Is it correct to say: Counting every corner atom of a cube as a whole atom (z = 8 for simple cubic).?

Each corner atom is shared by 8 cells and contributes 1/8. Simple cubic z = 1, bcc z = 2, fcc z = 4.

Is it correct to say: Using a = 2r for bcc and fcc lattices.?

a = 2r only for simple cubic. In bcc atoms touch along the body diagonal (√3a = 4r); in fcc along the face diagonal (√2a = 4r).

Is it correct to say: Putting the edge length in pm into ρ = zM/(a³N_A) and expecting g cm⁻³.?

Convert first: 1 pm = 10⁻¹⁰ cm, so a³ in pm³ must be multiplied by 10⁻³⁰ to get cm³.

Is it correct to say: Taking the number of tetrahedral voids equal to the number of spheres.?

N close-packed spheres give 2N tetrahedral voids and N octahedral voids.

Is it correct to say: Saying that a Frenkel defect lowers the density.?

In a Frenkel defect ions only shift to interstitial sites, so mass and volume are unchanged. Only the Schottky defect (ions missing) lowers density.

📄 Formula sheet (PDF)

Solid State

Unit cells, particles per unit cell, radius–edge relations, packing efficiency, density of cubic crystals and crystal defects — Maharashtra HSC Chemistry Ch 1

📐 13 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Learn the three radius–edge relations (sc: a = 2r, bcc: √3a = 4r, fcc: √2a = 4r) from the geometry of the cube — every packing-efficiency derivation and most numericals depend on them.
→A common long-answer question is the derivation of packing efficiency for sc, bcc or fcc. Write z, the r–a relation, the volume of spheres and the volume of the cell as separate steps.
→In density numericals convert the edge length to cm first (1 pm = 10⁻¹⁰ cm) so that ρ comes out in g cm⁻³.
→To identify the type of cubic lattice, compute z = ρa³N_A/M and round to 1 (sc), 2 (bcc) or 4 (fcc).
→For the formula of a compound from voids, remember N close-packed spheres give N octahedral and 2N tetrahedral voids.
→When comparing Schottky and Frenkel defects, state which ions are missing or displaced, give an example of each, and give the effect on density.

📐 Formulas(13)

Contribution of Particles to a Cubic Unit Cell

z = \frac{N _{c}}{8} + \frac{N _{f}}{2} + \frac{N _{e}}{4} + N_{b}

Symbol	Meaning
$z$	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

Particles per Unit Cell — sc, bcc, fcc★ Board fav

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

Radius–Edge Relation: Simple Cubic

a = 2 r \Rightarrow r = \frac{a}{2}

Symbol	Meaning
$a$	Edge length of the unit cell (pm)
$r$	Atomic radius (pm)

Radius–Edge Relation: Body-Centred Cubic

3 a = 4 r \Rightarrow r = \frac{3}{4} a = 0.433 a

Radius–Edge Relation: Face-Centred Cubic (ccp)

2 a = 4 r \Rightarrow r = \frac{a}{2 2} = 0.3536 a

Packing Efficiency (general)★ Board fav

Packing efficiency = \frac{z \times \frac{4}{3} π r ^{3}}{a ^{3}} \times 100%

Symbol	Meaning
$z$	Number of atoms per unit cell
$r$	Atomic radius
$a^{3}$	Volume of the cubic unit cell

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%

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%

Packing Efficiency: Face-Centred Cubic (ccp) and hcp★ Board fav

\frac{4 \times \frac{4}{3} π r ^{3}}{( 2 2 r ) ^{3}} \times 100 = \frac{π}{3 2} \times 100 = 74%

Density of a Cubic Crystal★ Board fav

ρ = \frac{z M}{a ^{3} N _{A}}

Symbol	Meaning
$ρ$	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

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}

Symbol	Meaning
$x$	Mass of the metal sample (g)
$ρ a^{3}$	Mass of one unit cell (g)

Voids in Close Packing

N_{tetrahedral} = 2 N, N_{octahedral} = N

Point Defects and Density

Schottky: ρ ↓ Frenkel: ρ unchanged

✏️ Solved Examples

1Solved Exampleeasy3 steps

Atoms of element Q form a ccp lattice and atoms of element P occupy one-third of the tetrahedral voids. Find the formula of the compound.

Let the number of Q atoms be N. A ccp lattice of N atoms has 2N tetrahedral voids.

