JEE Chemistry The Solid State MCQs Set 03

Answer: (d) 2 and 4

Explanation:
1. A BCC unit cell has 8 corner atoms and 1 body-centre atom: 8 \(\times\) 1/8 + 1 = 2.
2. An FCC unit cell has 8 corner atoms and 6 face-centre atoms: 8 \(\times\) 1/8 + 6 \(\times\) 1/2 = 4.
3. So Na has 2 atoms and Mg has 4 atoms per unit cell.

Teacher's Note:
1. Corner atoms are shared by 8 cells, face atoms by 2 cells, body-centre atoms by none.
2. Learn the standard values: simple cubic 1, BCC 2, FCC 4.

Answer: (c) \( 2.57 \times 10^{21} \)

Explanation:
1. Moles of NaCl = 1.00/58.5 = 0.01709 mol.
2. Formula units = 0.01709 \(\times\) 6.022 \(\times\) 10\(^{23}\) = 1.029 \(\times\) 10\(^{22}\).
3. A NaCl unit cell holds 4 formula units, so unit cells = 1.029 \(\times\) 10\(^{22}\)/4 = 2.57 \(\times\) 10\(^{21}\).

Teacher's Note:
1. Always divide the number of formula units by Z, the formula units per cell.
2. For rock salt Z = 4.

Answer: (b) Schottky defect

Explanation:
1. In the diagram equal numbers of Na⁺ and Cl⁻ ions are missing from their lattice sites.
2. No ion has moved to an interstitial position.
3. Missing cation-anion pairs make a Schottky defect.

Teacher's Note:
1. Schottky defect: pair of vacancies, density decreases.
2. Frenkel defect: an ion moves to an interstitial site, density unchanged.

Answer: (c) AB\(_3\)

Explanation:
1. A: 8 corners \(\times\) 1/8 = 1.
2. B: 6 faces \(\times\) 1/2 = 3.
3. Ratio A : B = 1 : 3, so the formula is AB\(_3\).

Teacher's Note:
1. Count each position by its share inside the cell.
2. The same cell with A at corners and B at face centres is the well-known AuCu\(_3\) type.

Answer: (d) \( \dfrac{16}{3}\pi r^3 \)

Explanation:
1. An FCC cell contains 4 atoms.
2. Volume of one atom = \( \tfrac{4}{3}\pi r^3 \).
3. Total = 4 \(\times\) \( \tfrac{4}{3}\pi r^3 \) = \( \tfrac{16}{3}\pi r^3 \).

Teacher's Note:
1. Packing fraction = (Z \(\times\) atom volume)/cell volume, so this total is the numerator.
2. With a = \(2\sqrt{2}\,r\) the fraction works out to 0.74.

Answer: (d) X\(_4\)Y\(_3\)

Explanation:
1. Let there be N atoms of Y in the ccp lattice.
2. Tetrahedral voids = 2N, and X occupies 2/3 of them = 4N/3.
3. X : Y = 4N/3 : N = 4 : 3, so the formula is X\(_4\)Y\(_3\).

Teacher's Note:
1. Remember: octahedral voids = N, tetrahedral voids = 2N for N close-packed atoms.
2. Multiply the fraction occupied by the number of voids before forming the ratio.

Answer: (a) 127 pm

Explanation:
1. In FCC the atoms touch along the face diagonal: \( 4r = \sqrt{2}\,a \).
2. \( r = \dfrac{a}{2\sqrt{2}} = \dfrac{361}{2.828} \) = 127.6 pm.
3. The nearest option is 127 pm.

Teacher's Note:
1. FCC: r = a/(2\(\sqrt{2}\)); BCC: r = \(\sqrt{3}\)a/4; simple cubic: r = a/2.
2. Check which direction the atoms touch before writing the relation.

Answer: (d) 144 pm

Explanation:
1. In the rock-salt type FCC arrangement, ions touch along the cell edge: a = 2(r\(_+\) + r\(_-\)).
2. r\(_+\) + r\(_-\) = 254 pm.
3. r\(_-\) = 254 − 110 = 144 pm.

Teacher's Note:
1. For NaCl-type crystals the edge equals twice the sum of the two ionic radii.
2. The anion is usually the larger ion, which supports 144 pm over 110 pm.

Answer: (b) 26% and 32%

Explanation:
1. Packing efficiency of ccp (FCC) is 74%, so free space is 26%.
2. Packing efficiency of BCC is 68%, so free space is 32%.
3. The pair is 26% and 32%.

Teacher's Note:
1. Packing efficiencies: simple cubic 52%, BCC 68%, FCC/HCP 74%.
2. Free space = 100 − packing efficiency.

