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Magazine Chapter 8: Problem 56
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
Expert verified
The N-O bond lengths are ordered as NO鈧� < NO鈧傗伝 < NO鈧冣伝 (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
01
Draw the Lewis structures for NO鈧�, NO鈧傗伝, and NO鈧冣伝
First, we'll draw the Lewis structures for each molecule/ion:
1. NO鈧�: Nitrogen (N) has 5 valence electrons, and oxygen (O) has 6 valence electrons. In NO鈧�, 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. NO鈧傗伝: Nitrogen (N) has 5 valence electrons, and each of the two oxygen atoms (O) has 6 valence electrons. In NO鈧傗伝, the molecule has a -1 charge overall, meaning it gains one electron. Thus, there are a total of 18 (5+2脳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. NO鈧冣伝: Nitrogen (N) has 5 valence electrons, and each of the three oxygen atoms (O) has 6 valence electrons. In NO鈧冣伝, the molecule has a -1 charge overall, meaning it gains one electron. Thus, there are a total of 24 (5+3脳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.
02
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鈧�: N forms a double bond with O; bond order is 2.
2. NO鈧傗伝: N forms one single bond with O and one double bond with O; average bond order is (1+2)/2 = 1.5.
3. NO鈧冣伝: N forms one double bond with O and two single bonds with O; average bond order is (2+1+1)/3 = 4/3 鈮� 1.33.
03
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鈧� (bond order = 2) - shortest bond length
2. NO鈧傗伝 (bond order = 1.5)
3. NO鈧冣伝 (bond order 鈮� 1.33) - longest bond length
Therefore, the N-O bond lengths are ordered as NO鈧� < NO鈧傗伝 < NO鈧冣伝 (from shortest to longest).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
NO鈧� bond length
In the context of Lewis structures, the bond length of the nitrogen-oxygen (NO) bond in the NO鈦� ion is influenced by the bond order, which is directly tied to the number of electrons shared between the atoms. For NO鈦�, 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鈦� is the shortest compared to its counterparts鈥擭O鈧傗伝 and NO鈧冣伝鈥攄ue 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鈦�, remember that:
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鈦� is the shortest compared to its counterparts鈥擭O鈧傗伝 and NO鈧冣伝鈥攄ue 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鈦�, remember that:
- Double bonds lead to higher bond orders.
- Higher bond orders result in shorter bond distances.
NO鈧傗伝 bond length
For the NO鈧傗伝 ion, bond length is calculated by considering the average bond order across the molecule. In NO鈧傗伝, 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 NO鈧傗伝 ion exhibits a bond length shorter than NO鈧冣伝, but longer than NO鈦�.
A helpful approach in handling molecules like NO鈧傗伝 involves imagining the resonance between different Lewis structures:
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 NO鈧傗伝 ion exhibits a bond length shorter than NO鈧冣伝, but longer than NO鈦�.
A helpful approach in handling molecules like NO鈧傗伝 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.
NO鈧冣伝 bond length
The nitrate ion, NO鈧冣伝, presents a fascinating case due to its resonance structures, which affect the bond order and consequently the bond length. In NO鈧冣伝, 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 NO鈧冣伝 are of equal length and show partial double bond character. This extended resonance reduces the bond's order compared to NO鈧傗伝 and NO鈦�, making the bond length the longest of the three. When evaluating NO鈧冣伝:
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 NO鈧冣伝 are of equal length and show partial double bond character. This extended resonance reduces the bond's order compared to NO鈧傗伝 and NO鈦�, making the bond length the longest of the three. When evaluating NO鈧冣伝:
- 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 NO鈧傗伝 and NO鈦�.
