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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. |
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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` |
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