JEE Chemistry The Solid State MCQs Set 14

Answer: (d) 5

Explanation:
1. Silicon has 4 valence electrons.
2. An n-type semiconductor is made by doping with an element having more than 4 valence electrons (group 15).
3. The extra electron becomes a free carrier, so the impurity has 5 valence electrons.

Teacher's Note:
1. n stands for negative charge carriers (electrons).
2. Group 15 dopants: P, As, Sb.

Answer: (a) \( \ce{MnO2} \)

Explanation:
1. \( \ce{MnO2} \) has Mn(IV) with unpaired d-electrons whose magnetic moments align in equal and opposite pairs, giving zero net moment.
2. \( \ce{TiO2} \) is diamagnetic and \( \ce{CrO2} \) is ferromagnetic.

Teacher's Note:
1. Antiferromagnetic: spins aligned in opposite directions and cancel.
2. Other examples are MnO and NiO.

Answer: (d) They exhibit directional properties of the bond

Explanation:
1. Ionic bonds arise from electrostatic attraction, which acts equally in all directions.
2. So the ionic bond is non-directional; statement (d) is wrong.
3. Statements (a), (b) and (c) are true for ionic solids.

Teacher's Note:
1. Directional bonding is a feature of covalent bonds.
2. Ionic crystals are hard, brittle and have high melting points.

Answer: (b) Graphite

Explanation:
1. In graphite each carbon is sp2 hybridised and uses three electrons in bonding.
2. The fourth electron is delocalised between the layers and can move freely.
3. Hence graphite conducts electricity.

Teacher's Note:
1. Diamond has all four electrons in localised bonds, so it is an insulator.
2. Silicon is only a semiconductor.

Answer: (c) In solid NaCl the ions are fixed and cannot move

Explanation:
1. Solid NaCl is built of \( \ce{Na+} \) and \( \ce{Cl-} \) ions held in fixed lattice positions.
2. Without mobile ions or electrons no current can pass.
3. It conducts when molten or dissolved because the ions become free.

Teacher's Note:
1. Do not say that there are no ions in the solid; they exist but are immobile.

Answer: (a) Silica

Explanation:
1. Quartz is crystalline \( \ce{SiO2} \).

Teacher's Note:
1. Quartz is a covalent network solid.
2. Glass is amorphous silica-based material, quartz is the crystalline form.

Answer: (b) Schottky defect

Explanation:
1. In a Schottky defect equal numbers of cations and anions are missing from the lattice.
2. Mass decreases while the volume stays the same, so the density decreases.
3. In a Frenkel defect an ion only shifts position, so the density is unchanged.

Teacher's Note:
1. Schottky defects occur in compounds with similar cation and anion size, such as NaCl and CsCl.
2. Frenkel defects occur in AgBr and ZnS.

Answer: (b) p-type semiconductor

Explanation:
1. Boron has 3 valence electrons, one fewer than silicon.
2. Each boron atom creates an electron hole.
3. Conduction by positive holes makes it a p-type semiconductor.

Teacher's Note:
1. Group 13 dopants give p-type, group 15 dopants give n-type.

Answer: (c) Presence of electrons trapped in the lattice

Explanation:
1. Excess sodium makes some anion sites vacant.
2. Electrons released from sodium atoms occupy these vacancies (F-centres).
3. Such trapped electrons absorb visible light and colour the crystal yellow.

Teacher's Note:
1. F stands for Farbe, the German word for colour.
2. This is a metal excess defect.

Answer: (a) \( 1\times10^{-3} \) mol %

Explanation:
1. Each \( \ce{Sr^2+} \) replaces two \( \ce{Na+} \) ions: one site is taken by Sr and the other becomes a vacancy.
2. So one vacancy is created per \( \ce{Sr^2+} \) ion.
3. Vacancies = \( 10^{-3} \) mol %.

Teacher's Note:
1. Aliovalent doping creates vacancies to keep the crystal neutral.

Answer: (c) 4

Explanation:
1. A ccp (fcc) unit cell has 4 atoms.
2. Number of octahedral voids = number of atoms in the packing = 4 (one at the body centre and 12 edge centres each shared by 4 cells: 1 + 3).

Teacher's Note:
1. Octahedral voids = N, tetrahedral voids = 2N for N atoms.

