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Chapter 9: Problem 100

Sodium azide is a shock-sensitive compound that releases \(\mathrm{N}_{2}\) upon physical impact. The compound is used in automobile airbags. The azide ion is \(\mathrm{N}_{3}\). (a) Draw the Lewis structure of the azide ion that minimizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central \(\mathrm{N}\) atom in the azide ion. (c) How many \(\sigma\) bonds and how many \(\pi\) bonds does the central nitrogen atom make in the azide ion?

Short Answer

Expert verified
The azide ion has a linear shape with a Lewis structure of \[ \dddot{N} \equiv \stackrel{.}{\ddot N} - \stackrel{.}{\ddot N} \]. The hybridization of the central nitrogen atom is sp, and it forms 2 sigma bonds (σ) and 2 pi bonds (π).

Step by step solution

01

Draw the Lewis Structure of the Azide Ion

To draw the Lewis structure, start by considering the number of valence electrons. Nitrogen has 5 valence electrons, so three nitrogen atoms contribute a total of 15 valence electrons (5 × 3 = 15). In the azide ion, two of the nitrogen atoms are bonded through a single bond, and the central nitrogen atom is bonded to the other terminal nitrogen atom via a triple bond: N ≡ N - N To show this linear configuration, we draw the azide ion as: \[ \dddot{N} \equiv \stackrel{.}{\ddot N} - \stackrel{.}{\ddot N} \] All three nitrogen atoms have a formal charge of zero, which minimizes the formal charge.
02

Determine Linear or Bent Shape

Since there are no lone pair electrons on the central nitrogen atom and it has two bonding areas (one single bonded terminal nitrogen and one triple bonded terminal nitrogen), the azide ion has a linear shape.
03

Identify Hybridization of Central Nitrogen Atom

To determine the hybridization of the central nitrogen atom, count its bonding regions and lone pairs. The central nitrogen atom has two bonding regions and no lone pairs. For two effective electron pairs around the central atom, the hybridization will be sp, as it includes one s and one p orbital. So, the hybridization of the central nitrogen atom in the azide ion is sp.
04

Calculate Sigma and Pi Bonds

With the azide ion's Lewis structure, we can now determine the number of sigma (σ) bonds and pi (π) bonds made by the central nitrogen atom. For each single bond, there is one sigma bond, and for every double or triple bond, there is only one sigma bond. In the azide ion, the central nitrogen atom is connected with one single bond and one triple bond. Thus, it has 1 single bond (1 σ bond) and 1 triple bond (1 σ bond + 2 π bonds). Therefore, the central nitrogen atom in the azide ion makes 2 sigma bonds (σ) and 2 pi-bonds (π).

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Most popular questions from this chapter

Consider the molecule \(\mathrm{PF}_{4}\) Cl. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl}\). How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

Benzaldehyde, \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}\), is a fragrant substance responsible for the aroma of almonds. Its Lewis structure is O=Cc1cccc(C=O)c1 (a) What is the hybridization at each of the carbonatoms of the molecule? (b) What is the total number of valence electrons in benzaldehyde? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals combine to form a \(\pi\) bond? Explain.

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the 3 d orbitals (Figure 6.23 ) is most likely to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and $\sigma_{3 d}^{*}$ antibonding MOs. (c) With reference to the "Closer Look" box on the phases oforbitals, explain why a node is generated in the \(\sigma_{3 d}^{*}\) MO. (d) Sketch the energylevel diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2} ?\)

How does a trigonal pyramid differ from a tetrahedron so far as molecular geometry is concerned?
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