To determine the correct Lewis structure for hydrogen peroxide, [tex]$H_2O_2$[/tex], we need to consider the bonding and the placement of electrons. Here's a detailed step-by-step process to derive the correct structure:
1. Count the total number of valence electrons:
- Each hydrogen (H) atom has 1 valence electron.
- Each oxygen (O) atom has 6 valence electrons.
Therefore, for [tex]$H_2O_2$[/tex]:
[tex]\[
(2 \times 1) \text{ H electrons} + (2 \times 6) \text{ O electrons} = 2 + 12 = 14 \text{ valence electrons}
\][/tex]
2. Determine the bonding between the atoms:
- Hydrogen atoms can each form one bond.
- Oxygen atoms each require two bonds to fulfill the octet rule.
For hydrogen peroxide, we can think of it as having a hydrogen atom on each end with the two oxygens in the middle, forming a chain-like structure:
[tex]\[
H - O - O - H
\][/tex]
3. Assign the electrons to form bonds and lone pairs:
- Start by placing single bonds between the atoms: H-O-O-H. This uses 4 electrons (2 per bond).
4. Distribute the remaining electrons to satisfy the octet rule:
- Each bond already has 2 electrons, so we have 10 remaining electrons to place around the atoms (14 total electrons - 4 used in bonds).
- Oxygen atoms typically have 8 electrons around them (including bonds), so after the single bonds are placed, we need to place additional electrons to complete the octets for oxygens:
- Each oxygen already has 2 electrons from the bonds, so they need 6 more electrons each (3 lone pairs per oxygen).
Therefore, the correct Lewis structure looks like this:
[tex]\[
H - \dot{O} - \dot{O} - H
\][/tex]
where each oxygen has two pairs of lone electrons (not shown in a simplified structure).
From the options provided, option (A) correctly depicts this arrangement.
Therefore, the correct answer is A.