Answer: (a) 32%

Explanation:
1. Packing efficiency of bcc = \( \frac{2\times\frac{4}{3}\pi r^3}{a^3} \) with \( a=\frac{4r}{\sqrt3} \), which equals 68%.
2. Vacant space = 100 - 68 = 32%.

Teacher's Note:
1. Packing efficiency: simple cubic 52.4%, bcc 68%, fcc and hcp 74%.

Answer: (c) \( 1.5\times10^{23} \)

Explanation:
1. 58.5 g NaCl = 1 mol = \( 6.022\times10^{23} \) formula units.
2. A unit cell contains 4 formula units.
3. Unit cells = \( 6.022\times10^{23}/4\approx1.5\times10^{23} \)

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (a) \( 2.57\times10^{21} \) unit cells

Explanation:
1. Moles of NaCl = 1/58.5 = 0.01709 mol.
2. Formula units = \( 0.01709\times6.022\times10^{23}=1.029\times10^{22} \).
3. Unit cells = \( 1.029\times10^{22}/4=2.57\times10^{21} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (b) 3.72 Å

Explanation:
1. In a bcc ionic crystal the ions touch along the body diagonal.
2. Shortest distance = \( \frac{\sqrt3}{2}a \) = \( 0.866\times4.3 \) = 3.72 Å.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (a) \( 1.857\times10^{-8} \) cm

Explanation:
1. For bcc, \( \sqrt3 a=4r \).
2. r = \( \frac{\sqrt3\times4.29}{4} \) = 1.857 Å = \( 1.857\times10^{-8} \) cm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (a) 219.25 pm

Explanation:
1. For fcc, face diagonal = \( \sqrt2 a=4r \).
2. r = \( \frac{620}{2\sqrt2} \) = 219.2 pm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (a) Four Fe2+ and four O2-

Explanation:
1. Z = \( \frac{\rho a^3 N_A}{M} \) = \( \frac{4.0\times(5.0\times10^{-8})^3\times6.022\times10^{23}}{72} \) = 4.2.
2. Z is nearly 4, so there are 4 formula units: four \( \ce{Fe^2+} \) and four \( \ce{O^2-} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (d) 908

Explanation:
1. In bcc the nearest-neighbour distance = \( \frac{\sqrt3}{2}a \) so a = \( \frac{2\times4.52}{\sqrt3} \) = 5.22 Å.
2. Density = \( \frac{2\times39}{6.022\times10^{23}\times(5.22\times10^{-8})^3} \) = 0.91 g \( cm^{-3} \) = about 908 kg \( m^{-3} \).

Teacher's Note:
1. Convert g/cm3 to kg/m3 by multiplying by 1000.

Answer: (b) \( 2.81\times10^{-23} \)

Explanation:
1. a = \( 3.04\times10^{-8} \) cm.
2. V = \( a^3=(3.04\times10^{-8})^3=2.81\times10^{-23} \) \( cm^3 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (b) \( \ce{NaWO3} \)

Explanation:
1. W: 8 corners \( \times\frac18 \) = 1.
2. O: 12 edge centres \( \times\frac14 \) = 3.
3. Na: 1 at the body centre.
4. Formula = \( \ce{NaWO3} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (d) \( \ce{XY3} \)

Explanation:
1. X: \( 8\times\frac18=1 \).
2. Y: \( 6\times\frac12=3 \).
3. Ratio X : Y = 1 : 3, so the formula is \( \ce{XY3} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (d) \( \ce{A3B4} \)

Explanation:
1. Original cell: A at corners and face centres = \( 1+3=4 \); B at edge centres and body centre = \( 3+1=4 \).
2. Face centres along one axis are two opposite faces: 2 atoms \( \times\frac12 \) = 1 A removed.
3. A = 3, B = 4, so the formula is \( \ce{A3B4} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Answer: (c) Octahedral

Explanation:
1. Radius ratio 0.414 to 0.732 corresponds to coordination number 6.
2. Coordination number 6 means an octahedral arrangement.

Teacher's Note:
1. 0.225-0.414: tetrahedral (CN 4); 0.732-1: cubic (CN 8).

Answer: (b) CsCl type

Explanation:
1. Radius ratio = 180/187 = 0.963.
2. This lies in the range 0.732-1.0, so the coordination number is 8, which is the CsCl structure.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.