# The Solid State

Question 165

## An atom has fcc crystal whose density is 10 gm–3 and cell edge is 100 pm. How many atoms are present in its 100 g?

Solution

Solution:
We have given that
Density = 10 gm-3
Mass = 100g
edge of unit cell ,a= 4 since it is a Fcc crystal

we have to find total number of atom, So by following relation we can get the result,

Total No. of atoms = $\frac{\mathrm{Z}×\mathrm{M}}{{\mathrm{a}}^{3}.\mathrm{d}}$

Therefore, number of atoms

thus the number of atoms is 4 x 1031

Question 166

## Analysis shows that a metal oxide has the empirical formula M0.98 O1.00. Calculate the percentage of M2+ and M3+ ions in this crystal.OrIn an ionic compound the anion (N) form cubic close type of packing. While the cation (M+) ions occupy one third of the tetrahedal voids. Deduce the empirical formula of the compound and the coordination number of (M) ions.

Solution

Solution:

M0.98 O1.00 is non-stoichiometric compound and is a mixture of M2+ and M3+ ions.
Let x atoms of M3+ are present in the compound. This means that x M2+ have been replaced by M3+ ions.
∴  Number of M2+ ions = 0.96 – x.
For electrical neutrality, positive charge on compound

= negative charge on compound.
∴ 2(0.96 – x) + 3x = 2
1.92 – 2x + 3x = 2
or x = 2 – 1.92 = 0.08

∴

Thus, M2+ is 92% and M3+ is 8% in given sample having formula, M0.96 O1.00.

Or

The number of tetrahedral voids formed is equal

to twice the number of atoms of element N and only $\frac{1}{3}\mathrm{rd}$ of these are occupied by the element M. Hence the ratio of number of atoms of M and is   So, the formula of the compound is M2N3.

Question 167

## An element E crystallizes in body centred cubic structure. If the edge length of the cell is 1.469 x 10–10 m and the density is 19.3 g cm–3, calculate the atomic mass of this element. Also calculate the radius of this element.

Solution
Solution:
We have given that
Z= 2 since it is bcc structure
edge length, a = 1.469 x 10-10
Density, d= 19.3 g cm-3
Mass = ?

mass of element is 18.4g mol-1

(ii) Cell edge,

$\mathrm{a}=1.469×{10}^{-10}\mathrm{m}$

Body diagonal = 4R

Then 4R = $\sqrt{39}$

or

Radius of element is 6.4 x 10-11m.
Question 168

## Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic (ii) body- centred cubic (iii) face - centred cubic . (With the assumptions that atoms are touching each other).

Solution

Solution:

(i) Simple Cubic: A simple cubic unit cell has one sphere (or atom) per unit cell. If r is the radius of the sphere, then volume occupied by one sphere present in unit cell = $\frac{4}{3}{\mathrm{\pi r}}^{3}$

Edge length of unit cell (a) = r + r = 2r
Volume of cubic
Volume of occupied by sphere  = $\frac{4}{3}{\mathrm{\pi r}}^{3}$
Percentage volume occupied = percentage of efficiency of packing

For simple cubic metal crystal the efficiency of packing = 52.4%.

(b) Body centred cubic: From the figure it is clear that the atom at the centre will be in touch with other two atoms diagonally arranged and shown with solid boundaries.

The length of the body diagonal c is equal to 4r where r is the radius of the sphere (atom). But c = 4r, as all the three spheres along the diagonal touch each other

∴

or                   $\mathrm{a}=\frac{4\mathrm{r}}{\sqrt{3}}$

or                    $\mathrm{r}=\frac{\sqrt{3}}{4}\mathrm{a}.$

Fig. BCC unit cell. In this sort of structure total number of atoms is two and their volume is $2×\left(\frac{4}{3}\right){\mathrm{\pi r}}^{3}.$

Volume of the cube, a
3 will be equal to

Therefore, Percentage of efficiency Volume occupied by four - spheres

(c) Face centred cubic: A face centred cubic cell (fcc) contains four spheres (or atoms) per unit volume occupied by one sphere of radius

r = 4/3 $\mathrm{\pi }$r3

Volume occupied by four spheres present in the unit cell

r = 4/3 $\mathrm{\pi }$r3 x 4 = 16/3 $\mathrm{\pi }$r3

Fig.  Face centred cubic unit cell. Figure indicates that spheres placed at the corners touches a face centred sphere. Length of the face diagonal
= r + 2r + r = 4r

Hence, for face centred cubic, efficiency of packing = 74%.