Answer: (d) A\(_2\)B\(_5\)

Explanation:
1. A: 8 \(\times\) 1/8 = 1.
2. B normally = 6 \(\times\) 1/2 = 3; one face centre is missing, so B = 5 \(\times\) 1/2 = 5/2.
3. A : B = 1 : 5/2 = 2 : 5, so the formula is A\(_2\)B\(_5\).

Teacher's Note:
1. Recount the faces that remain after removal; do not subtract a whole atom.
2. Each face-centre atom contributes only 1/2.

Answer: (d) 152 pm

Explanation:
1. In BCC the atoms touch along the body diagonal: \( 4r = \sqrt{3}\,a \).
2. \( r = \dfrac{\sqrt{3}\times 351}{4} \) = 151.98 pm.
3. So r is about 152 pm.

Teacher's Note:
1. Body diagonal of a cube = \(\sqrt{3}\)a, and it carries two radii plus one atom diameter, that is 4r.
2. Do not confuse this with the FCC face-diagonal relation.

Answer: (c) \( \dfrac{a}{\sqrt{2}} \)

Explanation:
1. Atoms touch along the face diagonal, whose length is \( \sqrt{2}\,a \).
2. The face diagonal carries 4r, so the closest approach 2r = \( \dfrac{\sqrt{2}\,a}{2} = \dfrac{a}{\sqrt{2}} \).

Teacher's Note:
1. Closest approach means the distance between the centres of two touching atoms, that is 2r.
2. For BCC it would be \(\sqrt{3}\)a/2.

Answer: (d) A\(_3\)B\(_4\)

Explanation:
1. In the NaCl structure A is at the 8 corners and 6 face centres: 1 + 3 = 4 per cell; B is at 12 edge centres and the body centre: 3 + 1 = 4 per cell.
2. Removing the two face-centre atoms along one axis removes 2 \(\times\) 1/2 = 1 atom of A.
3. A = 3, B = 4, so the formula is A\(_3\)B\(_4\).

Teacher's Note:
1. Only the A atoms were removed, so the count of B stays 4.
2. Always recompute both totals before taking the ratio.

Answer: (a) AB\(_3\)

Explanation:
1. A = 8 \(\times\) 1/8 = 1.
2. B = 6 \(\times\) 1/2 = 3.
3. Composition is AB\(_3\).

Teacher's Note:
1. This is the same count as the AIEEE problem with A at corners and B at face centres.
2. Shared-position counting works for any cubic cell.

Answer: (b) ZnS

Explanation:
1. In ZnS (zinc blende) S\(^{2-}\) ions form ccp, which has 8 tetrahedral voids per cell, and Zn\(^{2+}\) fills only 4 of them, that is alternate voids.
2. NaCl has cations in octahedral voids, CaF\(_2\) has cations in the ccp lattice with all tetrahedral voids filled by anions, and Na\(_2\)O (antifluorite) has all tetrahedral voids filled by cations.

Teacher's Note:
1. Zinc blende: half the tetrahedral voids. Fluorite and antifluorite: all tetrahedral voids.
2. Rock salt: all octahedral voids.

Answer: (b) 2 and 3 only

Explanation:
1. Statement 1 is false: Frenkel defect needs a small cation and a much larger anion.
2. Statement 2 is true: the ion is dislocated to an interstitial site.
3. Statement 3 is true: an electron trapped in an anion vacancy is an F-centre.
4. Statement 4 is false: Schottky defects lower the density.
5. Only 2 and 3 are correct.

Teacher's Note:
1. F-centres give colour to alkali halide crystals heated in alkali metal vapour.
2. Schottky defect lowers density; Frenkel defect does not.

Answer: (c) 8

Explanation:
1. A truncated octahedron is formed by cutting off the six corners of an octahedron.
2. It has 14 faces in all: 8 regular hexagons (from the original 8 triangular faces) and 6 squares (from the cut corners).
3. Hexagonal faces = 8.

Teacher's Note:
1. Each original triangular face becomes a hexagon, and each truncated vertex becomes a square.
2. Check with 8 + 6 = 14 faces.

Answer: (b) They belong to the same crystal system

Explanation:
1. Simple cubic and face-centred cubic are two Bravais lattices of the cubic crystal system.
2. Both have a = b = c and \(\alpha = \beta = \gamma = 90^\circ\), so they belong to the same crystal system.

Teacher's Note:
1. There are 7 crystal systems but 14 Bravais lattices.
2. The cubic system contains three lattices: simple, body-centred and face-centred.