Given below are the Solid State Class 12 Important Questions
a. Concepts questions
b. Calculation problems
c. Short Answer
d. Cubic cell problems

Question 1
A compound is formed by 2 elements M & N. N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?

Question 2
An element with molar mass $2.7 \times 10^{-2}$ Kg/mol. forms a cubic units cell with edge length of 405 p. m. If the density is $2.7 \times 10^3 \ Kg/m^3$. Find the nature of a cubic unit cell.

Answer

$d = \frac {(Z \times M)}{(N_A \times a^3 )}$ -(1)
Now $d = 2.7 \times 10^3 \ Kg/m^3 = = 2.7 g/cm^3$.
$M = 2.7 \times 10^{-2} \ Kg/mol =27 \ g/mol$
Volume=$a^3 = (405 \ pm)^3 = (405 \times 10^{-10} \ cm)^3$
$N_A=6.02 \times 10^{23}$
Z=?
Substituting these values in (1)
$2.7 \times 10^3=\frac { Z \times 27}{6.02 \times 10^{23} \times (405 \times 10^{-10} \ cm)^3}$
Z= 3.99 = 4.
Hence the Nature of the Cubic cell is CCP

Question 3
Gold (Atomic Radius = 0.144 nm) crystallizes in FCC. Find the length of the side of the cell.

Answer

In F. C. C.
$r = \frac {a}{2 \sqrt 2}$
Therefore
$a= r \times 2 \sqrt 2 = .144 \times 2 \times 1.414=.407 \ nm$

Question 4
Alluminium crystallizes in cubic close-packed structure. it's metallic radius is 125 pico meter.
(i) Find the length of the side of the unit cell.
(ii) How many unit cells are present in 1 cm^{3} of element.

Answer

(i) a = ?
for C. C. P. (F. C. C.)
$r = \frac {a}{2 \sqrt 2}$
Therefore
$a= r \times 2 \sqrt 2 = 125 \times 2 \times 1.414=354 \ pm$
(ii)
1 unit cell = $a^3 = (354 \ pm)^3 = 4.4 \times 10^{-23} \ cm^3$
Number of Unit cells= $ \frac {1}{4.4 \times 10^{-23}} = 2.27 \times 10^{22}$

Question 5
Non-stoichiometric cuprous oxide,($Cu_2O%) can be prepared in laboratory. In this oxide Cu to O ratio is slightly less than 2 : 1. Can you account for the fact that this substance is a P – type semi – conductor?

Answer

The crystal is having cation vacancy defect, having holes in the sample. Thus it’s a P – type semi conductor because the electricity conduction is due to holes.

Question 6
What type of stoichimetric defect is shown by
(i) ZnS
(ii) AgBr

Answer

(i) ZnS - Frankel defect
(ii) Ag Br -Short key Frankel defect

Question 7
Ionic solids are hard and brittle. Why? Solution

Ionic solids are hard due to strong electrostatic force or ionic force.
Ionic solids are brittle due to non – directional nature of ionic bonds. Question 8
Classify each of the following as either a P – type or N – type semi – conductor
(i) Germanium doped with indium
(ii) Boron doped with silicon

Answer

(i) Ge belongs to group 14 & In belongs to group 13, In has 3 valence electrons while Ge has 4 valence elecltron.On replacement an electron deficient hole was created ,therefore it is a p-type semi conductor.

(ii) B belongs to group 13 with 3 valence electrons & Si belongs to group 14 with 4 valence electrons.therefore on replacement there will be a free electron & it is n-type semiconductor.

Question 9
A metal crystallizes into two cubic system-face centred cubic (fcc) and body centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0 Å respectively. Calculate the ratio of densities of fcc and bcc.

Answer

We know that
$d = \frac {(Z \times M)}{(N_A \times a^3 )}$
For fcc, z=4
Therefore
$d_1 = \frac {4 \times M }{ N_A \times (3.5 \times 10^{-8})^3} \ g/cm^3$
For bcc, z=2
Therefore
$d_2 = \frac {2 \times M }{ N_A \times (3.0 \times 10^{-8})^3} \ g/cm^3$
Now
$\frac {d_1}{d_2} = (4/(3.5 \times 10^{-8})^3) / (2/(3.0 \times 10^{-8})^3)$
= 1.26:1

Question 10
Niobium crystallises in body-centred cubic structure. If the atomic radius is 143.1 pm, calculate the density of Niobium. (Atomic mass = 93u).

Question 11
(i)If the radius of the octahedral void is 'r' and radius of the atoms in close packing is 'R'. What is the relation between 'r' and 'R'
(ii) Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm. What is the radius of tungsten atom?

Answer

(i)r = 0.414 R
(ii) $r= \frac {\sqrt 3}{4} a= \frac {\sqrt 3}{4} \times 316.5 =136.88 \ pm$

Question 12
(i)A metal crystallizes in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is
the relationship between 'r' and 'a'?
(ii) An element with molar mass 63 g / mol forms a cubic unit cell with edge length of 360.8 pm. If its density is 8.92 g/ cm^{3}
What is the nature of the cubic unit cell?

Question 13
An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate thedensity if 300 g of this element contain $2 \times 10^{24}$ atoms.

Question 14
For Fe , a=286 pm, Density = 7.86 g/cm^{3}. Atomic Weight of Fe=55.85
(i) Find the nature of a cubic unit cell.
(ii) Calculate the Radius of Iron

Answer

Z=2, So it BCC
r=123.8 pm

Question 15
Tungsten has Density of 19.35 g/cm^{3} The length of the side of the unit cell is 316 pm and has Body centered cubic unit cell. How many atoms of the elements does 50 g of the element contain?

Answer

for BCC ,Z=2
$d = \frac {(Z \times M)}{(N_A \times a^3 )}$
Substituting the values of Z, Density and a, we get
M= 186.5 g
Now 186.5 g contains $6.023 \times 10^{23}$ atoms
So, 50 g will have= $6.023 \times 10^{23} \times \frac {50}{186.5}$
=$1.614 \times 10^{23}$ atoms

Question 16
A metallic element has a BCC lattice. Each edge of the unit cell is 288 pm. The density of the metal is 7.2 g/cm^{3}. How many atoms of this element are present in 100 g
Ans $1.158 \times 10^{24}$ atoms

Question 17
With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.

Question 18
Show that in a cubic close packed structure, eight tetrahedral voids are present per unit cell