(a) The nitrate ion, \(\mathrm{NO}_{3}^{-}\), has a trigonal planar structure with the \(\mathrm{N}\) atom as the central atom. Draw the Lewis structure(s) for the nitrate ion. (b) Given \(S=\mathrm{O}\) and \(\mathrm{S}-\mathrm{O}\) bond lengths are \(158 \mathrm{pm}\) and \(143 \mathrm{pm}\) respectively, estimate the sulphuroxygen bond distances in the ion.

Since this is a nitrate ion, it consists of one nitrogen atom and three oxygen atoms. The total number of valence electrons for this ion can be calculated as follows: - Nitrogen (N) has 5 valence electrons. - Oxygen (O) has 6 valence electrons, but there are three oxygen atoms in the nitrate ion, making 3*6 = 18 valence electrons in total for oxygen. - The ion has an extra electron due to its -1 charge. So, the total number of valence electrons for nitrate ion is 5 (N) + 18 (O) + 1 (extra electron) = 24 valence electrons.

Place the least electronegative atom in the center, which is the nitrogen atom, and attach each oxygen atom to the nitrogen using single bonds. This would assume 6 electrons from the total 24 valence electrons, leaving 18 electrons.

Now use the remaining 18 electrons to fill the octet of the oxygens. Each oxygen will require 6 more electrons since it already has one single bond (which counts as 2 electrons) in place. This will use up all the remaining electrons.

Now, check the charges on each atom: - Nitrogen: Formal charge = Valence electrons - (Non-bonding electrons + 0.5 * bonding electrons) = 5 - (0 + 0.5 * 6) = 5-3=+2 - Oxygen: Formal charge = 6 - (6 + 0.5 * 2) = 6 - 7 = -1. So, the Nitrogen has +2, each oxygen has -1, which results in a total of -1 formal charge for the whole ion. However, the formal charge should equal the ion charge, so the structure is not correct.

To achieve the right formal charge, we need to add more bonds to the central nitrogen atom. Let's replace one single bond between nitrogen and one of the oxygens with a double bond. This would decrease the formal charge of both nitrogen and oxygen by 1 unit. After this modification, the formal charge of the central nitrogen is +1, two oxygens have the formal charge of -1, and one oxygen has a formal charge of 0. The total formal charge for this ion is now -1, which matches the overall charge of the ion. As there are three oxygen atoms, we can distribute the double bond across the three, so the nitrate ion shows resonance with three possible structures. ##Part (b): Estimate the sulphuroxygen bond distances in the ion##

The sulfur (S) atom is in the same family as the oxygen atom (second period in group 16), which means they have similar properties. Given the bond lengths \(\mathrm{S}=\mathrm{O}\) (158 pm) and \(\mathrm{S}-\mathrm{O}\) (143 pm), we can estimate the average \(\mathrm{S}-\mathrm{O}\) bond length in the ion. The nitrate ion is composed of 1 nitrogen and 3 equivalent oxygen atoms. Therefore, we can assume: 1 double bond -> average length of the \(\mathrm{S}=\mathrm{O}\) bond (158 pm). 2 single bonds -> average length of the \(\mathrm{S}-\mathrm{O}\) bond (143 pm). Calculating the weighted average bond length: Average bond length = (1 * 158 pm + 2 * 143 pm) / 3 bonds. Average bond length = (158 + 286) / 3 = 444 / 3 = 148 pm. So, the estimated sulphuroxygen bond distance in the nitrate ion is 148 pm.