# Cationic vacancies created when non-stoichiometric iron oxide is doped with silicon(IV) ions

### Problem

It is believed that non-stoichiometric compound $$\ce{F_{0.93}O}$$ forms by doping of $$\ce{Fe^3+}$$ ions in $$\ce{FeO}$$ crystal by replacement of $$\ce{Fe^2+}$$. Calculate the number of cationic vacancies if all of the $$\ce{Fe^2+}$$ ions are replaced by $$\ce{Si^4+}$$ ions in $$\pu{0.1 mol}$$ of $$\ce{Fe_{0.93}O}$$. Express your answer as a multiple of Avogadro's Number.

0.0465

### My Solution

As the compound is electrically neutral I equated the net charges due to the cations and anions like this, assuming $$x$$ to be the fraction of $$\ce{Fe^2+}$$ in the compound.

$$[2x + 3(0.93-x)] - 2 = 0$$

This gave $$x = 0.79$$. Now, in $$\pu{0.1 mol}$$ of the compound there would be $$\pu{0.079 mol}$$ of $$\ce{Fe^2+}$$. As each silicon ion would replace two ions of $$\ce{Fe^2+}$$, one vacancy would be created for every replacement.

The number of replacements would be $$\left(\frac{0.079}{2}\right)\times N_\mathrm{A} = 0.0395 N_\mathrm{A}$$, and we would have as many vacancies. ($$N_\mathrm{A}$$ is Avogadro's Number)

Can anyone point out the error in my solution?

• You have not considered the vacancies formed when ferrate ions were doped with ferric ions. May 6, 2021 at 12:14
• @NisargBhavsar Thank you for improving so many posts. Please have a look at these guides for even better edits. And again, I am really, really not fond of this spoiler thing. May 6, 2021 at 18:03
• @ NisargBhavsar Thank you, that solves my problem. May 11, 2021 at 16:38