# Solid State

## InText Solution 2

Question: 1.13 - Explain how much portion of an atom located at (i) corner and (ii) bodycentre of a cubic unit cell is part of its neighbouring unit cell.

1. Atom located at the corner is shared among eight adjacent unit cell, thus only 1/8 th portion of the atom is located at corner.
2. Atom located at body center does not share any part of its neighbouring unit cell, thus whole portion of atom is located at body center of cubic unit cell.

Question: 1.14 - What is the two dimensional coordination number of a molecule in square close-packed layer?

Answer: The coordination number of a molecule in two dimensions in square close packed layer is 4.

Question: 1.15 - 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: Since number of particles present in 1 mol of compound =6.022xx10^(23)
Hence, number of particles in 0.5 mol =0.5xx6.022xx10^(23)=3.01
We know that number of octahedral voids = Number of atoms or particles
And number of tetrahedral voids =2xx Number of particles
Therefore, number of octahedral in the given compound =3.011xx10^23
Number of tetrahedral voids = 2xx3.011xx10^(23)=6.022xx10^(23)
Thus total number of voids =3.011xx10^(23)+6.022xx10^(23)=9.033xx10^(23)
And number of tetrahedral voids =6.022xx10^(23)

Question: 1.16 A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?

Answer: Let the number of octahedral voids occupied by element N = a

Therefore total number of tetrahedral voids = 2a

Since element M occupies 1/3 of tetrahedral voids

So, number of tetrahedral voids occupied by M=2a\xx1/3=(2a)/3

So, ratio of M and N =(2a)/3:a=2:3

Therefore, formula of compound wil be M2N2

Question: 1.17 Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?

Answer: The packing efficiency of simple cubic lattice is 52.4%, body centered is 68% and that of hexagonal close packed lattice is 74%.

Therefore, (iii) hexagonal close packed lattice has highest packing efficiency, i.e. 74%.

Question: 1.18 - An element with molar mass 2.7 xx 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 xx 103 kg m-3, what is the nature of the cubic unit cell?

Answer: By knowing the number of atom in the cubic unit cell of given lattice, its nature can be determined.

Given density (d) =2.7xx10^3 text(kg)m^(-3)

Molar mass (M) =2.7xx10^(-3)kg

Edge (a) =405 text(pm)=(405)/(10^(12))

=405xx10^(-12) m

Therefore, number of atom = ?

We know that d=(z\xx\M)/(a^3xx\N_A)

Or, 2.7xx10^3text(kg)m^(-3)=(z\xx2.7xx10^(-2) kg\text(mol)^(-1))

Or, z=(2.7xx10^3text(kgm)^(-3)xx(405xx10^(-12)m)^3xx6.022xx10^(23))/(2.7xx10^(-2) kg\text(mol)^(-1))

Since number of atoms in unit cell of the given element = 4

Thus lattice is cubic close packed (ccp)