Question

(a) Write one or more appropriate Lewis structures for the nitrite ion, \(\mathrm{NO}_{2}^{-} .\) (b) With what allotrope of oxygen is it isoelectronic? (c) What would you predict for the lengths of the bonds in \(\mathrm{NO}_{2}^{-}\) relative to \(\mathrm{N}-\mathrm{O}\) single bonds?

Short answer

The nitrite ion, \(\mathrm{NO}_{2}^{-}\), has two resonance structures: O-N=O and O=N-O. It is isoelectronic with ozone (O\(_3\)). The bond lengths of the N-O bonds in the nitrite ion are shorter than those in N-O single bonds due to the partial double bond character from resonance.

Step-by-step solution

Drawing the Lewis structures for the nitrite ion

To draw the Lewis structures for the nitrite ion, \(\mathrm{NO}_{2}^{-}\), first, determine the total number of valence electrons. Nitrogen has 5 valence electrons, each oxygen atom has 6 valence electrons, and the extra electron from the negative charge contributes 1 valence electron. Therefore, there are a total of 18 valence electrons. Now arrange the atoms and distribute the electrons: 1. Place the least electronegative atom (nitrogen) at the center. 2. Add single bonds between nitrogen and both oxygen atoms, using up 4 electrons. 3. Complete the octet of the oxygen atoms by adding lone pairs, using up 12 electrons. 4. Since we still have 2 more electrons, add a lone pair to the nitrogen atom (central atom). Now we need to check if the central nitrogen atom needs a double bond. Nitrogen prefers having 5 electrons in its outer shell. Removing one lone pair from the nitrogen atom and turning one of the N-O single bonds into a double bond would satisfy nitrogen's preferred electron configuration. The two possible Lewis structures for the nitrite ion are: Resonance Structure 1: O-N=O (with two lone pairs on the left oxygen atom and three lone pairs on the right oxygen atom) Resonance Structure 2: O=N-O (with three lone pairs on the left oxygen atom and two lone pairs on the right oxygen atom) Both structures are equivalent, and the actual structure of the nitrite ion is a resonance hybrid of these two structures.

Finding the isoelectronic allotrope of oxygen

The nitrite ion has a total of 18 electrons. To find the isoelectronic allotrope of oxygen, we need to find an allotrope with the same number of electrons. Here are the common allotropes of oxygen: 1. Dioxygen (O\(_2\)) has 12 valence electrons (each oxygen atom has 6 valence electrons). 2. Ozone (O\(_3\)) has 18 valence electrons (each oxygen atom has 6 valence electrons). Thus, the nitrite ion is isoelectronic with ozone (O\(_3\)).

Predicting bond lengths in the nitrite ion relative to N-O single bonds

In the nitrite ion, the N-O bond is a resonance hybrid between a single bond and a double bond, making it an intermediate bond; stronger and shorter than a single bond but weaker and longer than a double bond. This intermediate bond is commonly known as a "partial bond." In comparison to N-O single bonds, the N-O bonds in the nitrite ion would be shorter due to the partial double bond character caused by resonance.