Unit Cells, Lattices & Bragg's Equation (Old NCERT) - Free MCQ Practice

14 types of three-dimensional Bravais lattices (primitive and centred unit cells

Lattice sites always contain atoms/molecules/ions. If atom is replaced by electron, then imperfection arises in the crystal and it would not be pure crystal

Ionic compounds are held by electrostatic forces of attraction between anions and cations.  These compounds form an extensive a 3 dimensional network where the boundaries between molecules are difficult to determine. The formula of ionic compounds are the empirical formula and are called the formula units.

The number of Na+ ions present in one unit cell = 12×1/4 + 1 =4 atoms per unit cell.

The number of Cl- ions present in one unit cell = 6×1/2 + 8×1/8 = 4 atoms per unit cell.

So, each unit cell has 4 atoms of each Na+ and Cl-. Hence there are total 4 molecules of NaCl in each unit cell.

Hence B

(c) By Bragg's equation, spacing between two planes is given by

For first-order diffraction, n =1

∴             
If d = λ, then 2 sin θ =1

if d = λ/2, then 
sin θ = 1 => θ = 90^o

In a unit cell W atoms at the corner 

O-atoms at the centre of edges = 
Na-atoms at the centre of the cube = 1
W:O:Na = 1:3:1
Hence, formula is NaWO₃

 (b) Number of atoms in the unit cell

where, n_c = number of atoms at the corners n_f = number of atoms at the faces n_i = number of atoms inside 

(c) By using Bragg’s equation,
nλ = 2d sin θ
where, n =2  
(second order diffraction)
λ = d sin θ
or
log λ = log d + log sin θ
d is constant (distance between two planes) as θ changes, X also changes.
When θ = 90°, sin 90° =1, hence log sin 90°= 0

Correct Answer :- a,c,d

Explanation : orthorhombic : a ≠ b ≠ c, γ = 90

monoclinic : a ≠ b ≠ c, γ = 90, b ≠ 90

triclinic : a ≠ b ≠ c,  α ≠ β ≠ γ

Explanation: Maximum value of incident angle can be 90° for which sine is 1. Hence d = λ/2 ( nλ = 2.d.sinθ)

Orthorhombic lattice exists as all four types of unit cells. Besides primitive cell, monoclinic lattice exists only as end centered cell while tetragonal exists only as body centered lattices.

(b)
Calcium appears at eight corners, hence number of calcium atoms = 8/8 =1
Oxygen appears on six faces and two faces share one oxygen atom hence, number of oxygen atoms = 6/2 =3
Titanium appears at the centre, thus, number of Ti-atom = 1/1 =1
Thus, A is CaTi0₃.
Let the oxidation number of Ti = x
CaTi0₃
(+2) + x + 3 ( - 2 ) = 0
2 + x - 6 = 0
x = + 4

(c)
Calcium appears at eight corners,hence number of calcium atoms = 8/8 =1
Oxygen appears on six faces and two faces share one oxygen atom hence, number of oxygen atoms = 6/2 =3
Titanium appears at the centre, thus, number of Ti-atom = 1/1 =1
Thus, A is CaTi0₃.
Let the oxidation number of Ti = x CaTi0₃
(+2) + x + 3 ( - 2 ) = 0
2 + x - 6 = 0
x = + 4

(i) Triclinic, a b c and α β γ = 90° Thus, (i)→ (r)

(ii) Hexagonal, a = b c, α = β = 90°,γ = 120° Thus, (ii) → (s)

(iii) C u b ic ,a = b = c , α = β =γ = 90° Thus, (iii) → (p)

(iv) Tetragonal, a = b c, α = β =γ = 90^o 

Thus, (iv) → (q)

(6) Edge length : a, b, c and angles between the edges :  α,β,γ

 Primitive unit cells are

(i) Cubic

(ii) Tetragonal

(iii) Orthorhombic

(iv) Hexagonal

(v) Rhombohedral

(vi) Monoclinic

(vii) Triclinic

A rhombohedral is a prism whose base is shaped as a non-square rhombus. Hence two of its six faces are rhombuses while others are rectangles. This results in a geometry for which two of the axial angles are right angles but the third is not.

(1) By Bragg's equation,

n = 2 (second order diffraction)

(3)

Number of atoms in one unit cell of bcc structure =2
Thus, number of unit cells =  
Thus, y = 3