JEE Chemistry The Solid State MCQs Set 21

Answer: (d) 4.08%

Explanation:
1. Let x be the amount of M\(^{3+}\); then M\(^{2+}\) is 0.98 - x.
2. Charge balance: 3x + 2(0.98 - x) = 2, so x = 0.04.
3. Fraction = 0.04/0.98 = 4.08%.

Teacher's Note:
1. This is a metal-deficiency defect (as in FeO).
2. Some M\(^{2+}\) ions are replaced by M\(^{3+}\) to keep the crystal neutral.

Answer: (b) \( r_{Cs^+}+r_{Cl^-}=\dfrac{\sqrt{3}}{2}\,a \)

Explanation:
1. The cation at the body centre touches the anions at the corners.
2. Half the body diagonal = \( \dfrac{\sqrt3 a}{2} \) = r+ + r-.

Teacher's Note:
1. Be careful: it is the sum of radii, not each radius.
2. Body diagonal = \( \sqrt3\,a \).

Answer: (c) 1.86 Å

Explanation:
1. In bcc \( 4r=\sqrt3\,a \).
2. \( r=\dfrac{1.732\times4.29}{4}=1.86 \) Å.

Teacher's Note:
1. Lithium (3.51 Å) gives 1.52 Å.
2. Check the edge-to-radius ratio: a / r = 2.31 for bcc.

Answer: (b) \( \ce{CrO2} \)

Explanation:
1. Among these oxides, CrO2 conducts like a metal and is ferromagnetic.
2. It is used in magnetic recording tapes.

Teacher's Note:
1. VO2 is a metallic or semiconducting oxide depending on temperature but not ferromagnetic; TiO2 is an insulator.
2. CrO2 is the standard example asked in the exam.

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

Explanation:
1. Atoms touch along the face diagonal, \( \sqrt2 a=4r \).
2. Closest approach = 2r = \( \dfrac{\sqrt2 a}{2}=\dfrac{a}{\sqrt2} \).

Teacher's Note:
1. In bcc the closest approach is \( \sqrt3 a/2 \).
2. In simple cubic it is a.

Answer: (c) Each sodium ion is octahedrally surrounded by 6 chloride ions and each chloride ion is octahedrally surrounded by 6 sodium ions

Explanation:
1. NaCl has 6 : 6 coordination.
2. Cl\(^-\) forms the fcc lattice and Na\(^+\) fills all octahedral voids, so each ion has 6 neighbours of the opposite kind in an octahedron.

Teacher's Note:
1. The structure is symmetric under exchange of Na and Cl.
2. 4 : 4 coordination belongs to zinc blende.

Answer: (a) 14 atoms of the element, of which 4 belong to the unit cell

Explanation:
1. Count: 8 corners + 6 face centres = 14 atoms touch the cell.
2. Contribution: 8 x 1/8 + 6 x 1/2 = 4 atoms belong to it.

Teacher's Note:
1. Distinguish between atoms located on the cell and atoms belonging to it.
2. Effective atoms: sc 1, bcc 2, fcc 4.

Answer: (d) n-type semiconductor

Explanation:
1. Germanium has 4 valence electrons, arsenic has 5.
2. Each As atom donates one extra electron, giving an n-type semiconductor.

Teacher's Note:
1. Doping with group 13 (B, Al, Ga) gives p-type.
2. Intrinsic means undoped.

Answer: (b) Covalent (network) solid

Explanation:
1. SiC has a diamond-like three-dimensional network of Si-C covalent bonds.
2. It is very hard with a very high melting point.

Teacher's Note:
1. Other network solids: diamond, silica, AlN.
2. Carborundum is used as an abrasive.

Answer: (c) Monoclinic

Explanation:
1. Three unequal edges with exactly one angle (beta) different from 90 degrees is the monoclinic system.
2. Orthorhombic has all three angles equal to 90 degrees.

Teacher's Note:
1. Triclinic: all angles different from 90 degrees.
2. Monoclinic sulphur and Na2SO4.10H2O are examples.

Answer: (b) \( 3\times10^{23} \)

Explanation:
1. Moles = 117/58.5 = 2 mol.
2. Formula units = 2 x 6.022 x 10^23 = 1.2 x 10^24; unit cells = this / 4 = 3 x 10^23.

Teacher's Note:
1. Each NaCl cell holds 4 formula units.
2. 6 x 10^23 would be the number of formula units in 1 mol.

Answer: (d) Orthorhombic

Explanation:
1. A matchbox is a cuboid with three unequal edges and all angles 90 degrees.
2. That is the orthorhombic system.

