Solid State: JEE Advanced PYQ, 2007–2026

Answer: (B)

The key principle is that crystalline solids have long-range order and a sharp melting point, while amorphous solids lack long-range order and soften over a range of temperatures. Glass is an amorphous solid, so it softens gradually. Students who pick the sharp-melting option have assumed that every solid behaves like a crystal.

Answer: (A)

For fcc, Z = 4, so d = ZM/(N_A a³). This gives a³ = (4 × 108)/(6.0 × 10²³ × 10.5) = 6.86 × 10⁻²³ cm³, so a = 4.09 × 10⁻⁸ cm = 409 pm. In fcc the atoms touch along the face diagonal, √2a = 4r, so r = 409/(2√2) ≈ 145 pm.

Answer: (A)

List the angle conditions of the seven systems. Cubic, tetragonal and orthorhombic have α = β = γ = 90°, so all three are right angles. Rhombohedral has α = β = γ ≠ 90°, so none is a right angle. Triclinic has α ≠ β ≠ γ ≠ 90°, so none is a right angle. Hexagonal has α = β = 90° and γ = 120°, and monoclinic has α = γ = 90° and β ≠ 90°, so each has exactly two right angles. The Bravais lattices are 1 (primitive) for hexagonal and 2 (primitive and end-centred) for monoclinic, a total of 3.

Answer: (A)

Principle: heating an alkali halide in the vapour of its metal adds metal atoms. Each Na atom gives up its electron to the crystal, and the excess electron fills a Cl⁻ vacancy. This electron-in-vacancy is an F-centre. Its energy levels are spaced so that it absorbs visible light, which is why NaCl looks yellow. This is metal excess, so the electrons are the defect species, and the vacancy is an anion vacancy.

Answer: (A)

In ccp (ABCABC...) the close-packed layers are the (111) planes, perpendicular to the body diagonal of the fcc cell. The body diagonal has length a√3 and passes through three layer spacings per ABC repeat, so the spacing is d = a√3/3 = a/√3. Then d = 360/1.73 ≈ 208 pm. As a check, the nearest-neighbour distance a/√2 ≈ 255 pm is larger than d, as it must be, since layers sit in the hollows of the layer below.

Answer: (A)

Metal deficiency arises when cation sites are vacant and the missing positive charge is made up by raising the charge of some remaining cations of the same metal. So the metal must have a variable oxidation state, as Fe, Ni and Cu do. Ni₀.₉₈O fits (Ni²⁺ partly becomes Ni³⁺). Cs cannot become Cs²⁺, so Cs₀.₉₈Cl cannot exist this way. Na₁.₀₁Cl has excess metal, which is the F-centre type. Anion excess (NaCl₁.₀₁) is not found because the large anion cannot occupy interstitial sites.

Answer: (A)

A simple cubic cell has one sphere, since each of the 8 corners contributes 1/8. Spheres touch along edges, so a = 2R. Packing fraction = (4/3)πR³ / (2R)³ = π/6 ≈ 0.52.

Answer: (A)

In hcp (as in ccp), N close-packed atoms generate N octahedral voids and 2N tetrahedral voids. With N anions X there are N octahedral voids, and two-thirds of them hold A, so there are 2N/3 cations. A:X = (2/3):1 = 2:3, so the formula is A₂X₃.

Answer: (A)

In fcc, a = 2√2 r, so the body diagonal is √3 a = 2√6 r ≈ 4.90r. Along it, each corner atom contributes a length r inside the cell. Next come the tetrahedral void centre at a/4 along each axis and the octahedral void at the body centre, which are empty. The face-centre atoms lie at a perpendicular distance a/√6 ≈ 1.155r from the diagonal. This is more than r, so the diagonal misses them. The occupied length is 2r, so the fraction is 2r/(2√6 r) = 1/√6 ≈ 0.408, or 40.8%.

Answer: (B)

Close-packed layers are labelled A, B and C by their positions. In hcp the layers alternate between two positions, so the stacking is ABAB... In ccp (fcc) the layers follow the three positions in order, ABCABC...