hcp. A tetrahedral void is surrounded by 4 spheres, an octahedral void by 6. To find a formula, count ions in the lattice and in the fraction of voids occupied, then take the simplest whole-number ratio.
13.Point Defects and Density
$Schottky: ρ ↓ Frenkel: ρ unchanged$
Schottky: equal numbers of cations and anions missing (NaCl, KCl) — density decreases. Frenkel: the smaller ion, usually the cation, moves to an interstitial site (ZnS, AgCl) — density unchanged. Both keep the crystal electrically neutral. AgBr shows both.