Which of the following sets shows the correct resonance structures for $S_e{O}_2$?

• A. $O-\mathrm{Se}-O$ and $O-\mathrm{Se}-O$
• B. $O=\mathrm{Se}=O$ and $O=\mathrm{Se}-O-$
• C. $-O-\mathrm{Se}\equiv O$ and $O\equiv \mathrm{Se}-O-$
• D. $O=\mathrm{Se}=O$ and $O-\mathrm{Se}\equiv O$

Answer: (C)

Verified step by step guidance

1. Step 1: Identify the central atom and its valence electrons. Selenium (Se) is the central atom in SeO\_2. Selenium is in group 16, so it has 6 valence electrons, and each oxygen atom also has 6 valence electrons.
2. Step 2: Determine the total number of valence electrons available for bonding. Since there are two oxygen atoms, total valence electrons = 6 (Se) + 2 × 6 (O) = 18 electrons.
3. Step 3: Draw a basic Lewis structure with single bonds between Se and each O atom, then distribute the remaining electrons to satisfy the octet rule for each atom. This initial structure will have lone pairs on oxygen and selenium.
4. Step 4: Generate resonance structures by moving electrons to form double or triple bonds between Se and O atoms. Use formal charge calculations to evaluate the stability of each resonance form. The best resonance structures minimize formal charges and place negative charges on the more electronegative atoms (oxygen).
5. Step 5: Recognize that the correct resonance structures for SeO\_2 involve one Se=O double bond and one Se–O single bond with a negative charge on the singly bonded oxygen, and the resonance form where the double and single bonds are switched. This corresponds to resonance between a structure with a double bond and a single bond with a negative charge on oxygen, and vice versa, often represented with a resonance arrow (\(\equiv\)) between them.