JEE Chemistry The Solid State MCQs Set 17

Answer: (b) Isotropic nature

Explanation:
1. Crystalline solids are anisotropic: properties such as refractive index and electrical conductivity depend on the direction of measurement.
2. Isotropy is a property of amorphous solids, so it is not a characteristic of crystalline solids.

Teacher's Note:
1. Remember: crystalline means anisotropic with a sharp melting point; amorphous means isotropic with no sharp melting point.

Answer: (b) Quartz glass (SiO\(_2\))

Explanation:
1. Quartz glass is a supercooled liquid with no long-range order, so it is amorphous.
2. Graphite, chrome alum and SiC are crystalline.

Teacher's Note:
1. Glass is the classic amorphous solid. Quartz (crystalline SiO\(_2\)) and quartz glass (amorphous SiO\(_2\)) are different forms of the same compound.

Answer: (c) Diamond

Explanation:
1. In diamond every carbon atom is joined to four others by covalent bonds throughout the crystal, giving a giant network.
2. I\(_2\), ice and dry ice are molecular solids.

Teacher's Note:
1. Diamond, SiC and quartz are the standard network solids. They are very hard with very high melting points.

Answer: (c) an electron

Explanation:
1. A lattice point is occupied by an atom, an ion or a molecule.
2. Electrons are not placed at lattice points; they move within or around the lattice (except in F-centres, which are defects).

Teacher's Note:
1. Constituent particles of a crystal are atoms, ions or molecules. Electrons trapped in anion vacancies are a defect, not a normal lattice occupant.

Answer: (b) equal numbers of cations and anions are missing from the lattice

Explanation:
1. In a Schottky defect, equal numbers of cations and anions leave the lattice, so the crystal stays electrically neutral.
2. The displacement of an ion into an interstitial site describes a Frenkel defect.

Teacher's Note:
1. Schottky: vacancies of both ions, density decreases. Frenkel: ion shifts to an interstitial site, density unchanged.

Answer: (b) electrically neutral

Explanation:
1. Doping silicon with a group 13 element creates holes, but the dopant atoms and silicon atoms are all neutral.
2. Hence the crystal as a whole is neutral.

Teacher's Note:
1. Both n-type and p-type semiconductors are electrically neutral. The names refer to the charge carriers.

Answer: (c) 32

Explanation:
1. The packing efficiency of bcc is 68%.
2. Empty space = 100 \(-\) 68 = 32%.

Teacher's Note:
1. Do not confuse packing efficiency (68%) with empty space (32%).

Answer: (a) A, B and C only

Explanation:
1. hcp has coordination number 12 and 74% efficiency, so A and B are true.
2. In the ABAB... pattern the third layer sits over the tetrahedral voids of the second layer, so C is true.
3. The fourth layer is aligned with the second layer, not the first, so D is false.

Teacher's Note:
1. hcp is ABAB..., while ccp is ABCABC.... In hcp the third layer repeats the first.

Answer: (a) A and D

Explanation:
1. A is rock salt, with 6:6 coordination.
2. B is fluorite, with 8:4. C is antifluorite, with 4:8.
3. D is zinc blende, with 4:4.
4. So A and D have equal coordination numbers.

Teacher's Note:
1. Equal coordination numbers occur when cation and anion numbers are equal (1:1 compounds such as NaCl, ZnS, CsCl).

Answer: (d) Electronic defect

Explanation:
1. Doping adds extra electrons or creates holes.
2. These are defects in the electronic structure of the crystal, called electronic defects.

Teacher's Note:
1. Doping makes impurity defects that change the number of charge carriers, which are called electronic defects.

Answer: (a) A, B and C only

Explanation:
1. A, B and C are correct statements.
2. The type of void occupied depends on the radius ratio, but the fraction of voids occupied is fixed by the formula (stoichiometry) of the compound, not by radii, so D is false.

Teacher's Note:
1. Radius ratio decides which kind of void; stoichiometry decides how many of them are filled.

Answer: (d) 8 tetrahedral voids within the unit cell

Explanation:
1. Tetrahedral voids = 2 \(\times\) 4 atoms = 8.
2. They lie on the four body diagonals, entirely inside the cell, so none is shared.

Teacher's Note:
1. Octahedral voids (4) are at the body centre and the edge centres. Tetrahedral voids (8) are wholly inside the cell.

