JEE Chemistry The Solid State MCQs Set 05

Answer: (d) AB

Explanation:
1. B: 8 \(\times\) 1/8 = 1.
2. A: 1 at the body centre.
3. A : B = 1 : 1, so the formula is AB.

Teacher's Note:
1. Corner atoms count 1/8 each, the body-centre atom counts 1.
2. This is the CsCl arrangement.

Answer: (c) AuCu₃

Explanation:
1. Au: 8 \(\times\) 1/8 = 1.
2. Cu: 6 \(\times\) 1/2 = 3.
3. Formula = AuCu₃.

Teacher's Note:
1. Name the lattice position first and then use the sharing fraction.
2. Face-centre atoms are shared by two cells.

Answer: (a) A₂BC

Explanation:
1. Let there be N atoms of A.
2. Tetrahedral voids = 2N, so B = 2N/4 = N/2.
3. Octahedral voids = N, so C = N/2.
4. A : B : C = N : N/2 : N/2 = 2 : 1 : 1, so the formula is A₂BC.

Teacher's Note:
1. The unit cell gives A₄B₂C₂, which reduces to A₂BC.
2. An empirical formula is always in simplest whole-number ratio.

Answer: (c) 59.9% and +4

Explanation:
1. For N oxide ions there are N octahedral holes, and Ti fills N/2, so Ti : O = 1 : 2 and the formula is TiO₂.
2. Mass % of Ti = 47.87/(47.87 + 32) \(\times\) 100 = 59.9%.
3. Oxidation number of Ti = +4.

Teacher's Note:
1. Charge balance: 2 oxide ions give −4, so Ti is +4.
2. Mass percent uses atomic masses, not counts of atoms.

Answer: (b) ZnAl₂O₄

Explanation:
1. For 4 O²⁻ per cell: tetrahedral holes = 8, octahedral holes = 4.
2. Zn²⁺ = 8/8 = 1.
3. Al³⁺ = 4/2 = 2.
4. Formula ZnAl₂O₄; charge check: +2 + 6 = +8 = 4 \(\times\) 2.

Teacher's Note:
1. Always check electroneutrality after finding the ratio.
2. MgAl₂O₄ is the parent spinel.

Answer: (d) 267 pm, 534 pm, 378 pm

Explanation:
1. Shortest K-F distance = r\(_+\) + r\(_-\) = 132 + 135 = 267 pm.
2. Edge a = 2 \(\times\) 267 = 534 pm.
3. Closest K-K distance = face diagonal/2 = \( a/\sqrt{2} \) = 534/1.414 = 377.6 pm.

Teacher's Note:
1. Along the cell edge of NaCl there is one cation, one anion and the next cation, so a = 2(r\(_+\) + r\(_-\)).
2. The nearest same-kind neighbours lie along the face diagonal.

Answer: (b) 267 pm, 231.2 pm, 188.8 pm

Explanation:
1. Simple cubic: r = a/2 = 267 pm.
2. BCC: r = \( \sqrt{3}a/4 \) = 1.732 \(\times\) 534/4 = 231.2 pm.
3. FCC: r = \( a/(2\sqrt{2}) \) = 534/2.828 = 188.8 pm.

Teacher's Note:
1. Radius for a fixed edge decreases as packing gets denser: sc > bcc > fcc.
2. Write the touching direction first: edge, body diagonal, face diagonal.

Answer: (b) 3.51 g cm⁻³

Explanation:
1. Two atoms per lattice point and 4 lattice points give Z = 8.
2. \( \rho = \dfrac{Z M}{N_A a^3} = \dfrac{8 \times 12}{6.022\times10^{23} \times (3.569\times10^{-8})^3} \).
3. a\(^3\) = 4.546 \(\times\) 10\(^{-23}\) cm\(^3\), so \( \rho \) = 3.51 g cm\(^{-3}\).

Teacher's Note:
1. Do not forget the second atom at each lattice point.
2. Using Z = 4 would give the wrong value of 1.75.

Answer: (c) \( 3.47 \times 10^{24} \)

Explanation:
1. Atoms per cell = 8 \(\times\) 1/8 + 2 = 3.
2. Mass of one cell = 7.2 \(\times\) 24 \(\times\) 10\(^{-24}\) = 1.728 \(\times\) 10\(^{-22}\) g.
3. Number of cells in 200 g = 200/(1.728 \(\times\) 10\(^{-22}\)) = 1.157 \(\times\) 10\(^{24}\).
4. Atoms = 3 \(\times\) 1.157 \(\times\) 10\(^{24}\) = 3.47 \(\times\) 10\(^{24}\).

