Trying to figure out where peroxide gains 2 electrons and why it has a 2- charge? On the Lewis dot structure it shows each O has 6 electrons and that it has a single bond between the two Oxygen atoms.

To me, each O atom would have 4 electrons and have a double bond between them?


1 Answer 1


Well, I might take a different approach at an answer, and try to get to a reasonable Lewis structure with data we already should have had from drawing structural diagrams! It's kinda upside down, since none of what's written below is happening in real-life chemistry.

The structure you have in mind is the diatomic oxygen molecule, with 12 electrons in its system overall.

$\hspace{33ex}$ $\hspace{45ex}$Source

While the peroxide ion overall has 14 electrons in its system, since each oxygen has 6 valence electrons and the system also has two extra electrons indicated by the $2-$ charge of the ion. Now, let's oversimplify this a bit: Imagine that in peroxide also, the oxygen atoms need to fulfill their octet configuration, hence each of them would have 8 electrons in its outer shell.

Let $N_b$ be the number of bonding electrons and $N_{nb}$ the number of non-bonding electrons of each oxygen atom. We know that

  1. A single oxygen atom would have non-bonding electrons and bonding electrons that fill the octet.
  2. All of the electrons in the system, which are the non-bonding electrons for each of the two oxygens, and the bonding electrons, make 14 electrons.

Hence we solve:

$$\Bigg\lbrace\begin{array}\\ N_b + N_{nb} = 8 \\ N_b + 2N_{nb} = 14 \end{array}\Bigg\rbrace \Rightarrow \boxed{N_b = 2},~\boxed{N_{nb} = 6} $$

Hence we could figure out how many bonding and non-bonding electrons there are in compounds we can draw Lewis structures for, by finding out the overall number of electrons (taking into account the number of electrons in the valence of each atom and the number of electrons lost or gained due to the charge of the ion) with a rudimentary approach.

Try to find out how many bonding and non-bonding electrons there are for the azide, permanganate, manganate, sulfate, oxalate, salicylate, and superoxide anions and thionyl, sulfuryl, nitrosyl and methanium cations.

  • $\begingroup$ Thank you so much. I still dont see where the ion picks up the additional two electrons however.. Are the two oxygen atoms that come together to form peroxide are they already ions ? I understand how its 14 and how to calculate the formal charge but how exactly do they pick up the additional 2? U $\endgroup$ Dec 11, 2015 at 22:08
  • $\begingroup$ @Atticus283blink The same question applies to any other ion. $\ce{Na+}$, for example - where did it lose that missing electron? Or $\ce{Cl-}$ - where did it get that extra electron? Chemistry is full of mysteries, you know. $\endgroup$ Dec 11, 2015 at 22:32
  • $\begingroup$ @ivan Neretin @M A R How is peroxide actually formed? Do two oxygen atoms that have already lost 1 electron each bind together or? $\endgroup$ Dec 12, 2015 at 1:44
  • $\begingroup$ @Atticus283blink Like any other ion or molecule, it can be actually formed in a variety of ways. Losing electrons, however, is definitely not one of them. Losing electrons would get you a particle with positive charge. $\endgroup$ Dec 12, 2015 at 21:19
  • $\begingroup$ I meant gain electrons. It says each oxygen in peroxide has an oxidation state of -1 what does this entail? $\endgroup$ Dec 13, 2015 at 0:51

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