In Fe_{0.96}O,Fe is present in +2 and +3 oxid | vedclass

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(D) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.96 - x)$.Since the compound is electrically neutral,the total posi

Vedclass · Accepted Answer

$11/12$

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(D) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.96 - x)$.Since the compound is electrically neutral,the total positive charge must equal the total negative charge.The charge of $O^{2-}$ is $-2$.Therefore,$2x + 3(0.96 - x) - 2 = 0$.$2x + 2.88 - 3x - 2 = 0$.$-x + 0.88 = 0$.$x = 0.88$.So,the number of $Fe^{2+}$ ions is $0.88$ and the number of $Fe^{3+}$ ions is $0.96 - 0.88 = 0.08$.The mole fraction of $Fe^{2+}$ is the ratio of the number of $Fe^{2+}$ ions to the total number of $Fe$ ions.Mole fraction of $Fe^{2+} = \frac{0.88}{0.96} = \frac{88}{96} = \frac{11}{12}$.

In $Fe_{0.96}O$,$Fe$ is present in $+2$ and $+3$ oxidation states. What is the mole fraction of $Fe^{2+}$ ions in the compound?

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