Teacher's Note:
1. The 2 atoms on the body diagonal are inside the cell and count fully.
2. Multiply cells by atoms per cell to get atoms.

Answer: (c) face-centred cubic

Explanation:
1. Test FCC: a = 4r/\(\sqrt{2}\) = 407 pm.
2. \( \rho = \dfrac{4 \times 107.9}{6.022\times10^{23}\times(4.07\times10^{-8})^3} \) = 10.6 g cm\(^{-3}\), which matches.
3. The same radius would give only 9.74 g cm\(^{-3}\) for BCC (a = 332.6 pm) and 7.50 g cm\(^{-3}\) for simple cubic (a = 288 pm).
4. So silver is FCC.

Teacher's Note:
1. Compute the density for each possible lattice and compare with the data.
2. Close-packed metals such as Ag, Au, Cu are FCC.

Answer: (b) 0.24%

Explanation:
1. Cell volume = (5.55 \(\times\) 10\(^{-8}\))\(^3\) = 1.71 \(\times\) 10\(^{-22}\) cm\(^3\).
2. Actual number of formula units per cell \( Z = \dfrac{\rho a^3 N_A}{M} = \dfrac{5.561 \times 1.71\times10^{-22} \times 6.023\times10^{23}}{143.5} \) = 3.99.
3. Ideal Z = 4, so vacant fraction = (4 − 3.99)/4 \(\approx\) 0.24%.

Teacher's Note:
1. Density lower than the ideal value means some ion sites are empty, which is a Schottky-type defect.
2. The calculation is very sensitive, so keep extra digits in the intermediate steps.

Answer: (c) 438.4 pm and 219.2 pm

Explanation:
1. Nearest-neighbour distance = \( a/\sqrt{2} = 620/1.414 \) = 438.4 pm.
2. Radius = half of that = 219.2 pm.

Teacher's Note:
1. The nearest-neighbour distance is 2r.
2. It is shorter than the cell edge in FCC.

Answer: (c) 6

Explanation:
1. Radius ratio = 88/200 = 0.44.
2. This lies between 0.414 and 0.732, the range of octahedral coordination.
3. Coordination number of A⁺ = 6.

Teacher's Note:
1. Ranges: 0.155-0.225 (CN 3), 0.225-0.414 (CN 4), 0.414-0.732 (CN 6), 0.732-1 (CN 8).
2. Divide the smaller radius by the larger.

Answer: (c) 346.4 pm

Explanation:
1. The Cs⁺ and Cl⁻ ions touch along the body diagonal.
2. Interionic distance = half the body diagonal = \( \dfrac{\sqrt{3}}{2}a = 0.866 \times 400 \) = 346.4 pm.

Teacher's Note:
1. Body diagonal = \(\sqrt{3}\)a.
2. The cation-anion contact is half of it.

Answer: (b) 19.4 g cm⁻³ and 143.9 pm

Explanation:
1. \( \rho = \dfrac{4 \times 197}{6.022\times10^{23}\times(4.07\times10^{-8})^3} = \dfrac{788}{6.022\times10^{23}\times6.742\times10^{-23}} \) = 19.4 g cm\(^{-3}\).
2. \( r = \dfrac{a}{2\sqrt{2}} = \dfrac{407}{2.828} \) = 143.9 pm.

Teacher's Note:
1. Gold is about 19 times denser than water.
2. Use Z = 4 for FCC.

Answer: (c) Face-centred cubic (Z = 4)

Explanation:
1. \( Z = \dfrac{\rho a^3 N_A}{M} = \dfrac{2.75 \times (6.54\times10^{-8})^3 \times 6.023\times10^{23}}{119} \).
2. a\(^3\) = 2.797 \(\times\) 10\(^{-22}\) cm\(^3\), so Z = 2.75 \(\times\) 2.797 \(\times\) 10\(^{-22}\) \(\times\) 6.023 \(\times\) 10\(^{23}\)/119 = 3.89 \(\approx\) 4.
3. Four formula units per cell means a face-centred cubic (NaCl-type) lattice.

Teacher's Note:
1. Z comes out close to a whole number; round it to the nearest integer.
2. Z = 4 for an ionic compound means the rock-salt arrangement.