N_{tetrahedral} = 2 N

2Solved Exampleboard3 steps

Copper crystallises in an fcc lattice with edge length 361 pm. Calculate the density of copper. (Cu = 63.5 g mol⁻¹, N_A = 6.022 × 10²³ mol⁻¹)

For fcc, z = 4. Convert the edge length to cm and find the cell volume.

a = 361 pm = 3.61 \times 1 0^{- 8} cm, a^{3} = 4.705 \times 1 0^{- 23} cm^{3}

3Solved Exampleboard3 steps

Iron has a bcc unit cell and density 7.87 g cm⁻³. Calculate the edge length of the unit cell and the atomic radius of iron. (Fe = 55.85 g mol⁻¹, N_A = 6.022 × 10²³ mol⁻¹)

For bcc, z = 2. Rearrange the density formula for a³.

a^{3} = \frac{z M}{ρ N _{A}} = \frac{2 \times 55.85}{7.87 \times 6.022 \times 1 0 ^{23}} = 2.357 \times 1 0^{- 23} cm^{3}

4Solved ExampleHOTS3 steps

A metal (molar mass 27 g mol⁻¹) crystallises in a cubic lattice with edge length 405 pm. Its density is 2.70 g cm⁻³. Identify the type of unit cell and find the atomic radius. (N_A = 6.022 × 10²³ mol⁻¹)

Find the cell volume in cm³.

a^{3} = (4.05 \times 1 0^{- 8} cm)^{3} = 6.643 \times 1 0^{- 23} cm^{3}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes5

1
Counting every corner atom of a cube as a whole atom (z = 8 for simple cubic).
✓Each corner atom is shared by 8 cells and contributes 1/8. Simple cubic z = 1, bcc z = 2, fcc z = 4.
2
Using a = 2r for bcc and fcc lattices.
✓a = 2r only for simple cubic. In bcc atoms touch along the body diagonal (√3a = 4r); in fcc along the face diagonal (√2a = 4r).
3
Putting the edge length in pm into ρ = zM/(a³N_A) and expecting g cm⁻³.
✓Convert first: 1 pm = 10⁻¹⁰ cm, so a³ in pm³ must be multiplied by 10⁻³⁰ to get cm³.
4
Taking the number of tetrahedral voids equal to the number of spheres.
✓N close-packed spheres give 2N tetrahedral voids and N octahedral voids.
5
Saying that a Frenkel defect lowers the density.
✓In a Frenkel defect ions only shift to interstitial sites, so mass and volume are unchanged. Only the Schottky defect (ions missing) lowers density.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Silver crystallises in an fcc lattice with edge length 408 pm. Calculate the atomic radius of silver.
Q2
What percentage of the volume of a bcc unit cell is empty space?
Q3
Polonium crystallises in a simple cubic lattice with edge length 335 pm. Calculate its density. (Po = 209 g mol⁻¹, N_A = 6.022 × 10²³ mol⁻¹)
Q4
Nickel has an fcc lattice with edge length 352 pm and density 8.9 g cm⁻³. How many unit cells and how many atoms are present in 10 g of nickel?
Q5
Anions B form a ccp lattice and cations A occupy half of the octahedral voids. What is the formula of the compound?
Q6
NaCl is doped with 10⁻³ mol % SrCl₂. Each Sr²⁺ ion replaces two Na⁺ ions and creates one cation vacancy. Find the number of cation vacancies per mole of NaCl.

📝 Notes

Solid State — Maharashtra HSC Overview

Chapter 1 of the Maharashtra State Board Std XII Chemistry textbook moves from the types of solids to cubic unit cells, packing, density and defects. Almost every numerical in the chapter uses just three ideas: the number of atoms per unit cell $z$ , the radius–edge relation, and the density equation.

One cube, three lattices

Start every problem by writing $z$ and the r–a relation for the lattice:

Simple cubic: $z = 1$ , $a = 2r$ , coordination number 6, packing 52.36%.
Body-centred cubic: $z = 2$ , $\sqrt{3}a = 4r$ , coordination number 8, packing 68%.
Face-centred cubic (ccp): $z = 4$ , $\sqrt{2}a = 4r$ , coordination number 12, packing 74%.

The packing-efficiency derivation is the same in all three cases: volume of $z$ spheres divided by $a^3$ , after replacing $a$ with its value in terms of $r$ .

Density numericals

$\rho = zM/(a^3N_A)$ links five quantities. A question gives four and asks for the fifth. If it asks for the type of lattice, solve for $z$ and round to 1, 2 or 4. Keep the edge length in cm: a slip in the power of ten is a common source of a wrong answer.

Voids and formulas

In ccp or hcp, $N$ spheres create $N$ octahedral and $2N$ tetrahedral voids. Multiply the number of voids by the fraction occupied, compare with $N$ , and reduce to the simplest whole-number ratio.

Defects

Stoichiometric defects keep the formula unchanged: vacancy, interstitial, Schottky and Frenkel. Non-stoichiometric defects (metal deficiency, metal excess with F-centres) change the cation–anion ratio. Impurity defects such as SrCl₂ in NaCl create cation vacancies, one for each Sr²⁺ ion.