How to find the formula of an binary oxide based on the ionisation energies? I am a little confused about this multiple choice question and what the requirements are to answer it. It is for a year 11 chemistry student that I am helping and I think I am going in a little too much detail to try to solve it to help them. I am sure that I am overthinking this. 


  The energies needed to remove electrons in succession from a neutral
  atom of a certain element $\ce{A}$ are given ($\mathrm{kJ\cdot  mol^{-1}}$): $1010$, $2032$, $7056$, $8120$, $9300$.
   The formula  of the oxide of $\ce{A}$ is:
   $\ce{AO}$, $\ce{A2O}$, $\ce{A2O3}$, $\ce{A2O4}$ or $\ce{A3O5}$. 


My understanding of the question and what I was trying to do to solve the problem by looking at the ratios between the energies and I was trying to apply
1st ionization energy: $$\ce{X -&gt; X+ + e−},$$ 
2nd ionization energy: $$\ce{X+ → X^2+ + e−},$$ 
and 3rd ionization energy: $$\ce{X^2+ -&gt; X^3+ + e−}$$ 
to the energies below and find a ratio. I thought the ratio was going to be c. I understood that oxide is $2-$, however, I couldn't calculate the $\ce{A}$ value as I didn't know what the element was.
The point of this problem is to test if you can see where there is a jump in ionization energies. The third ionization energy is considerably larger than the others. Although ionization energies will naturally increase in succession, this is the most profound increase, which would suggest that this atom will tend to lose 2 electrons before obtaining a stable octet structure.

This would mean that its ion would be $\ce{A^{2+}}$. Since the oxide ion is $\ce{O^{2-}}$, the charges will be balanced if the ratio of the ions is 1:1, meaning the answer is $\ce{AO}$.