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.

Answer stated by the book


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?

  • $\begingroup$ You have not considered the vacancies formed when ferrate ions were doped with ferric ions. $\endgroup$ May 6 '21 at 12:14
  • $\begingroup$ @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. $\endgroup$ May 6 '21 at 18:03
  • $\begingroup$ @ NisargBhavsar Thank you, that solves my problem. $\endgroup$ May 11 '21 at 16:38

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!

  • $\begingroup$ On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. (Btw: If it doesn't help, don't post it; if it does, don't post the tagline.) $\endgroup$ May 6 '21 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.