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Chapter 8: Problem 56

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

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

Step by step solution

01

Drawing the Lewis Structures

For each molecule, we need to draw the correct Lewis structure: 1. NO鈦�: 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鈦� 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鈮鈧�, where there is a triple bond between N and O. 2. NO鈧傗伝: 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 NO鈧傗伝 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. NO鈧冣伝: 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.
02

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鈦�: There is no resonance structure for NO鈦�. It simply has a triple bond between N and O, resulting in a relatively short bond length. 2. NO鈧傗伝: In the resonance structures of NO鈧傗伝, 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. NO鈧冣伝: 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 NO鈧冣伝 will have bond lengths between a single bond and a double bond, and likely be slightly longer than those in NO鈧傗伝 due to the extra oxygen atom sharing the double bond character.
03

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鈦� has a triple bond, resulting in the shortest bond length. 2. The N-O bonds in NO鈧傗伝 have resonance, making them shorter than a typical single bond but longer than a double bond. 3. The N-O bonds in NO鈧冣伝 have a double bond character shared among all the bonds, and their bond lengths are likely slightly longer than those in NO鈧傗伝. So, the ordering of the N-O bond lengths is NO鈦� < NO鈧傗伝 < NO鈧冣伝.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈦� 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 NO鈧傗伝 and NO鈧冣伝 ions are affected by resonance, distributing the bond character and consequently affecting the bond length. In NO鈧冣伝, bonds are longer than in NO鈧傗伝 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 NO鈧傗伝, 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 NO鈧冣伝, 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鈦� 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 NO鈧傗伝, we count 18 valence electrons in total after accounting for additional electrons from the negative charge. Lastly, NO鈧冣伝 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.

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Most popular questions from this chapter

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+}\) and is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-}\). Why, then, is \(\mathrm{CaO}\) stable relative to the free elements?

Predict the ordering of the \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{CO}, \mathrm{CO}_{2}\), and \(\mathrm{CO}_{3}^{2-}\).

Some chemists believe that satisfaction of the octet rule should be the top criterion for choosing the dominant Lewis structure of a molecule or ion. Other chemists believe that achieving the best formal charges should be the top criterion. Consider the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) in which the \(\mathrm{H}\) atoms are bonded to \(\mathrm{O}\) atoms. (a) What would be the predicted dominant Lewis structure if satisfying the octet rule is the top criterion? (b) What would it be if achieving the best formal charges is the top criterion? (c) Is there another Lewis structure you can draw that satisfies neither of these criteria?

An ionic substance of formula MX has a lattice energy of \(6 \times 10^{3} \mathrm{k} \mathrm{J} / \mathrm{mol}\). Is the charge on the ion \(\mathrm{M}\) likely to be \(1+\), \(2+\) or \(3+?\) Explain your reasoning.

The \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Are there any differences in the electron configurations of \(\mathrm{Ti}^{2+}\) and Ca? (b) With reference to Figure 6.24 , comment on the changes in the ordering of the \(4 s\) and \(3 d\) subshells in Ca and \(\mathrm{Ti}^{2+}\), (c) Will Ca and \(\mathrm{Ti}^{2+}\) have the same number of unpaired electrons? Explain.
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