Answer: (c) \( r_{Cs^+} + r_{Cl^-} = \frac{\sqrt{3}a}{2} \)

Explanation:
1. Cs\(^+\) at the body centre touches Cl\(^-\) at the corners.
2. Half of the body diagonal is \( \frac{\sqrt{3}a}{2} \), and this equals the sum of the two radii.

Teacher's Note:
1. In CsCl the ions touch along the body diagonal, but the cation-anion distance is only half of it.

Answer: (b) 4.08%

Explanation:
1. Let x be the amount of M\(^{3+}\). Charge balance: \( 3x + 2(0.98 - x) = 2 \), so \( x + 1.96 = 2 \) and \( x = 0.04 \).
2. Percentage = \( \frac{0.04}{0.98} \times 100 = 4.08\% \).

Teacher's Note:
1. Write the charge balance for one formula unit, then divide by the total metal content.

Answer: (d) 152 pm

Explanation:
1. For bcc, \( \sqrt{3}a = 4r \).
2. \( r = \frac{\sqrt{3} \times 351}{4} = 152 \) pm.

Teacher's Note:
1. For bcc use \( r = \frac{\sqrt{3}}{4}a \), for fcc \( r = \frac{a}{2\sqrt{2}} \).

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

Explanation:
1. Number of formula units = \( \frac{1.00}{58.5} \times 6.022 \times 10^{23} = 1.03 \times 10^{22} \).
2. NaCl is fcc with 4 formula units per unit cell.
3. Unit cells = \( \frac{1.03 \times 10^{22}}{4} = 2.57 \times 10^{21} \).

Teacher's Note:
1. Always divide the number of formula units by Z (4 for NaCl) to get the number of unit cells.

Answer: (b) 26% and 32%

Explanation:
1. ccp efficiency = 74%, so free space = 26%.
2. bcc efficiency = 68%, so free space = 32%.

Teacher's Note:
1. Free space = 100% minus packing efficiency.

Answer: (c) 2, 4

Explanation:
1. bcc: 8 \(\times \frac{1}{8}\) + 1 = 2.
2. fcc: 8 \(\times \frac{1}{8}\) + 6 \(\times \frac{1}{2}\) = 4.

Teacher's Note:
1. Atoms per cell: sc 1, bcc 2, fcc 4.

Answer: (b) AB\(_3\)

Explanation:
1. A = 8 \(\times \frac{1}{8}\) = 1.
2. B = 6 \(\times \frac{1}{2}\) = 3.
3. Formula = AB\(_3\).

Teacher's Note:
1. Corner atoms count 1/8 each and face-centre atoms count 1/2 each.

Answer: (c) 32%

Explanation:
1. Packing efficiency of bcc = 68%.
2. Vacant space = 32%.

Teacher's Note:
1. bcc 32% vacant, fcc and hcp 26% vacant, simple cubic 48% vacant.

Answer: (d) \( \frac{\sqrt{3}}{2}a \)

Explanation:
1. The body diagonal of the cube is \( \sqrt{3}a \).
2. The body centre is at its midpoint, so the distance to a corner is \( \frac{\sqrt{3}}{2}a \).

Teacher's Note:
1. Body diagonal \( = \sqrt{3}a \), face diagonal \( = \sqrt{2}a \).

Answer: (d) 8

Explanation:
1. Diamond has carbon atoms at the fcc positions (8 \(\times \frac{1}{8}\) + 6 \(\times \frac{1}{2}\) = 4).
2. Four more occupy alternate tetrahedral voids.
3. Total = 8.

Teacher's Note:
1. Diamond has the zinc blende arrangement with C in place of both Zn and S.

Answer: (a) 1

Explanation:
1. In ccp there are 4 atoms and 4 octahedral voids in a unit cell.
2. So there is 1 octahedral void per atom.

Teacher's Note:
1. Octahedral voids = N and tetrahedral voids = 2N for N atoms in close packing.

Answer: (a) 8

Explanation:
1. Diameter of one atom = 5.2 Å = \( 5.2 \times 10^{-8} \) cm.
2. Number of atoms = \( \frac{2.50}{5.2 \times 10^{-8}} = 4.81 \times 10^{7} \).
3. Moles = \( \frac{4.81 \times 10^{7}}{6.022 \times 10^{23}} = 7.98 \times 10^{-17} \approx 8 \times 10^{-17} \).

Teacher's Note:
1. Atoms touch along the row, so each atom takes up one diameter, not one radius.