Draw the dominant Lewis structure for NO2+. Note, that this ion is isoelectronic with CO2. How many double bonds does this resonance structures have?
To draw the dominant Lewis structure for NO2+, we need to follow a certain set of rules:
1. Determine the total number of valence electrons: For NO2+, the nitrogen atom contributes 5 valence electrons, while each oxygen atom contributes 6 valence electrons. Since there is a positive charge on NO2+, we subtract one electron. So, the total number of valence electrons is 5 + 2(6) - 1 = 16.
2. Identify the central atom: Nitrogen (N) is the central atom because it is less electronegative than oxygen.
3. Connect the atoms using single bonds: Start by placing a nitrogen atom in the center and connecting it to the two oxygen atoms using single bonds. This uses 4 valence electrons (8 electrons remaining).
4. Distribute the remaining valence electrons around the atoms: Distribute the remaining 8 electrons as lone pairs on each oxygen atom, giving each oxygen atom a full octet (8 electrons used, 0 remaining).
5. Check if the central atom has an octet: In this case, the nitrogen atom has only 6 electrons around it, so it does not have an octet.
Now, to create resonance structures:
6. Move a lone pair from an oxygen atom to form a double bond with the nitrogen atom.
7. Check if the central atom now has an octet: The nitrogen atom now has an octet, and each oxygen atom has 6 electrons.
8. Count the number of double bonds in each resonance structure: In the original Lewis structure (before resonance), there are zero double bonds. In the resonance structure formed by moving a lone pair to form a double bond with the nitrogen atom, there is one double bond.
Therefore, the dominant Lewis structure for NO2+ has one resonance structure with one double bond.