# 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 '21 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 '21 at 18:03
• @ NisargBhavsar Thank you, that solves my problem. May 11 '21 at 16:38

## 1 Answer

Your approach is correct, however, you forgot to account for the vacancies created when Fe3+ replaces Fe2+. First, calculate the cationic vacancies formed due to the replacement of Fe2+ by Fe3+. Assuming 𝑥 to be the fraction of Fe2+ in the compound, the fraction of Fe3+ in the compound is (0.93 - x).The total charge on x Fe2+ and (0.93 – x) Fe3+ should be equal to the charge due to O2- ions.

Hence, 2x + 3(0.93 – x) = 2
=> 2x + 2.79 – 3x = 2.00 or x = 0.79. So, fraction of Fe3+ ions in the compound is 0.14

Now, 2 Fe3+ ions replace 3 Fe2+ ions, so there's one vacancy created for every 2 Fe3+ ions. That means for 0.1 mole of F0.93O, the number of vacancies will be 0.007𝑁A due to Fe3+ replacing Fe2+ in FeO. This is the part you missed.

Now, as you mentioned, each Si4+ ion will replace two Fe2+ ions in the compound, so again, one vacancy per replacement. And, as you already solved, The number of replacements would be (0.079/2)𝑁A = 0.0395𝑁A. So, the total number of vacancies are 0.0395NA + 0.007NA = 0.0465NA.

Hope this helps!

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