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Solid State Chemistry

Description: Chemistry Solid StateSolid State
Number of Questions: 15
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Which of the following solid state materials are examples of group 12 – 16 compounds?

  1. InSb, AlP, GaAs

  2. InSb, CdSe, GaAs

  3. ZnS, AlP, HgTe

  4. ZnS, CdSe, HgTe

  5. ZnS, CdSe, GaAs


Correct Option: D
Explanation:

ZnS, CdSe and HgTe are examples of groups 12 – 16 compounds.

Hexagonal crystal system is present in

  1. Graphite, ZnO and CdS

  2. NaCl, HgS and Graphite

  3. KNO3, BaSO4 and CaCO3

  4. ZnO, HgS and CdS

  5. CdS, NaCl and Graphite


Correct Option: A
Explanation:

Hexagonal crystal system is present in Graphite, ZnO and CdS.

The ionic compound, which crystallises in corundum structure is

  1. CaO

  2. SrCl2

  3. Li2O

  4. ZnS

  5. Al2O3


Correct Option: E
Explanation:

Corundum structure: Anions form hexagonal closest packing and cations are present in 2 / 3 of the octahedral holes. The general formula of the compound is M2O3. Examples: Fe2O3, Al2O3, Cr2O3, etc.

A compound is formed by two elements M and N. Atoms of the element N (as anions) make ccp and those of the element M (as cations) occupy all the octahedral voids. What is the formula of the compound?

  1. MN3

  2. M3N2

  3. MN2

  4. M2N

  5. MN


Correct Option: E
Explanation:

The ccp lattice is formed by the element N. The number of octahedral voids generated would be equal to the number of atoms of Y present in it. Since all the octahedral voids are occupied by the atoms of M, so their number would also be equal to that of the element N. Thus, the atoms of elements M and N are present in equal numbers or 1 : 1 ratio. Therefore, the formula of the compound is MN.

Which of the following is an example of n-type semiconductor?

  1. Silicon doped with boron

  2. Silicon doped with gallium

  3. Silicon doped with aluminium

  4. Germanium doped with aluminium

  5. Germanium doped with phosphorus


Correct Option: E
Explanation:

Germanium doped with phosphorus is an example of n-type semiconductor. Silicon and germanium are group 14 elements having four valence electrons. In their crystals, each atom forms four covalent bonds with its neighbours. When doped with a group 15 element like P or As, which contains five valence electrons, they occupy some of the lattice sites in silicon or germanium crystal. Four out of five electrons are used in the formation of four covalent bonds with the four neighbouring silicon atoms. The fifth electron is extra and becomes delocalised. These delocalised electrons increase the conductivity of doped silicon or germanium. Here the increase in conductivity is due to the negatively charged electron. Hence, silicon or germanium doped with electron-rich impurity are called n-type semiconductor.

Atoms of element Q form hcp lattice and those of the element P occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements P and Q?

  1. P3Q4

  2. P4Q3

  3. P2Q3

  4. P3Q2

  5. PQ


Correct Option: B
Explanation:

The number of tetrahedral voids formed is equal to twice the number of atoms of element Q and only 2 / 3rd of these are occupied by the atoms of element P. Hence, the ratio of the number of atoms of P and Q is 2 × (2 / 3) : 1 or 4 : 3 and the formula of the compound is P4Q3.

An element has a body-centred cubic (BCC) structure with a cell edge of 210 pm. The density of the element is 4.2 g/cm3. Number of atoms present in 210 g of the element is

  1. 5.03 x 1024

  2. 10.06 ×1024

  3. 20.12 × 1024

  4. 30.18 ×1024

  5. 40.24 ×1024


Correct Option: B
Explanation:

Volume of the unit cell = (210 pm)3= (210 ×10-12 m)3 = (210 × 10-10 cm)3 = 9.94 × 10-24 cm3 Volume of 208 g of the element = mass / density = 210 g / 4.2 g cm-3 = 50 cm3 Number of unit cells in this volume = 50 cm3 / 9.94 x 10-24 cm3 = 5.03 x 1024 Since each bcc cubic unit cell contains 2 atoms, so the total number of atoms in 210 g of the element = 2 (atoms / unit cell) × 5.03 x 1024 unit cells = 10.06×1024 atoms

In which of the following crystal structures are cations arranged into cubic close packing and the anions occupy all the tetrahedral voids?

  1. Rock-salt structure

  2. Fluorite structure

  3. Anti-fluorite structure

  4. Sphalerite structure

  5. Corundum structure


Correct Option: B
Explanation:

In fluorite structure, cations are arranged into cubic close packing and the anions occupy all the tetrahedral voids.

Which of the following statements are incorrect regarding rock-salt structure? P. Cations occupy the lattice points of expanded face centred cubic lattice and the anions occupy all the octahedral holes. Q. The limiting radius ratio is in between 0.414 and 0.732. R. The coordination numbers of cation and anion are 4 and 8 respectively. S. The general formula of ionic compounds having rock-salt structure is AB. Select the correct answer using codes given below:

  1. P and Q

  2. P and R

  3. P and S

  4. Q and R

  5. R and S


Correct Option: B
Explanation:

Rock-salt structure: Anions occupy the lattice points of expanded face centred cubic lattice and the cations occupy all the octahedral holes. The limiting radius ratio is in between 0.414 and 0.732. The co-ordination numbers for both cation and anion is 6 : 6. The general formula of ionic compounds having this structure is AB. Examples: NaCl, KCl, MgO, CaO etc.

