The ratio `Fe^(3+) and Fe^(2+)` ions in `Fe_(0.9)S_(1.0)` is.

Options

A. 0.28
B. 0.5
C. 2
D. 4

Answer

A

Explanation

Let the number of `Fe^(2+)` ions present be x. then `Fe^(3+)=90-x`
Charge on `Fe^(2+)` and `Fe^(3+)` should be equal to `100S^(2-)` ions
So, `2x+3(90-x)=2xx100impliesx=70`
Therefore, Number of `Fe^(2+)=70 and Fe^(3+)=20`, Their ratio is `Fe^(3+)//Fe^(2+)=20//70=0.28`