Teacher's Note:
1. Tetragonal has two equal edges.
2. Cubic needs all three equal.

Answer: (c) 2 and 8

Explanation:
1. bcc: 8 x 1/8 + 1 = 2 atoms.
2. The central atom touches the 8 corner atoms, so the coordination number is 8.

Teacher's Note:
1. fcc: 4 atoms, CN 12.
2. Simple cubic: 1 atom, CN 6.

Answer: (a) 173 pm

Explanation:
1. \( r=\dfrac{\sqrt3}{4}a=\dfrac{1.732\times400}{4}=173 \) pm.

Teacher's Note:
1. 141 pm would be \( a/(2\sqrt2) \), the fcc value.
2. 200 pm is a/2, the simple cubic value.

Answer: (d) 4

Explanation:
1. Volume of the hexagonal cell = \( \dfrac{\sqrt3}{2}a^2c=0.866\times(4.53)^2\times7.41=131.7 \) Å\(^3\) = \( 1.317\times10^{-22} \) cm\(^3\).
2. Mass = 0.92 x 1.317 x 10^-22 = 1.21 x 10^-22 g.
3. Molecules = mass / 18 x N_A = 4.05, i.e. 4.

Teacher's Note:
1. The base of the hexagonal cell is a rhombus of area \( \dfrac{\sqrt3}{2}a^2 \).
2. The number of molecules must be a whole number, so the calculated 4.05 is rounded to 4.

Answer: (b) Schottky defect

Explanation:
1. Ideal density = \( \dfrac{4\times119}{6.022\times10^{23}\times(6.58\times10^{-8})^3}=2.77 \) g cm\(^{-3}\).
2. Observed density 2.44 g cm\(^{-3}\) is smaller, so about 3.5 formula units per cell instead of 4: some ions are missing.
3. Equal numbers of M\(^+\) and X\(^-\) missing is a Schottky defect.

Teacher's Note:
1. Frenkel defect leaves the density unchanged.
2. Interstitial atoms would raise the density.

Answer: (c) \( \ce{La^{3+}},\ \ce{2I-}\ \text{and}\ \ce{e-} \)

Explanation:
1. Lanthanum prefers the +3 state, so the solid is La\(^{3+}\)(I\(^-\))\(_2\)(e\(^-\)).
2. The extra electron is delocalised and gives metallic conduction.

Teacher's Note:
1. Similar delocalisation occurs in LaI2 and CeI2.
2. Electrons are not in a separate I2 molecule.

Answer: (a) Particle size

Explanation:
1. Both forms have the same composition; the red form has larger particles and the yellow form finer particles.
2. Particle size changes how light is scattered and absorbed.

Teacher's Note:
1. Defects do not cause this colour difference.
2. Hg(II) is d10 so there is no d-d transition.

Answer: (d) NiO becomes a p-type semiconductor

Explanation:
1. Li\(^+\) replaces Ni\(^{2+}\) in the lattice, so to keep the charge balance some Ni\(^{2+}\) is oxidised to Ni\(^{3+}\).
2. The Ni\(^{3+}\) acts like a positive hole, so conduction is p-type.

Teacher's Note:
1. Doping with a lower-valent ion creates positive holes.
2. The same idea applies to ZnO doped with Ga2O3 (opposite).

Answer: (c) Vacant K\(^+\) sites and a lower density than pure KCl

Explanation:
1. Each Ca\(^{2+}\) replaces two K\(^+\) ions to keep the crystal neutral, so one cation site becomes vacant.
2. Ca (40) and K (39) have almost equal mass, so removing one ion per Ca lowers the mass per cell and hence the density.

Teacher's Note:
1. This is an impurity defect (cation vacancy).
2. The number of Cl\(^-\) sites is unchanged.

Answer: (a) A p-type semiconductor

Explanation:
1. Boron has 3 valence electrons, carbon has 4.
2. Each B leaves an electron vacancy (hole) in the bonding network, so the doped diamond is p-type.

Teacher's Note:
1. Nitrogen has 5 electrons and would give an n-type material.
2. Diamond itself is an insulator.

Answer: (d) \( \dfrac{a^3}{8} \)

Explanation:
1. Take cube 1 from 0 to a on each axis; cube 2 has a corner at (a/2, a/2, a/2) and extends to 3a/2.
2. The overlap along each axis is from a/2 to a, a length a/2.
3. Common volume = (a/2)^3 = a^3/8.

Teacher's Note:
1. The overlap is itself a cube.
2. One of its corners is the body centre and the opposite one is a corner of cube 1.