NEET The Solid State Previous Year Questions with Solutions

In a cubic close packed (ccp / face-centered cubic) lattice of spheres, let the number of close-packed constituent atoms forming the lattice be N. Geometrical packing dictates that the number of octahedral voids is N, and the number of tetrahedral voids is 2N:

Atoms of element A occupy one-third (1/3) of the available tetrahedral voids. Therefore, the total number of atoms of A per N atoms of B is:

Find the simplest whole-number stoichiometric ratio between elements A and B:

Comparing the empirical formula A₂B₃ with A_x B_y identifies x = 2 and y = 3. Computing their sum:

In a three-dimensional cubic lattice, four unit cells share a common edge (two in the same horizontal layer and two in the layer directly above or below):

Because an edge is divided equally among four unit cells, only 1/4 of any point, sphere, or void positioned at an edge centre belongs to any single unit cell:

A cube has 12 edges. The total contribution of all edge-centred octahedral voids to the unit cell is 12 × (1/4) = 3. Combined with 1 body-centred void (1 × 1 = 1), an fcc unit cell has 4 total octahedral voids, but the fraction of ONE edge void is strictly 1/4:

A hexagonal primitive (hcp) unit cell contains: - 12 corner atoms, each shared among 6 unit cells: $12 \times \frac{1}{6} = 2$ - 2 face-centered atoms, each shared between 2 unit cells: $2 \times \frac{1}{2} = 1$ - 3 completely enclosed internal atoms: $3 \times 1 = 3$ Total effective constituent particles per unit cell:

For any close-packed arrangement of $N$ spheres: - Number of Octahedral Voids ($O_V$) $= N$ - Number of Tetrahedral Voids ($T_V$) $= 2N$

Substituting $N = 6$:

Auguste Bravais demonstrated that only 14 space lattices are mathematically possible in three dimensions across the 7 crystal systems: 1. Cubic (3): Primitive, Body-centred, Face-centred 2. Tetragonal (2): Primitive, Body-centred 3. Orthorhombic (4): Primitive, Body-centred, Face-centred, End-centred 4. Hexagonal (1): Primitive 5. Rhombohedral / Trigonal (1): Primitive 6. Monoclinic (2): Primitive, End-centred 7. Triclinic (1): Primitive

Summing the body-centred lattices: - Cubic body-centred: 1 - Tetragonal body-centred: 1 - Orthorhombic body-centred: 1 No other crystal system allows a body-centred Bravais lattice without reducing to a simpler existing lattice.

In a body-centered cubic (BCC) crystal lattice, the central atom touches all eight corner atoms. Therefore, sphere contact occurs along the body diagonal of the cube.

Let the edge length of the unit cell be $$a$$. The face diagonal of any face is $$b = \sqrt{a^2 + a^2} = \sqrt{2}a$$. The body diagonal $$c$$ running from one corner through the central sphere to the opposite corner is: $$c = \sqrt{b^2 + a^2} = \sqrt{(\sqrt{2}a)^2 + a^2} = \sqrt{3a^2} = \sqrt{3}a$$

Along the body diagonal, there is one radius from the starting corner atom, two radii (diameter) of the central atom, and one radius from the ending corner atom: $$\text{Body Diagonal} = r + 2r + r = 4r$$ Equating the two expressions: $$4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4}a$$

With $$a = 288\text{ pm}$$: $$r = \frac{\sqrt{3}}{4} \times 288\text{ pm}$$ This corresponds precisely to Option (1).