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Determine the percentages of Fe^(2+)andF...

Determine the percentages of `Fe^(2+)andFe^(3+)` ions in the crystal of `Fe_(0.88)O`.

Text Solution

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Suppose, the number of `Fe^(2+)` ions in one formula unit of `Fe_(0.88)O` = x
So, number of `Fe^(3+)` ions per formula unit = (0.88 - x).
The number of O atoms per formula unit of `Fe_(0.88)O` = 1
In one formula unit of `Fe_(0.88)O`, total charge of cations = `2xx x+3(0.88-x)` and total charge of anions = -2.
Since crystal of `Fe_(0.88)O` is electrically neutral, the sum total positive charge and total negative charge is equal to zero.
`therefore" "2x+3(0.88-x)-2=0or,x=0.64`
Hence, in the crystal of `Fe_(0.88)O`, the percentage of `Fe^(2+)` ions = `0.64/0.88xx100=72.72%` and the percentage of `Fe^(3+)` ions = `(0.88-0.64)//0.88xx100=27.28%`
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