The unit cell of an element of atomic mass 96, and density 10.3 g cm–3 is a cube with edge length of 314 pm. Find the structure of crystal lattice (simple cubic, F.C.C. or B.C.C.) Avogadro’s constant. NA = 6.023 x 1023 mol–1?

solution:
we have given

Density of element,  ρ=10.3 g cm-3
Cell edge  a= 314pm or 3.14 x 10-10 cm          

     NA =6.023×1023 mol-1

Atomic mass = 96 g mol-1

                 P = Z×Ma3×NAZ = P×a3×NAM

= 10.g cm-3×(3.14)3×10-30cm3×6.023×1023 mol-196 g mol-1=2

The structure of the crystal lattice is B.C.C.

2325 Views

The ionic radius of CI ion is 181 pm. Consider the closest packed structure in which all anions are just touching:
(i) Calculate the radius of the cation that just fits into the octahedral holes of this lattice of anions. (ii) Calculate the radius of the cation that just fits into the tetrahedral holes of this lattice of anions.

given:
Radius = 181 pm
thus,
Radius of octahedral void = 0.414 r = 0.414 x 181 pm = 74.934 pm
Cation having radius 74.934 pm will just fit into octahedral voids.

Radius of tetrahedral void

= 0.225 r = 0.225 x 181 pm = 40.725 pm
Cation having radius 40.725 pm will just fit into tetrahedral void.
217 Views

Advertisement

How would you account for the following:
Impurity doped silicon is a semiconductor.


Silicon forms four covalent bonds thus two types of impurities can be added to silicon.
for example : boron can be added or phosphorus can be added.
In case of boron : boron form three bonds with silicon, it will result in an electron deficient bond andwill create a hole. These holes can move through crystal like positive charge giving rise electrical conductivity.
In case of phosphorus: when silicon forms four bonds with phosphorus one electron of phosphorous atom will remain unbonded due to which it become delocalizedn and contributes to electrical conductivity.
462 Views

Advertisement
A metallic element exists as a cubic lattice. Each edge of the unit cell is 2.88 A°. The density of the metal is 7.20 g cm–3. How many unit cells will be there in 100 g of the metal?

solution: 
we have given a= 2.88A0
Density = 7.20g cm-3
mass = 100g

Volume of the unit cell = (2.88 A°)3
                          = (2.88×10-8cm)3= 23.9 × 10-24 cm3

Volume of 100g of the metal

                 = WeightDensity=1007.20= 13.9 cm3

No. of unit cell in 13.9 cm3
   
                   = 13.9 cm323.9 × 10-24 cm3= 5.82 × 1023.

2490 Views

Iron (II) oxide has a cubic structure and each unit cell has side 5A. If the density of the oxide is 4 g cm–3, calculate the number of Fe2+ and O2– ions present in each unit cell. (Molar mass of FeO = 72 g mol–1, NA = 6.02 x 1023 mol–1).

Solution:
We  have given
volume of the unit cell = (5 x 10-8cm)3 = 1.25 x 10-22cm3

Density of FeO = 4g cm-3

Density,

ρ = z×Ma3×NA4= z×72(5×10-8)3×6.02×1023z= 4×1.25×10-22 ×6.02×102372 = 4.18 4

Each unit cell has four units of FeO. So it has four Fe2+ and four O2– ions.

2407 Views

Advertisement