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Chapter 8: Problem 6

Incomplete Lewis structures for the nitrous acid molecule, \(\mathrm{HNO}_{2},\) and the nitrite ion, \(\mathrm{NO}_{2}^{-},\) are shown below. (a) Complete each Lewis structure by adding electron pairs as needed. (b) Is the formal charge on \(\mathrm{N}\) the same or different in these two species? (c) Would either \(\mathrm{HNO}_{2}\) or \(\mathrm{NO}_{2}^{-}\) be expected to exhibit resonance? (d) Would you expect the \(\mathrm{N}=\mathrm{O}\) bond in \(\mathrm{HNO}_{2}\) to be longer, shorter, or the same length as the \(\mathrm{N}-\mathrm{O}\) bonds in \(\mathrm{NO}_{2}^{-}\) ? Explain. [Sections 8.5 and \(\left.8.6\right]\)

Short Answer

Expert verified
The completed Lewis structures for HNO鈧� and NO鈧傗伝 are: HNO鈧�: H-O-N=O with lone pairs on O atoms. NO鈧傗伝: (-)O-N=O(-) with resonance. The formal charge on N is +1 in HNO鈧� and 0 in NO鈧傗伝. Resonance is only exhibited in the NO鈧傗伝 ion. The N=O bond in HNO鈧� is expected to be shorter than the average N-O bond length in NO鈧傗伝 due to resonance.

Step by step solution

01

Complete Lewis structures

First, we will complete the Lewis structures for HNO鈧� and NO鈧傗伝. Recall that N has 5 valence electrons, O has 6, and H has 1. Here are the completed Lewis structures: For HNO鈧�: H-O-N=O with the following electron pairs (lone pairs in parentheses): )O-N=O( \(\\ \) For NO鈧傗伝: O-N=O with the following electron pairs (lone pairs in parentheses): (-)O-N=O(-)
02

Compare formal charges

Next, we calculate the formal charge on the nitrogen atom for both HNO鈧� and NO鈧傗伝. Formal charge = (valence electrons) - (nonbonding electrons) - (1/2 * bonding electrons) For HNO鈧�: Formal charge on N = 5 - 0 - (1/2 * 8) = 5 - 4 = +1 For NO鈧傗伝: Formal charge on N = 5 - 0 - (1/2 * 10) = 5 - 5 = 0 The formal charge on nitrogen is different in these two species.
03

Resonance structures

A molecule exhibits resonance when there are multiple valid Lewis structures with the same arrangement of atoms but different electronic configurations. In HNO鈧�, there is only one valid Lewis structure, so it does not exhibit resonance. However, for NO鈧傗伝, there is another valid Lewis structure: (-)O-N=O<->O=N-O(-) Therefore, NO鈧傗伝 exhibits resonance.
04

Compare bond lengths

HNO鈧� has a double bond between N and O (N=O), while NO鈧傗伝 has a combination of single and double bonds (N-O and N=O) due to resonance. The actual bond length in NO鈧傗伝 will be an average of single and double bond lengths. A double bond is shorter than a single bond, so we expect the N=O bond in HNO鈧� to be shorter than the average N-O bond length in NO鈧傗伝 due to resonance.

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Key Concepts

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

Formal Charge
Understanding formal charge is crucial for determining the most stable Lewis structure of a molecule. A formal charge on an atom is an artificial charge assigned to it as a way of bookkeeping. It's calculated using the formula:
\[\text{Formal Charge} = (\text{Valence Electrons}) - (\text{Nonbonding Electrons}) - \frac{1}{2} \times (\text{Bonding Electrons})\]For example, in the nitrous acid molecule, \(\text{HNO}_2\), the nitrogen atom has 5 valence electrons. After accounting for the shared and non-shared electrons, the formal charge on nitrogen in \(\text{HNO}_2\) is calculated to be +1.
In contrast, in the nitrite ion, \(\text{NO}_2^-\), the nitrogen atom carries a formal charge of 0. This distinct difference in formal charge helps predict and explain the differing chemical behaviors of \(\text{HNO}_2\) and \(\text{NO}_2^-\). This calculation is essential for determining molecular geometry and reactivity as molecules tend to arrange themselves to minimize formal charges across their atoms.
Resonance Structures
Resonance structures are different ways of depicting a molecule where the array of atoms stays constant, but the distribution of electrons varies. Such structures are crucial for molecules like \(\text{NO}_2^-\), which can exhibit resonance. This reflects the potential for electrons to be shared between different oxygen atoms in different configurations.
While \(\text{HNO}_2\) has only one Lewis structure, indicating no resonance, the nitrite ion \(\text{NO}_2^-\) can resonate between two forms.
  • In one structure, the negative charge is on the "left" oxygen, while in the other, it is on the "right" oxygen.
  • This ability for resonance implies more stability for \(\text{NO}_2^-\) due to equal distribution of charge and electron density.
Understanding resonance is essential because it explains why certain molecule or ions are more stable, makes calculations for predicted bond lengths possible, and affects how a molecule might react with others.
Bond Length Comparison
Bond length can tell us a lot about the strength and nature of bonds within a molecule. In general, double bonds are shorter than single bonds due to additional electron sharing strengthening the bond. In our exercise, \(\text{HNO}_2\) has a fixed double bond between nitrogen and oxygen, implying a shorter bond length.
Conversely, \(\text{NO}_2^-\) features resonance between its two structures, which results in neither a pure single nor double bond, but rather an intermediate bond length.
  • Resonance blurs the division between single and double bond lengths.
  • This results in bonds being shorter than a typical single bond but not as short as a typical double bond.
Thus, the \(\text{N=O}\) bond in \(\text{HNO}_2\) should be shorter than the bond lengths in \(\text{NO}_2^-\) due to its definitive double bond, as opposed to the averaged bond length in \(\text{NO}_2^-\) resulting from resonance.

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Most popular questions from this chapter

Use Table 8.4 to estimate the enthalpy change for each of the following reactions: (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{Cl}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12}(g)\) (the six carbon atoms form a six-membered ring with two \(\mathrm{H}\) atoms on each \(\mathrm{C}\) atom \()\)

Draw the Lewis structures for each of the following ions or molecules. Identify those that do not obey the octet rule, and explain why they do not: (a) \(\mathrm{SO}_{3}^{2-}\), (b) \(\mathrm{AlH}_{3}\) (c) \(\mathrm{N}_{3}^{-}\) (d) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (e) \(\mathrm{SbF}_{5}\)

Based on data in Table 8.2 , estimate (within \(30 \mathrm{~kJ} / \mathrm{mol}\) ) the lattice energy for (a) LiBr, (b) CsBr, (c) \(\mathrm{CaCl}_{2}\)

A common form of elemental phosphorus is the tetrahedral \(\mathrm{P}_{4}\) molecule, where all four phosphorus atoms are equivalent: At room temperature phosphorus is a solid. (a) Do you think there are any unshared pairs of electrons in the \(\mathrm{P}_{4}\) molecule? (b) How many \(\mathrm{P}-\mathrm{P}\) bonds are there in the molecule? (c) Can you draw a Lewis structure for a linear \(\mathrm{P}_{4}\) molecule that satisfies the octet rule? (d) Using formal charges, what can you say about the stability of the linear molecule versus that of the tetrahedral molecule?

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