# 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.

**Answer:**

- 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.
- 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 M_{2}N_{2}

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` 10^{3} 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)