(a) Draw the best Lewis structure(s) for the nitrite ion, NO \(_{2}^{-}\) (b) With what allotrope of oxygen is it isoelectonic? (c) What would you predict for the lengths of the bonds in \(\mathrm{NO}_{2}^{-}\) relative to \(\mathrm{N}-\mathrm{O}\) single bonds and double bonds?

To draw the Lewis structure for the nitrite ion, we need to consider the total number of valence electrons and arrange the atoms accordingly. 1. Count Valence Electrons: Nitrogen (N) has 5 valence electrons, Oxygen (O) has 6 valence electrons, and there is one additional electron due to the negative charge of the ion. So, the total number of valence electrons is (5 + 2*6 + 1) = 18. 2. Arrange the Atoms: Put the least electronegative atom in the center (N), and the remaining atoms (O) around it. Connect the central atom to surrounding atoms with single bonds. 3. Add Lone Pairs: Complete the octets for the surrounding atoms (O) by adding lone pairs. 4. Add Multiple Bonds: If the central atom (N) does not have a complete octet, create multiple bonds with one or more surrounding atoms. In the case of NO₂⁻, there are two possible resonance structures, which are equally important for the description of the bonding in the ion. The resulting two resonance structures can be represented as follows: O-N=O <-----> O=N-O

An allotrope of oxygen that is isoelectronic with NO₂⁻ should have the same number of total electrons. Since NO₂⁻ has 5+6+6+1=18 electrons, the oxygen allotrope must also have 18 electrons. Ozone (O₃) is found to be isoelectronic with NO₂⁻, as each oxygen atom has 6 electrons, and hence O₃ has 3*6 = 18 electrons.

Since NO₂⁻ has two resonance structures, O-N=O <-----> O=N-O, it shows that the N-O bonds are delocalized, which means they are somewhere between a single bond (N-O) and a double bond (N=O). A single bond (N-O) will typically be longer than a double bond (N=O), because there is only one shared pair of electrons in a single bond, causing less electrostatic attraction between the atoms. In NO₂⁻, the delocalized bonds mean that the electrons are shared over both N-O bonds, giving each bond a bond order of about 1.5. This partial double bond character will make the bond lengths in NO₂⁻ shorter than N-O single bonds, but longer than N=O double bonds.