Question

Based on Lewis structures, predict the ordering, from shortest to longest, of \(N-O\) bond lengths in \(N O^{+}, N O_{2}^{-},\) and \(N O_{3}^{-} .\)

Short answer

The N-O bond lengths are ordered as NO₊ < NO₂⁻ < NO₃⁻ (from shortest to longest), based on their Lewis structures and corresponding bond orders of 2, 1.5, and approximately 1.33, respectively.

Step-by-step solution

Draw the Lewis structures for NO₊, NO₂⁻, and NO₃⁻

First, we'll draw the Lewis structures for each molecule/ion: 1. NO₊: Nitrogen (N) has 5 valence electrons, and oxygen (O) has 6 valence electrons. In NO₊, the molecule has a +1 charge overall, meaning it loses 1 electron. Thus, there are a total of 10 (5+6-1) valence electrons available. Nitrogen forms a double bond with oxygen, and both atoms have complete octets. 2. NO₂⁻: Nitrogen (N) has 5 valence electrons, and each of the two oxygen atoms (O) has 6 valence electrons. In NO₂⁻, the molecule has a -1 charge overall, meaning it gains one electron. Thus, there are a total of 18 (5+2×6+1) valence electrons available. Nitrogen forms a single bond with one oxygen and a double bond with the other oxygen, and all atoms have complete octets. 3. NO₃⁻: Nitrogen (N) has 5 valence electrons, and each of the three oxygen atoms (O) has 6 valence electrons. In NO₃⁻, the molecule has a -1 charge overall, meaning it gains one electron. Thus, there are a total of 24 (5+3×6+1) valence electrons available. Nitrogen forms a double bond with one oxygen and a single bond with the other two oxygen atoms, and all atoms have complete octets.

Determine the bond orders for N-O bonds in each molecule/ion

Now, let's determine the bond order for each N-O bond in these molecules/ions: 1. NO₊: N forms a double bond with O; bond order is 2. 2. NO₂⁻: N forms one single bond with O and one double bond with O; average bond order is (1+2)/2 = 1.5. 3. NO₃⁻: N forms one double bond with O and two single bonds with O; average bond order is (2+1+1)/3 = 4/3 ≈ 1.33.

Order the N-O bond lengths based on bond order

As a general rule, a higher bond order implies a shorter bond length. With the bond orders calculated in Step 2, we can now order the N-O bond lengths from shortest to longest: 1. NO₊ (bond order = 2) - shortest bond length 2. NO₂⁻ (bond order = 1.5) 3. NO₃⁻ (bond order ≈ 1.33) - longest bond length Therefore, the N-O bond lengths are ordered as NO₊ < NO₂⁻ < NO₃⁻ (from shortest to longest).

Key concepts

NO₊ bond length

In the context of Lewis structures, the bond length of the nitrogen-oxygen (NO) bond in the NO⁺ ion is influenced by the bond order, which is directly tied to the number of electrons shared between the atoms. For NO⁺, nitrogen and oxygen form a double bond. This double bond means that the bond order is 2.

This is significant because there is a correlation between the bond order and bond length: higher bond orders typically equate to shorter bond lengths. This is due to the fact that more electron sharing means stronger attraction between atoms and therefore a closer proximity between them. As such, the NO bond in NO⁺ is the shortest compared to its counterparts—NO₂⁻ and NO₃⁻—due to its highest bond order among them.

Understanding the relationship between bond order and bond length is pivotal. For students considering the structure of NO⁺, remember that:

• Double bonds lead to higher bond orders.
• Higher bond orders result in shorter bond distances.

This knowledge will help predict and rationalize the bond length hierarchy when comparing multiple compounds.

NO₂⁻ bond length

For the NO₂⁻ ion, bond length is calculated by considering the average bond order across the molecule. In NO₂⁻, nitrogen interacts with two oxygen atoms. One bond is a single bond, and the other is a double bond.

Combining these differing types of bonds gives an average bond order of 1.5. This is achieved by adding the bond orders from each nitrogen-oxygen bond (1 + 2) and dividing by two, given that two bonds exist. With this average bond order, the NO₂⁻ ion exhibits a bond length shorter than NO₃⁻, but longer than NO⁺.

A helpful approach in handling molecules like NO₂⁻ involves imagining the resonance between different Lewis structures:

• A resonance structure represents a valid arrangement of electrons across bonds.
• Resonance contributes equally to the average bond order calculation.
• A presence of resonance stabilizes the bond length into an intermediate value.

This intermediate bond length results from the partial double bond character distributed over its structure. Such an idea assists in explaining why NO₂⁻ does not have the shortest or the longest bond length, but is instead placed in between.

NO₃⁻ bond length

The nitrate ion, NO₃⁻, presents a fascinating case due to its resonance structures, which affect the bond order and consequently the bond length. In NO₃⁻, nitrogen is bonded to three oxygen atoms. It forms one double bond and two single bonds.

To understand the NO bond length here, note that these bonds are not fixed, but rather exist as resonance hybrids. When considering its multiple resonance forms, each contributes equally allowing the calculation of an average bond order of (2+1+1)/3 = 1.33.

This results in a situation where all NO bonds in NO₃⁻ are of equal length and show partial double bond character. This extended resonance reduces the bond's order compared to NO₂⁻ and NO⁺, making the bond length the longest of the three. When evaluating NO₃⁻:

• Consider the effects of resonance in equalizing bond lengths.
• Recognize that increased composition of single bond character results in longer bond lengths.
• Acknowledge that the calculated average bond order is less than 1.5, hence resulting in longer bonds compared to NO₂⁻ and NO⁺.

This demonstrates how resonance is not merely an abstract concept but has practical implications for molecular size and structure.