Assign an oxidation state for each nitrogen atom in \(\mathrm{N}_{3}^{-} .\) (Hint: Begin with a dot diagram for the ion.)

First, we will draw a Lewis dot structure for the azide ion to help visualize the bonding and sharing of electrons. The azide ion consists of three nitrogen atoms and has a net charge of -1. Since each nitrogen atom has 5 valence electrons and the overall charge is -1, there are a total of 16 valence electrons (5+5+5+1) in the azide ion. The Lewis dot structure for the azide ion is: \(\mathrm{\hspace{1cm} N = N \equiv N \:{}^-}\) In this structure, there is: - A single bond between the first and second nitrogen atoms - A triple bond between the second and third nitrogen atoms - One lone pair on the first nitrogen atom - A negative formal charge on the third nitrogen atom

To assign the oxidation states, we must consider the number of valence electrons and the electrons involved in bonding for each nitrogen atom. 1. For the first nitrogen atom in \(\mathrm{N}_{3}^{-}\): - It has 5 valence electrons (all nitrogen atoms start with 5 valence electrons) - It forms a single bond with the second nitrogen atom, sharing two electrons, so we will assign one electron to the first nitrogen. - It also has one lone pair, which means it has two more electrons. - In total, the first nitrogen atom has 5 - 1 - 2 = 2 electrons less than its original valence electrons, giving it an oxidation state of +2. 2. For the second nitrogen atom in \(\mathrm{N}_{3}^{-}\): - It also has 5 valence electrons. - The second nitrogen atom forms a single bond with the first nitrogen atom, sharing two electrons, so we will assign one electron to the second nitrogen atom. - It also forms a triple bond with the third nitrogen atom, sharing six electrons, so we will assign three electrons to the second nitrogen atom. - In total, the second nitrogen atom has 5 - 1 - 3 = 1 electron less than its original valence electrons, giving it an oxidation state of +1. 3. For the third nitrogen atom in \(\mathrm{N}_{3}^{-}\): - It has 5 valence electrons. - It forms a triple bond with the second nitrogen atom, sharing six electrons, so we will assign three electrons to the third nitrogen atom. - The third nitrogen atom also carries the negative charge, which means it gains an extra electron. - In total, the third nitrogen atom has 5 - 3 + 1 = 3 electrons more than its original valence electrons, giving it an oxidation state of -3. In conclusion, the oxidation states for each nitrogen atom in \(\mathrm{N}_{3}^{-}\) are +2, +1, and -3.