Solid State Density Numericals
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Direct answer
One equation decides essentially every solid-state numerical: density ρ = (Z × M)/(a^3 × N_A), where Z is the number of formula units per unit cell (simple cubic 1, bcc 2, fcc 4, hcp 6), M the molar mass in grams, a the edge length in centimetres and N_A Avogadro's number. Geometry enters through the sphere-touching relations — simple cubic a = 2r, bcc 4r = √3 a along the body diagonal, fcc 4r = √2 a along the face diagonal — so questions freely convert between density, edge, radius and lattice type. The only real discipline is unit hygiene: 1 pm = 10^-10 cm.
What you must remember
- Master equation with units: ρ (g/cm3) = Z M/(a^3 N_A) with a in centimetres; rearrangements give a from ρ, or Z from ρ and a.
- Z values: sc 1, bcc 2, fcc 4, hcp 6 per conventional hexagonal cell; ionic lattices — NaCl 4, CsCl 1, ZnS 4, CaF2 4.
- Radius-edge geometry: sc r = a/2; bcc 4r = √3 a; fcc 4r = √2 a; in NaCl the chloride sublattice is fcc, so a = 2√2 r(Cl−).
- Conversion discipline: 1 pm = 10^-10 cm and 1 Å = 10^-8 cm; metal densities land between 2 and 22 g/cm3, so sanity-check magnitudes.
- Counting from mass: unit cells in mass m number (m/M)(N_A/Z); total atoms simply (m/M)N_A for a metal.
- Void census: per close-packed sphere one octahedral and two tetrahedral voids; per fcc cell, 4 octahedral (1 body centre plus 12 edges at quarter share) and 8 tetrahedral.
- Packing fractions for cross-checks: sc 52.4%, bcc 68%, fcc/hcp 74% — density differences between metals track both packing and atomic mass.
- Reference computations: copper (fcc, r = 127.8 pm) gives 8.94 g/cm3; iron (bcc, r = 124 pm) gives about 7.9 g/cm3 — both NCERT-scale results worth reproducing by hand.
A density calculation from radius to grams
Compute copper's density from r = 127.8 pm. Copper is fcc, so atoms touch along the face diagonal: 4r = √2 a gives a = 4 × 127.8/1.414 = 361.5 pm = 3.615 × 10^-8 cm, and a^3 = 4.72 × 10^-23 cm3. Then ρ = (4 × 63.55)/(4.72 × 10^-23 × 6.022 × 10^23) = 254.2/28.43 = 8.94 g/cm3 — the tabulated value, reproduced exactly because fcc copper is a perfect closest packing. Reverse problems run the same film backwards: given iron's density 7.90 g/cm3 and bcc candidacy, a^3 = ZM/(ρN_A) yields a = 286 pm, and r = √3 a/4 = 124 pm. A third template asks for the lattice itself: from measured ρ and a, compute Z = ρ a^3 N_A/M and match the integer 1, 2 or 4 to simple cubic, bcc or fcc. Rounding belongs only at the final step — the nearest integer is the answer, and fractions en route mean an arithmetic slip upstream.
Syllabus status and the classic slips
Solid state appears on the JEE Main 2024 deleted-topics list after the NTA's syllabus rationalisation, but JEE Advanced retains it in full — unit cells, packing efficiency, voids, ionic structures and density numericals all remain Advanced territory, which is why the chapter keeps a seat in an Advanced aspirant's schedule while Main-only candidates skim it for concept. The errors are almost always mechanical. Plugging picometres directly into a^3 inflates the density a millionfold. Quoting hcp's Z as 2 or 4 instead of 6 per conventional cell. Using oxygen's atomic mass where the oxide's formula mass belongs. And in ionic lattices, forgetting that Z counts formula units — four NaCl pairs, not four sodiums plus four chlorides as separate cells. Each slip is avoidable by writing the unit chain before touching the calculator.
Frequently asked questions
What is Z for simple cubic, bcc and fcc unit cells?
1, 2 and 4 respectively — corners contribute one-eighth each, the body centre adds a whole atom, and fcc collects 8 × 1/8 plus 6 × 1/2 = 4.
How is copper's density derived from its atomic radius?
For fcc, 4r = √2 a gives a = 361.5 pm; then ρ = ZM/(a^3 N_A) = (4 × 63.55)/(4.72 × 10^-23 × 6.022 × 10^23) = 8.94 g/cm3.
How many octahedral and tetrahedral voids does an fcc cell contain?
Four octahedral voids (one at the body centre plus twelve edge-centres at quarter share) and eight tetrahedral voids — two tetrahedral per close-packed sphere.
Why must the edge length be converted to centimetres?
Because grams and Avogadro's number make ρ emerge in g/cm3; using picometres or angstroms directly shifts the answer by powers of ten that bury the mark.
How is a metal's lattice identified from density and edge length?
Compute Z = ρ a^3 N_A/M and match the integer: 1 means simple cubic, 2 means bcc, 4 means fcc.