Answer: (c) 41.67 g cm⁻³

Explanation:
1. Mass of one atom = 200/(24 \(\times\) 10\(^{23}\)) = 8.33 \(\times\) 10\(^{-23}\) g.
2. Mass of the cell (4 atoms) = 3.33 \(\times\) 10\(^{-22}\) g.
3. Volume = (2 \(\times\) 10\(^{-8}\) cm)\(^3\) = 8 \(\times\) 10\(^{-24}\) cm\(^3\).
4. \(\rho\) = 3.33 \(\times\) 10\(^{-22}\)/8 \(\times\) 10\(^{-24}\) = 41.67 g cm\(^{-3}\).

Teacher's Note:
1. You can find the mass of an atom without knowing the element.
2. Edge length is converted from pm to cm by multiplying with 10\(^{-10}\).

Answer: (a) 5.74 g cm⁻³

Explanation:
1. \( a = 2\sqrt{2}\,r = 2.828 \times 1.42 \) = 4.016 \(\mathring{A}\).
2. a\(^3\) = 6.477 \(\times\) 10\(^{-23}\) cm\(^3\).
3. \( \rho = \dfrac{4 \times 56}{6.022\times10^{23}\times6.477\times10^{-23}} \) = 5.74 g cm\(^{-3}\).

Teacher's Note:
1. Get a from r using the face-diagonal relation.
2. The true density of iron differs because this question uses a hypothetical FCC iron.

Answer: (a) 594 pm and \( 2.10 \times 10^{-22}\ \text{cm}^3 \)

Explanation:
1. In NaCl type a = 2 \(\times\) (shortest cation-anion distance) = 594 pm.
2. V = a\(^3\) = (5.94 \(\times\) 10\(^{-8}\) cm)\(^3\) = 2.096 \(\times\) 10\(^{-22}\) cm\(^3\).

Teacher's Note:
1. The shortest cation-anion distance is half the edge in rock salt.
2. Convert to cm before cubing.

Answer: (b) 1.81 \(\mathring{A}\)

Explanation:
1. Body diagonal = 2r₊ + 2r₋ = 7 \(\mathring{A}\).
2. r₊ + r₋ = 3.5 \(\mathring{A}\).
3. r₋ = 3.5 − 1.69 = 1.81 \(\mathring{A}\).

Teacher's Note:
1. The body diagonal contains two radii of each ion.
2. The answer matches the known radius of Cl⁻, 181 pm.

Answer: (d) XY₃ and 4.38 g cm⁻³

Explanation:
1. X = 8 \(\times\) 1/8 = 1 and Y = 6 \(\times\) 1/2 = 3, so the formula is XY\(_3\).
2. Mass of the cell = (60 + 3 \(\times\) 90)/6.022 \(\times\) 10\(^{23}\) = 5.48 \(\times\) 10\(^{-22}\) g.
3. Volume = (5 \(\times\) 10\(^{-8}\))\(^3\) = 1.25 \(\times\) 10\(^{-22}\) cm\(^3\), so \(\rho\) = 4.38 g cm\(^{-3}\).

Teacher's Note:
1. First get the formula, then use the molar mass of the whole formula unit.
2. Z = 1 here.

Answer: (c) 3.53 \(\mathring{A}\) and 1.25 \(\mathring{A}\)

Explanation:
1. \( a^3 = \dfrac{4 \times 58.89}{8.9 \times 6.022\times10^{23}} = 4.395\times10^{-23}\ \text{cm}^3 \), so a = 3.53 \(\times\) 10\(^{-8}\) cm = 3.53 \(\mathring{A}\).
2. \( r = \dfrac{a}{2\sqrt{2}} = \dfrac{3.53}{2.828} \) = 1.25 \(\mathring{A}\).

Teacher's Note:
1. Solve for a first, then for r.
2. Check: the known atomic radius of nickel is about 1.24 \(\mathring{A}\).

Answer: (c) Z \(\approx\) 4 (about 3.9); face-centred cubic (zinc blende type)

Explanation:
1. \( Z = \dfrac{\rho a^3 N_A}{M} = \dfrac{4.63 \times (5.87\times10^{-8})^3 \times 6.022\times10^{23}}{144.46} \).
2. a\(^3\) = 2.023 \(\times\) 10\(^{-22}\) cm\(^3\), so Z = 3.90.
3. The ideal value for the zinc blende structure is 4; the slight shortfall is due to a few vacancies.

Teacher's Note:
1. Z slightly below 4 points to Schottky defects.
2. The ideal Z would give a density of 4.74 g cm\(^{-3}\).