Question

a mixed oxide of calcium, an unknown metal and oxygen has cubic crystals. in a primitive cubic unit cell of calcium ions the unknown metal ( at wt -48) ions is present at the centre and oxide ions are present in some lattice points the edge length of the cube is 500 pm and density is 1.8g/c . find the oxidation state of metal ion ​

Answer 1

Answer:

The length of edge of the unit cell is 5 A

o

or 5×10

−8

cm

The volume of the unit cell is (5×10

−8

cm)

3

=1.25×10

−22

cm

3

The density is 4.0 g cm

−3

The mass of the unit cell is 4.0 g cm

−3

×1.25×10

−22

cm

3

=5.0×10

−22

g......(1)

The mass of one mole of FeO is 55.8+16=71.8 g.

The mass of one molecule of FeO will be

6.023×10

23

71.8

=1.192×10

−22

g.....(2)

Divide (1) with (2) to get number of FeO formula units per unit cell.

1.192×10

−22

5.0×10

−22

g

≃4

Then the number of Fe

2+

and O

2−

ions

present in each unit cell will be four Fe

2+

and four O

2−