To draw the Lewis structure for the peroxide ion (O₂²⁻), we start by determining the total number of valence electrons available. Each oxygen atom has 6 valence electrons, and since there are two oxygen atoms, that gives us:
6 electrons x 2 = 12 electrons
However, since the ion carries a -2 charge, we add 2 more electrons, resulting in a total of:
12 + 2 = 14 electrons
Next, we place the two oxygen atoms next to each other. We then connect them with a single bond, which uses 2 electrons, leaving us with:
14 – 2 = 12 electrons
We distribute the remaining electrons to complete the octets for both oxygen atoms. Placing 6 electrons (3 pairs) around each oxygen gives us:
12 – 12 = 0 electrons remaining
This results in:
O: (6 electrons) ↔ O: (6 electrons)
Since each oxygen is surrounded by 8 electrons (including the bonding pair), they achieve a full octet. However, we must consider the overall charge of the ion. In the O₂²⁻ structure, we can represent the bonding electron pairs as:
:O:↔O: The double bond between the oxygen atoms suggests that they are sharing two pairs of electrons, which is characteristic of the O₂²⁻ ion. It’s important to note that the peroxide ion has a single bond, but sometimes the Lewis structure displays a resonance hybrid representation if we consider the bond order.
Final summary:
- The two oxygen atoms in O₂²⁻ are bonded by a single bond, and each has 3 lone pairs of electrons.
- This configuration ensures that both oxygen atoms have access to a total of 8 electrons around them, satisfying the octet rule.