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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 HNO₂ and NO₂⁻ are: HNO₂: H-O-N=O with lone pairs on O atoms. NO₂⁻: (-)O-N=O(-) with resonance. The formal charge on N is +1 in HNO₂ and 0 in NO₂⁻. Resonance is only exhibited in the NO₂⁻ ion. The N=O bond in HNO₂ is expected to be shorter than the average N-O bond length in NO₂⁻ due to resonance.

Step by step solution

01

Complete Lewis structures

First, we will complete the Lewis structures for HNO₂ and NO₂⁻. Recall that N has 5 valence electrons, O has 6, and H has 1. Here are the completed Lewis structures: For HNO₂: H-O-N=O with the following electron pairs (lone pairs in parentheses): )O-N=O( \(\\ \) For NO₂⁻: 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 HNO₂ and NO₂⁻. Formal charge = (valence electrons) - (nonbonding electrons) - (1/2 * bonding electrons) For HNO₂: Formal charge on N = 5 - 0 - (1/2 * 8) = 5 - 4 = +1 For NO₂⁻: 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 HNO₂, there is only one valid Lewis structure, so it does not exhibit resonance. However, for NO₂⁻, there is another valid Lewis structure: (-)O-N=O<->O=N-O(-) Therefore, NO₂⁻ exhibits resonance.
04

Compare bond lengths

HNO₂ has a double bond between N and O (N=O), while NO₂⁻ has a combination of single and double bonds (N-O and N=O) due to resonance. The actual bond length in NO₂⁻ 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 HNO₂ to be shorter than the average N-O bond length in NO₂⁻ due to resonance.

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Most popular questions from this chapter

In the following pairs of binary compounds determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2}\), (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\), (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\).

With reference to the "Chemistry Put to Work" box on explosives, (a) use bond enthalpies to estimate the enthalpy change for the explosion of \(1.00 \mathrm{~g}\) of nitroglycerin. (b) Write a balanced equation for the decomposition of TNT. Assume that, upon explosion, TNT decomposes into \(\mathrm{N}_{2}(g), \mathrm{CO}_{2}(g)\), \(\mathrm{H}_{2} \mathrm{O}(g),\) and \(\mathrm{C}(s)\)

Average bond enthalpies are generally defined for gas-phase molecules. Many substances are liquids in their standard state. coo (Section 5.7) By using appropriate thermochemical data from Appendix C, calculate average bond enthalpies in the liquid state for the following bonds, and compare these values to the gas-phase values given in Table 8.4: (a) \(\mathrm{Br}-\mathrm{Br}\), from \(\mathrm{Br}_{2}(l) ;\) (b) \(\mathrm{C}-\mathrm{Cl},\) from \(\mathrm{CCl}_{4}(l) ;\) (c) \(\mathrm{O}-\mathrm{O},\) from \(\mathrm{H}_{2} \mathrm{O}_{2}(I)\) (assume that the \(\mathrm{O}-\mathrm{H}\) bond enthalpy is the same as in the gas phase). (d) What can you conclude about the process of breaking bonds in the liquid as compared to the gas phase? Explain the difference in the \(\Delta H\) values between the two phases.

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