Chemistry

Solid State and Crystal Structures

151 Questions

Solid state chemistry focuses on the structural properties of crystalline solids. Questions cover Bravais lattices, unit cells, and close packed structures. This topic is essential for chemistry papers in UPSC and state PSC examinations.

Bravais lattice typesUnit cell calculationsClose packed structuresCrystal densityBragg's LawWurtzite structure

Solid State and Crystal Structures Questions

Multiple choice
  1. P only

  2. Q only

  3. R only

  4. P, Q

  5. Q, R

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

P. Pentaborane-11 (B5H11) has an unsymmetrical square pyramidal geometry, in which five boron atoms occupy the five corners of a square pyramid. Q. The crystal of CsBr has body centred cubic structure. R. MgAl2O4 and Fe3O4 have spinel structures. Hence, statement P and Q both are incorrect.

Multiple choice
  1. amorphous boron

  2. β-rhombohedral boron

  3. α-rhombohedral boron

  4. γ-boron

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

α-rhombohedral boron has a unit cell of twelve boron atoms. The structure consists of B12 icosahedra in which each boron atom has five nearest neighbors within the icosahedron. If the bonding were the conventional covalent type then each boron would have donated 5 electrons.

Multiple choice
  1. Cubic: 3 - face centred cubic

  2. Orthorhombic: 2 - body centred rectangular prism

  3. Tetragonal: 2 - body centred tetragonal prism

  4. Monoclinic: 2 - face centred parallelepiped

  5. Triclinic: 1 - triclinic parallelopiped

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Orthorhombic crystals have four space lattices with body centred rectangular prism arrangement.

Multiple choice
  1. 125.5 pm

  2. 128.3 pm

  3. 160.6 pm

  4. 187.35 pm

  5. 530 pm

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Edge length of the NaCl crystal = 2 x distance between Na+ and Cl- ions = 2 x 265 = 530 pm Now, for face centred cubic unit cell, a√2 = 4r r = a√2 / 4 = 530 x 1.414/4 = 187.35 pm. Hence, the apparent radius of NaCl crystal is 187.35 pm.

Multiple choice
  1. body centered cubic

  2. face centered cubic

  3. hexagonal closed packed

  4. body centered tetragonal

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Austenite is a solid solution of carbon in $\gamma - iron$. It has a solid solubility of upto 2% C at $1130^0C$

Multiple choice
  1. MNO

  2. MN3O

  3. MNO3

  4. M3NO

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Atom M is shared by 8 corners; 1 atom of O is shared by 4 unit cells. Atom N is present at centre of the unit cell. Hence, the effective number of atoms of M per unit cell = 8 X 1/8 = 1. The effective number of atoms of O per unit cell = 12/4 = 3.The effective number of atoms of N per unit cell = 1. Hence, the formula of the compound is MNO3.

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

In a solid $AB$ of $NaCl$ structure, A atoms occupy the corners of the cubic unit cell. If all the corner atoms are removed then the formula of the unit cell will be

  1. $A _{4}B _{4}$
  2. $B$
  3. $A _{3}B _{4}$
  4. $AB$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In an NaCl structure, A atoms are at corners and face centers (total 4). If all corner atoms (8 corners * 1/8 = 1 atom) are removed, 3 A atoms remain. B atoms are at edge centers and body center (total 4). The formula becomes A3B4.

Multiple choice chemistry atomic theory, periodic classification and properties of elements introduction to periodic table necessity of classification need for classification

A metal crystallizes with a face-centred cubic lattice. The edge length of the unit cell is 408 pm. The diameter of the metal atom is:

  1. 228 pm

  2. 288 pm

  3. 144 pm

  4. 326 pm

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a face-centered cubic (FCC) lattice, the relation between edge length (a) and atomic radius (r) is a = 2 * sqrt(2) * r. The diameter (d) is 2r. Thus, d = a / sqrt(2). Given a = 408 pm, d = 408 / 1.414 = 288.5 pm, which rounds to 288 pm.