Which of the following statements correctly describes the Lewis dot structure of the neutral compound O2F2 (dioxygen difluoride)?

• A. Each oxygen atom is bonded to two fluorine atoms, with no O–O bond present.
• B. Each oxygen atom has three lone pairs and is not bonded to any other atom.
• C. The two fluorine atoms are bonded directly to each other, with each also bonded to an oxygen atom.
• D. Each oxygen atom is bonded to one fluorine atom and to the other oxygen atom, with each oxygen having one lone pair.

Answer: (D)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons for O2F2. Oxygen has 6 valence electrons and fluorine has 7 valence electrons. Since there are 2 oxygen atoms and 2 fluorine atoms, calculate the total as: \$2 \times 6 + 2 \times 7$.
2. Step 2: Arrange the atoms to reflect the most likely bonding pattern. Since fluorine typically forms only one bond and oxygen can form two bonds, consider connecting the two oxygen atoms together and bonding each oxygen to one fluorine atom.
3. Step 3: Draw single bonds between the atoms: O–O and each O–F. Each single bond represents 2 shared electrons. Subtract these bonding electrons from the total valence electrons calculated in Step 1.
4. Step 4: Distribute the remaining electrons as lone pairs to satisfy the octet rule for each atom. Start by completing the octets on the fluorine atoms (which usually have three lone pairs) and then place the remaining electrons on the oxygen atoms, ensuring each oxygen has one lone pair after bonding.
5. Step 5: Verify the Lewis structure by checking that each atom has a complete octet (8 electrons around it) and that the total number of electrons used equals the total valence electrons calculated initially. This confirms that each oxygen atom is bonded to one fluorine atom and to the other oxygen atom, with each oxygen having one lone pair.