Question

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

Short answer

Based on the analysis of resonance structures and bond character in the given molecules, the ordering of N-O bond lengths is NO⁺ < NO₂⁻ < NO₃⁻. This is because NO⁺ has a triple bond, making it the shortest bond length, while NO₂⁻ has resonance structures with shared double bond character, making its bond length longer than a double bond but shorter than a single bond. Finally, NO₃⁻ has double bond character shared among all the bonds, resulting in the longest bond lengths among the three molecules.

Step-by-step solution

Drawing the Lewis Structures

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.

Analyze Bond Lengths Based on Resonance Structures

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.

Predict the Ordering of Bond Lengths

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⁻.

Key concepts

Bond Lengths

Bond lengths refer to the distance between the nuclei of two bonded atoms. When we analyze bond lengths in relation to Lewis structures, we find that the type of bond is critical in determining this distance. A single bond is longer than a double bond, and a double bond is longer than a triple bond.

• Single bond: Involves one pair of shared electrons (longest length).
• Double bond: Involves two pairs of shared electrons (shorter than a single bond).
• Triple bond: Involves three pairs of shared electrons (shortest length).

In the context of the exercise, the NO⁺ ion has a triple bond between nitrogen and oxygen. This leads to the shortest bond length compared to the other molecules. On the other hand, the bond lengths in the NO₂⁻ and NO₃⁻ ions are affected by resonance, distributing the bond character and consequently affecting the bond length. In NO₃⁻, bonds are longer than in NO₂⁻ due to the presence of additional resonance.

Resonance Structures

Resonance is a key concept in understanding certain molecules' structures. It occurs when more than one valid Lewis structure can be drawn for a molecule. These structures, known as resonance structures, distribute electrons differently, but the actual molecule is a hybrid of these forms.

• Each resonance structure must obey the rules of molecular bonding, such as the octet rule.
• The true form of the molecule is a resonance hybrid which is an average of all possible structures.

For NO₂⁻, the resonance involves two structures, where the position of the double bond alternates between two oxygen atoms. This makes each bond shorter than a single bond, but longer than a double bond. Similarly, in NO₃⁻, resonance structures spread the double bond character across all three nitrogen-oxygen bonds, affecting their lengths uniformly. This creates equivalent bond lengths that are consistently among single and double bond lengths throughout the molecule.

Valence Electrons

Valence electrons are the electrons found in the outermost shell of an atom. They are crucial in chemical bonding because they are the electrons shared, lost, or gained during chemical reactions. Understanding valence electrons is essential for determining a molecule's Lewis structure.

• Nitrogen typically has 5 valence electrons.
• Oxygen usually has 6 valence electrons.
• Charges on ions influence the total count of valence electrons (add for negative charges and subtract for positive charges).

In the case of the NO⁺ ion, it starts with 11 valence electrons (5 from N and 6 from O) and loses one due to its positive charge, resulting in 10. For NO₂⁻, we count 18 valence electrons in total after accounting for additional electrons from the negative charge. Lastly, NO₃⁻ has 24 valence electrons, also incorporating one extra due to its negative charge. This calculation helps to draw the correct structure, showcasing the distribution of bonds and resonance within the molecule.