JEE Chemistry The Solid State MCQs Set 20

Answer: (c) \( \ce{X7Y8} \)

Explanation:
1. Corner atoms contribute 1/8 each: X = 7 x 1/8 = 7/8.
2. Body centre gives Y = 1.
3. X : Y = 7/8 : 1 = 7 : 8.

Teacher's Note:
1. Always count the removed atom as 1/8, not as a whole atom.
2. Multiply by 8 to clear the fraction.

Answer: (a) F-centres, i.e. electrons trapped in anion vacancies created by missing Cl\(^-\) ions

Explanation:
1. Sodium atoms deposit on the surface and Cl\(^-\) ions diffuse to it, leaving anion vacancies.
2. The sodium atoms lose electrons to become Na\(^+\); the electrons diffuse into the crystal and occupy the anion vacancies.
3. These trapped electrons absorb visible light, giving the colour (F-centres).

Teacher's Note:
1. F comes from the German Farbe, colour.
2. The result is a metal-excess defect.

Answer: (d) \( 2x+2y=a \)

Explanation:
1. In NaCl the ions touch along the cell edge: Cl\(^-\), Na\(^+\), Cl\(^-\).
2. Along the edge the sequence is Cl-, Na+, Cl- sharing corner anions, so the edge equals two cation-anion distances: a = 2(r+ + r-).

Teacher's Note:
1. Half the edge is the cation-anion distance, a/2 = x + y.
2. The same relation holds for every rock-salt crystal.

Answer: (c) The positive and negative ions are of nearly the same size

Explanation:
1. Schottky defects need both a cation and an anion to leave their sites.
2. This is easy when the ions are similar in size and the coordination number is high, as in NaCl, KCl and CsCl.

Teacher's Note:
1. Frenkel defect, in contrast, needs a small cation and a large anion (AgCl, ZnS).
2. Schottky defects lower the density.

Answer: (b) 0.225 to 0.414

Explanation:
1. Radius ratio 0.155 - 0.225: triangular (CN 3). 0.225 - 0.414: tetrahedral (CN 4).
2. 0.414 - 0.732: octahedral (CN 6). 0.732 - 1: cubic (CN 8).

Teacher's Note:
1. The limits come from geometry: a tetrahedral void has \( r/R=0.225 \), an octahedral one 0.414.
2. ZnS has r+/r- about 0.40.

Answer: (a) The ability of layers of metal ions to slide over one another in the sea of electrons

Explanation:
1. In a metal the delocalised electrons bind all the kernels together without fixed bonds.
2. When a layer slides, the bonding environment stays the same, so the metal deforms without breaking.

Teacher's Note:
1. In an ionic crystal sliding brings like charges together and the crystal cleaves.
2. Malleable means hammered into sheets; ductile means drawn into wires.

Answer: (d) \( \ce{XYZ3} \)

Explanation:
1. X = 8 x 1/8 = 1; Y = 1 (body centre); Z = 6 x 1/2 = 3.
2. Formula XYZ3 (this is the perovskite type).

Teacher's Note:
1. Corner 1/8, edge 1/4, face 1/2, body centre 1.
2. CaTiO3 is the classic example.

Answer: (c) Both Schottky and Frenkel defects

Explanation:
1. Ag\(^+\) is small, so it can move to an interstitial site (Frenkel).
2. AgBr also has a fairly high coordination number and the ions are not too different in size, so Schottky defects occur too.

Teacher's Note:
1. AgBr is the standard example of a crystal showing both defects.
2. Frenkel defect does not change the density.

Answer: (a) \( \dfrac{4}{\sqrt{2}}\,r \)

Explanation:
1. Atoms touch along the face diagonal: \( \sqrt{2}a=4r \).
2. \( a=\dfrac{4r}{\sqrt{2}}=2\sqrt{2}\,r \).

Teacher's Note:
1. For bcc the touching direction is the body diagonal, \( a=4r/\sqrt3 \).
2. For simple cubic a = 2r.

Answer: (b) \( \dfrac{\sqrt{3}}{2}\,a \)

Explanation:
1. CsCl is body-centred cubic: the Cs\(^+\) ion lies at the body centre and touches Cl\(^-\) at the corners.
2. The contact is along half the body diagonal: \( r_++r_-=\dfrac{\sqrt{3}a}{2} \).

Teacher's Note:
1. The body diagonal of a cube is \( \sqrt{3}a \).
2. For NaCl the same distance is a/2.