Which of the following metals crystallise(s) in fcc arrangement?

  1. Copper and nickel

  2. Chromium and molybdenum

  3. Potassium and rubidium

  4. Cadmium and cobalt

  5. Barium and tungsten


Correct Option: A
Explanation:

Both copper and nickel crystallise in face centered cubic (fcc) arrangement.

Which of the following ionic compounds crystallises in fluorite structure?

  1. NaCl and KCl

  2. CaF2 and PbCl2

  3. Na2S and Cl2O

  4. ZnS and BeO

  5. Fe2O3 and Cr2O3


Correct Option: B
Explanation:

CaF2 and PbCl2 crystallise in fluorite structure.

Match the column – I (crystal system) with column – II (example) and select the correct option.

Column – I Column – II
P. Tetragonal 1. SnO2
Q. Orthorhombic 2. BaSO4
R. Hexagonal 3. ZnO
S. Trigonal 4. CaCO3
  1. P - 1, Q - 2, R - 3, S - 4

  2. P - 1, Q - 3, R - 2, S - 4

  3. P - 2, Q - 3, R - 4, S - 1

  4. P - 4, Q - 2, R - 3, S - 1

  5. P - 4, Q - 3, R - 2, S - 1


Correct Option: A
Explanation:

SnO2 crystallises to form tetragonal lattice structure. BaSO4 crystallises to form orthorhombic lattice structure. ZnO crystallises to form hexagonal lattice structure. CaCO3 crystallises to form trigonal lattice structure.

Match the entries in Group – I with molecules in Group – II and select the correct option.

Group – I Group – II
A. Rock-salt structure 1. MgO
B. Fluorite structure 2. Na2S
C. Anti-fluorite structure 3. CaF2
D. Sphalerite structure 4. ZnS
  1. A - 1, B - 2, C - 3, D - 4

  2. A - 1, B - 3, C - 2, D - 4

  3. A - 2, B - 1, C - 4, D - 3

  4. A - 3, B - 2, C - 4, D - 1

  5. A - 3, B - 2, C - 1, D - 4


Correct Option: B
Explanation:

Rock-salt structure: Anions occupy the lattice points of expanded face centred cubic lattice and cations occupy all the octahedral holes. The limiting radius ratio is in between 0.414 and 0.732. The coordination numbers of cation and anion are (6 : 6). The general formula of ionic compounds having this structure is AB. Examples: NaCl, KCl, MgO, CaO, etc. Fluorite structure: Cations are arranged into cubic close packing and anions occupy all the tetrahedral voids. Thus, there are four cations and eight anions per unit cell. The formula of ionic compound having this structure is A4B8 or AB2. The coordination numbers of cation and anion are (8 : 4). Examples: CaF2, ZrO2, ThO2, BaF2, BaCl2, SrCl2, PbCl2, etc. Anti-fluorite structure: Anions are arranged in cubic closest packing and cations occupy all the tetrahedral voids. There are four anions and eight cations per unit cell of this structure and hence, the general formula of an ionic compound is A8B4 or A2B. The ideal radius ratio is between 0.225 - 0.414 (but this ratio is not always maintained). The coordination numbers of cation and anion are (4 : 8). Examples: Na2O, K2O, Li2O, Rb2O, K2S, Cl2O, Na2S, etc. Zinc-blende or sphalerite structure: Anions occupy the face centred cubic lattice points and cations occupy half of the tetrahedral holes. The ideal radius ratio is in between 0.225 and 0.414. There are four anions and four cations in the unit cell. Therefore, the formula is A4B4 or AB. The coordination numbers of cation and anion are (4 : 4). Examples: ZnS, BeO, etc.

Which of the following statements is/are false?

  1. Packing efficiency of bcc lattice structure is 68%.
  2. Hcp and bcc lattice structures have same packing efficiency.
  3. Coordination number of atoms in bcc crystal lattice is 8. Select the correct answer using codes given below:
  1. 1 only

  2. 2 only

  3. 3 only

  4. 1 and 2

  5. 2 and 3


Correct Option: B
Explanation:

Packing efficiency in hcp lattice is 74%. Packing efficiency of bcc lattice structure is 68%.

Match the column – I (type of solid) with column – II (example) and select the correct option.

Column – I
Column – II
P. Molecular solid 1. AlN
Q. Ionic solid 2. Cu
R. Metallic solid 3. CaF2
S. Network solid 4. CCl4
  1. P - 1, Q - 2, R - 3, S - 4

  2. P - 1, Q - 3, R - 2, S - 4

  3. P - 2, Q - 3, R - 4, S - 1

  4. P - 4, Q - 2, R - 3, S - 1

  5. P - 4, Q - 3, R - 2, S - 1


Correct Option: E
Explanation:

Correct answer. CCl4 is an example of molecular solid. CaF2 is an example of ionic solid. Cu is an example of metallic solid. AlN is an example of network solid.

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