Based on Lewis structures, predict the ordering of \(\mathrm{N}-\mathrm{O}\) bond lengths in \(\mathrm{NO}^{+}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{NO}_{3}^{-}\).

For each molecule, we need to draw the correct Lewis structure: 1. NO⁺: In this molecule, nitrogen (N) has 5 valence electrons and oxygen (O) has 6. However, the molecule has a +1 charge, which means one electron is missing. So, the total number of valence electrons in NO⁺ is (5+6)-1 = 10. Place N and O next to each other and arrange the electrons in such a way that each atom obeys the octet rule, which gives you the following Lewis structure: N≡O₊, where there is a triple bond between N and O. 2. NO₂⁻: Nitrogen has 5 valence electrons, and each oxygen has 6. This molecule has a -1 charge, which means we need to add one extra electron. So, the total number of valence electrons in NO₂⁻ is 5+6+6+1 = 18. Arrange the electrons to follow the octet rule, giving you the following Lewis structure: O-N=O, with a single bond to one oxygen and a double bond to the other. This molecule has resonance, meaning that the double bond can be in either position. 3. NO₃⁻: Nitrogen has 5 valence electrons, and each oxygen has 6. Since this molecule has a -1 charge, an extra electron must be added, for a total of 5+6+6+6+1 = 24 valence electrons. Arrange them to satisfy the octet rule, giving you the Lewis structure with resonance: O-N=O, O=N-O, and O-N=O, where there is a double bond between N and different O atoms in each resonance structure.

Now we analyze the influence of the resonance structures on the bond lengths in the molecules: 1. NO⁺: There is no resonance structure for NO⁺. It simply has a triple bond between N and O, resulting in a relatively short bond length. 2. NO₂⁻: In the resonance structures of NO₂⁻, there is a single bond to one oxygen atom and a double bond to the other. Due to the presence of resonance, all the N-O bonds share the double bond character. Consequently, they will be shorter than a typical single bond but longer than a double bond. 3. NO₃⁻: In the resonance structures, there is one double bond and two single bonds between N and the three O atoms. Due to resonance, the double bond character is equally distributed among all the bonds, which makes them equivalent. Thus, the N-O bonds in NO₃⁻ will have bond lengths between a single bond and a double bond, and likely be slightly longer than those in NO₂⁻ due to the extra oxygen atom sharing the double bond character.

Based on the analysis of resonance structures and bond character, we can now predict the ordering of N-O bond lengths for the molecules: 1. The N-O bond in NO⁺ has a triple bond, resulting in the shortest bond length. 2. The N-O bonds in NO₂⁻ have resonance, making them shorter than a typical single bond but longer than a double bond. 3. The N-O bonds in NO₃⁻ have a double bond character shared among all the bonds, and their bond lengths are likely slightly longer than those in NO₂⁻. So, the ordering of the N-O bond lengths is NO⁺ < NO₂⁻ < NO₃⁻.