Answer: (c) \( 5\times10^{24} \)

Explanation:
1. Volume = \( (2\times10^{-8})^3=8\times10^{-24} \) cm\(^3\); mass of a cell = 8 x 10^-23 g.
2. Cells in 100 g = 100 / (8 x 10^-23) = 1.25 x 10^24.
3. Atoms = 4 x 1.25 x 10^24 = 5 x 10^24.

Teacher's Note:
1. fcc has 4 atoms per cell.
2. Equivalent molar mass: 12.04 g/mol, so 100 g is 8.3 mol.

Answer: (b) \( \dfrac{1}{4} \) th of the distance from a corner

Explanation:
1. Zn\(^{2+}\) ions occupy alternate tetrahedral voids.
2. A tetrahedral void is at (1/4, 1/4, 1/4) from a corner, i.e. one quarter of the way along the body diagonal.

Teacher's Note:
1. The four Zn\(^{2+}\) ions sit on the four body diagonals, each at 1/4 (or 3/4) of the length.
2. Distance from corner = \( \sqrt3 a/4 \).

Answer: (a) F, F, T

Explanation:
1. S1 is false: cubic has three Bravais lattices (simple, body-centred, face-centred).
2. S2 is false: diamagnetic substances are weakly repelled by a magnetic field.
3. S3 is true: covalent bonding within layers and van der Waals forces between layers.

Teacher's Note:
1. Paramagnetic substances are attracted; diamagnetic ones are repelled.
2. Graphite is soft because of its layered structure.

Answer: (d) 32%

Explanation:
1. Packing efficiency of bcc = 68%.
2. Vacant space = 100 - 68 = 32%.

Teacher's Note:
1. Efficiency: simple cubic 52.4%, bcc 68%, fcc and hcp 74%.
2. Vacant space in fcc is 26%.

Answer: (c) 100

Explanation:
1. Octahedral voids = number of atoms in the packing; tetrahedral voids = twice the number of atoms.
2. So 100 atoms give 100 octahedral voids (and 200 tetrahedral voids).

Teacher's Note:
1. The statement holds for both ccp and hcp.
2. Per atom: 1 octahedral, 2 tetrahedral.

Answer: (d) 1.11

Explanation:
1. For rock salt, a = 2(r+ + r-). Taking \( r_{Cl^-}=1 \) unit: a(NaCl) = 2(1.55), a(KCl) = 2(1.72).
2. Ratio = 1.72 / 1.55 = 1.11.

Teacher's Note:
1. The edge grows with the size of the cation because the anion is the same.
2. Actual edges: 629 pm for KCl and 564 pm for NaCl, ratio 1.12.

Answer: (b) 2 and 4

Explanation:
1. bcc: 8 x 1/8 + 1 = 2.
2. fcc: 8 x 1/8 + 6 x 1/2 = 4.

Teacher's Note:
1. Mg actually has hcp structure; the question treats it as fcc.
2. Simple cubic has 1 atom per cell.

Answer: (c) \( 2.57\times10^{21} \)

Explanation:
1. Moles of NaCl = 1/58.5; formula units = \( 6.022\times10^{23}/58.5=1.03\times10^{22} \).
2. A rock-salt unit cell has 4 formula units, so cells = \( 1.03\times10^{22}/4=2.57\times10^{21} \).

Teacher's Note:
1. Do not forget to divide by Z = 4.
2. Dividing by 8 (ions) would give half of this value.

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

Explanation:
1. fcc has 4 atoms per cell.
2. Volume = \( 4\times\dfrac{4}{3}\pi r^3=\dfrac{16}{3}\pi r^3 \).

Teacher's Note:
1. Check: cell volume \( (2\sqrt2 r)^3=16\sqrt2 r^3 \), efficiency = \( \pi/(3\sqrt2)=0.74 \).
2. bcc would give \( \dfrac{8}{3}\pi r^3 \).

Answer: (a) \( \ce{X4Y3} \)

Explanation:
1. Take N atoms of Y: tetrahedral voids = 2N.
2. X = (2/3)(2N) = 4N/3.
3. X : Y = 4/3 : 1 = 4 : 3.

Teacher's Note:
1. Tetrahedral voids are twice the number of atoms.
2. Octahedral voids equal the number of atoms.

Answer: (d) 127 pm

Explanation:
1. In fcc \( 4r=\sqrt{2}a \).
2. \( r=\dfrac{361}{2\sqrt{2}}=127.6\approx127 \) pm.

Teacher's Note:
1. Dividing by 2 only gives 180.5 pm, which is half the edge.
2. 108 pm is \( a/(2\sqrt3) \), wrong geometry.

Answer: (a) 26% and 32%

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

Teacher's Note:
1. Simple cubic has 48% free space.
2. Efficiency of bcc = \( \sqrt3\pi/8=0.68 \).