The significance of a lattice point is that each lattice point represents one constituent particle of a solid which may be an atom, a molecule (group of atom), or an ion.

Question - 13 : -
Name the parameters that characterize a unit cell.

Answer - 13 : -

The six parameters that characterise a unit cell are as follows.

(i) Its dimensions along the three edges, a, b, and c

These edges may or may not be equal.

(ii) Angles between the edges

These are the angle ∝ (between edges b and c), β (between edges a and c), and γ (between edges a and b).

Question - 14 : -
Distinguish between

(i)Hexagonal and monoclinic unit cells

(ii) Face-centred and end-centred unit cells.

Answer - 14 : -

(i) Hexagonal unit cell

For a hexagonal unit cell,

Monoclinic unit cell

For a monoclinic cell,

(ii) Face-centred unit cell

In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face.

End-centred unit cell

An end-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.

Question - 15 : -
Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.

Answer - 15 : -

(i)An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells.

Therefore, portion of the atom is shared by one unit cell.

(ii)An atom located at the body centre of a cubic unit cell is not shared by its neighbouring unit cell. Therefore, the atom belongs only to the unit cell in which it is present i.e., its contribution to the unit cell is 1.

Question - 16 : -
What is the two dimensional coordination number of a molecule in square close packed layer?

Answer - 16 : -

In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two-dimensional coordination number of a molecule in square close-packed layer is 4.

Question - 17 : -
A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

Answer - 17 : -

Number of close-packed particles = 0.5 × 6.022 × 10_23 = 3.011 × 10_23

Therefore, number of octahedral voids = 3.011 × 10_23

And, number of tetrahedral voids = 2 × 3.011 × 10_23 = 6.022 ×10_23

Therefore, total number of voids = 3.011 × 10_23 + 6.022 × 10_23 = 9.033 × 10_23

Question - 18 : -
A compound is formed by two elements M and N.The element N forms ccp and atoms of M occupy 1/3^rd oftetrahedral voids. What is the formula of the compound?

Answer - 18 : -

The ccp lattice is formed bythe atoms of the element N.

Here, the number of tetrahedral voids generatedis equal to twice the number of atoms of the element N.

According to the question, the atoms of element M occupyof the tetrahedral voids.

Therefore, the number ofatoms of M is equal to

of the number

of atoms of N.

Therefore, ratio of thenumber of atoms of M to that of N is M: N

Thus, the formula of the compound is M₂ N₃.

Question - 19 : -
Which of the following lattices has the highestpacking efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonalclose-packed lattice?

Answer - 19 : -

Hexagonal close-packed lattice has the highestpacking efficiency of 74%. The packing efficiencies of simple cubic andbody-centred cubic lattices are 52.4% and 68% respectively.

Question - 20 : -
An element with molar mass 2.7 × 10^-2 kgmol^-1 forms a cubic unit cell with edge length 405 pm. If itsdensity is 2.7 × 103 kg m⁻³, what is the nature of the cubic unitcell?

Answer - 20 : -

It is given that density of the element, d =2.7 × 10³ kg m⁻³

Molar mass, M = 2.7 × 10⁻² kgmol⁻¹

Edge length, a = 405 pm =405 × 10⁻¹² m

= 4.05 × 10⁻¹⁰ m

It is known that, Avogadro’s number, N_A =6.022 × 1023 mol⁻¹

Applying the relation,

This implies that four atoms of the element arepresent per unit cell. Hence, the unit cell is face-centred cubic (fcc)or cubic close-packed (ccp).

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Chapter 1 The Solid State Solutions

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