The Solid State

• Question 145

## Copper crystallizes into a fcc lattice with edge length 3.61 x 10–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm–3.

Solution

We have given that
Length of edge, a = 3.61x10–8 cm
Atomic mass of cupper, M = 63.55 g/mol
Avogadro constant, NA = 6.022x1023 per mol
Number of atoms in unit cell of FCC, Z= 4
Use formula of density,

Question 146

## Analysis shows that nickel oxide has formula Ni 0.98 O1.00 What fractions of Nickel exist as Ni2+ and Ni3+ ions?

Solution

Formula is Ni0.98O1.00
So the ration of Ni : O = 98:100
So if there are 100 atoms of oxygen then 98 atoms of Ni,
Let number of atoms of Ni+2 = x
Then number of atoms of Ni+3 = 98–x
Charge on Ni = charge on O
So that oxygen has charge –2
3(98–x) + 2x = 2 (100)
294 –3x +2x = 200
–x = – 94
x = 94
Percentage of Ni+2 = (atom of Ni+2/total number of atoms of Ni)100
=100(94/98)x100
= 96%
Percentage of Ni+3
=100 – Ni+2
=100 – 96
= 4 %

Question 147

## Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Solution
Unit cell is fcc, edge of unit cell, a = ?
In face centered unit cell, diagonal of face = 4r = $\mathrm{a}\sqrt{2}$
[∵ face diagonal = $\mathrm{a}\sqrt{2}$]

Therefore,

Length of side of unit cell,

a = 0.4073 nm.

Question 148

## Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is 125 pm.(a) What is the length of the side of the unit cell?(b) How many unit cells are there in 1.00 cm3 of aluminium?

Solution

Given, radius of atom (r) = 125 pm
(a) For ccp structure, we know that,