1.

Ferric oxide crystallises in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Answer»

SOLUTION :Number of oxide ION per unit cell = 6
`therefore` Total number of OCTAHEDRAL VOIDS = 6
Number of ferric ion present `= 2/3 xx 6 = 4`
`therefore ` Ratio of `Fe^(3+)` ions to `O_("ions")^(2-)= 4 : 6 = 2:3`
Hence formula of oxide is `Fe_2O_3`


Discussion

No Comment Found