Which choice presents the Lewis dot structure for diatomic oxygen
The Correct Answer and Explanation is:
The Lewis dot structure for diatomic oxygen (O2) can be drawn as follows:
O=O with each oxygen atom having 6 valence electrons, and each shared pair of electrons forming a covalent bond between the two oxygen atoms. Here’s a breakdown:
Step 1: Count the total valence electrons
Each oxygen atom has 6 valence electrons (since oxygen is in Group 16 of the periodic table). So for O2, the total number of valence electrons is:6 electrons×2 atoms=12 electrons.6 \text{ electrons} \times 2 \text{ atoms} = 12 \text{ electrons}.6 electrons×2 atoms=12 electrons.
Step 2: Connect the two oxygen atoms
Oxygen atoms form a double bond in the O2 molecule. We place the two oxygen atoms next to each other and draw two shared electron pairs between them, symbolized as a double line (O=O). This uses 4 of the 12 available electrons.
Step 3: Distribute the remaining electrons
After forming the double bond, we have 8 electrons left (12 total – 4 for the bond). Each oxygen atom has 4 electrons left to be distributed as lone pairs. Each oxygen will have two lone pairs (each lone pair is 2 electrons), so that both oxygen atoms now have 8 electrons around them: 2 electrons from the double bond and 4 electrons as lone pairs.
Final Lewis structure:
:O=O:\text{:O=O:}:O=O:
Where the colon (:) represents lone pairs of electrons. Each oxygen atom has 2 lone pairs (on top and bottom), and the two oxygen atoms share 2 pairs of electrons in the double bond.
Explanation:
- Each oxygen atom is now stable with 8 electrons in its valence shell (following the octet rule).
- The double bond between the oxygen atoms represents the sharing of 4 electrons (2 pairs), which is common in molecules like O2.
- Oxygen, being highly electronegative, forms a stable molecule in this way.
This is the correct Lewis dot structure for diatomic oxygen, O2, and ensures that both oxygen atoms satisfy the octet rule.
