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Chapter 10: Problem 107

Sodium azide, \(\mathrm{NaN}_{3}\), is the nitrogen gas-forming substance used in automobile air-bag systems. It is an ionic compound containing the azide ion, \(\mathrm{N}_{3}^{-}\). In this ion, the two nitrogen-to-nitrogen bond lengths are \(116 \mathrm{pm} .\) Describe the resonance hybrid Lewis structure of this ion.

Short Answer

Expert verified
The azide ion, \(\mathrm{N}_{3}^{-}\), has a resonance hybrid structure in which the central nitrogen atom is interchangeably double-bonded to each of the other two nitrogen atoms and each nitrogen-to-nitrogen bond is a 1.5 bond due to the equal bond lengths of 116 pm.

Step by step solution

01

Sketch Lewis Structures

To start with, each Nitrogen atom has 5 valence electrons and we have 3 Nitrogen atoms, so we have 15 valence electrons. When we add the negative charge, that gives us one more electron, so we have a total of 16 electrons to distribute. There are several viable Lewis structures for the \(\mathrm{N}_{3}^{-}\) ion. One has a nitrogen atom in the center singly bonded (2 electrons each) to the two other nitrogen atoms and one lone pair on the center nitrogen atom, while the other nitrogen atoms each carry a negative charge and three lone pairs of electrons (6 electrons each). The other structure is similar, but with the negative charge and the additional bond on the other side.
02

Show the resonance

The actual structure of the azide ion is a resonance hybrid of these two Lewis structures. This is indicated by using a double-sided arrow between the two structures. The center nitrogen is doubly bonded to either of the other nitrogen atoms interchangeably, showing the non-localized nature of the pi electrons.
03

Represent the Resonance Hybrid

To represent the equivalent bond lengths in the resonance hybrid, we can draw a dashed line between the nitrogen atoms in the Lewis structure to show that each nitrogen-to-nitrogen bond is actually 1.5 bonds in the hybrid. This explains the equal bond lengths of 116 pm.

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Most popular questions from this chapter

Each of the following is either linear, angular (bent), planar, tetrahedral, or octahedral. Indicate the correct shape of (a) \(\mathrm{H}_{2} \mathrm{S} ;\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4} ;\) (c) \(\mathrm{HCN} ;\) (d) \(\mathrm{SbCl}_{6}^{-}\); (e) \(\mathrm{BF}_{4}^{-}\)

Each of the following molecules contains at least one multiple (double or triple) covalent bond. Give a plausible Lewis structure for (a) \(\mathrm{OCS} ;\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{F}_{2} \mathrm{CO} ;\) (d) \(\mathrm{Cl}_{2} \mathrm{SO} ;\) (e) \(\mathrm{C}_{2} \mathrm{H}_{2}\).

Both oxidation state and formal charge involve conventions for assigning valence electrons to bonded atoms in compounds, but clearly they are not the same. Describe several ways in which these concepts differ.

Draw Lewis structures for two different molecules with the formula \(\mathrm{C}_{3} \mathrm{H}_{4}\). Is either of these molecules linear? Explain.

Alternative strategies to the one used in this chapter have been proposed for applying the VSEPR theory to molecules or ions with a single central atom. In general, these strategies do not require writing Lewis structures. In one strategy, we write (1) the total number of electron pairs \(=[\) (number of valence electrons) \(\pm\) (electrons required for ionic charge) \(] / 2\) (2) the number of bonding electron pairs \(=\) (number of atoms) -1 (3) the number of electron pairs around central atom \(=\) total number of electron pairs \(-3 \times[\) number of terminal atoms (excluding \(\mathrm{H}\) )] (4) the number of lone-pair electrons = number of central atom pairs - number of bonding pairs After evaluating items \(2,3,\) and \(4,\) establish the VSEPR notation and determine the molecular shape. Use this method to predict the geometrical shapes of the following: (a) \(\mathrm{PCl}_{5} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{ClF}_{3} ;\) (d) \(\mathrm{SO}_{2} ;\) (e) \(\mathrm{ClF}_{4}^{-}\); (f) \(\mathrm{PCl}_{4}^{+}\). Justify each of the steps in the strategy, and explain why it yields the same results as the VSEPR method based on Lewis structures. How does the strategy deal with multiple bonds?
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