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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 minizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central Natom 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
a) The Lewis structure of the azide ion is either N鈮=N鈦� or N鈦�=N鈮, and its molecular geometry is linear. b) The hybridization of the central nitrogen atom in the azide ion is sp. c) The central nitrogen atom forms 2 sigma (蟽) bonds and 2 pi (蟺) bonds in the azide ion.

Step by step solution

01

Calculate the total number of valence electrons

In the azide ion, N3-, there are three nitrogen atoms, and since nitrogen has 5 valence electrons, the total valence electrons are 3 * 5 = 15. Additionally, as there is a negative charge on the ion, one extra electron must be considered, resulting in a total of 16 valence electrons.
02

Draw the skeleton Lewis structure

Place the least electronegative atom (central nitrogen) in the center and connect the other two nitrogen atoms to it: N - N - N
03

Distribute the remaining valence electrons to minimize formal charge

In this step, we will place the remaining valence electrons on the nitrogen atoms as lone pairs to minimize formal charges while considering the octet rule. N鈮=N鈦� 鈫� N鈦�=N鈮 The structure on the left has a formal charge of -1 on the left nitrogen atom, +1 on the central nitrogen, and 0 on the right nitrogen atom. The structure on the right has a formal charge of 0 on the left nitrogen atom, +1 on the central nitrogen, and -1 on the right nitrogen atom. As both structures have the same overall formal charge, either structure can represent the azide ion. So, the final Lewis structure for the azide ion is either: N鈮=N鈦� or N鈦�=N鈮
04

Determine the molecular geometry

The central nitrogen atom has two atoms directly attached and no lone pairs. Hence, it is linear.
05

Determine the hybridization of the central nitrogen atom

The central nitrogen atom is involved in a single bond and a triple bond; therefore, it has three sigma bonds in total. According to the VSEPR theory, the hybridization on the central nitrogen atom is sp, as it has two electron domains.
06

Identify the number of sigma and pi bonds for the central nitrogen atom

Since the central nitrogen is involved in a single bond (1 sigma bond) and a triple bond (1 sigma and 2 pi bonds), there is a total of: -2 sigma (蟽) bonds: 1 from the single bond and 1 from the triple bond -2 pi (蟺) bonds: These are the remaining 2 bonds from the triple bond To summarize: a) The Lewis structure of the azide ion is either N鈮=N鈦� or N鈦�=N鈮, and its molecular geometry is linear. b) The hybridization of the central nitrogen atom in the azide ion is sp. c) The central nitrogen atom forms 2 sigma (蟽) bonds and 2 pi (蟺) bonds in the azide ion.

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Key Concepts

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

Lewis Structure
Understanding the Lewis structure of a molecule is a foundation in studying its chemical behavior. In the Lewis structure, electrons are represented as dots, and bonds between atoms are represented by lines. For the azide ion, (N_3^{-}), the process of drawing the structure begins with accounting for the valence electrons. Nitrogen, with five valence electrons, contributes a total of fifteen for three of its atoms, and an additional electron comes from the negative charge, making up sixteen valence electrons.

To minimize formal charges, which helps in predicting the most stable structure, we have two resonance forms: N鈮=N鈦� and N鈦�=N鈮. The negative charge can be on the terminal nitrogen atoms, not on the central nitrogen, to ensure each nitrogen atom has a complete octet. By fulfilling the octet rule and reducing formal charges, the Lewis structure gives a depiction of stable electron distribution in the azide ion.
Molecular Geometry
The molecular geometry is determined by the spatial arrangement of atoms in a molecule or ion. The VSEPR (Valence Shell Electron Pair Repulsion) theory is invaluable in predicting these structures. For the azide ion, we observe that the central nitrogen atom has two bonding domains and no lone pairs, leading to a linear shape. This linear geometry is typical for species with a hybridization of sp, where two electron domains are present. A linear structure also implies a bond angle of 180 degrees, a characteristic trait of molecules with a 'straight line' appearance, which includes the azide ion.
Sigma and Pi Bonds
In chemical bonding, we differentiate between sigma (蟽) and pi (蟺) bonds. Sigma bonds are the first bonds formed between two atoms and involve head-on overlap of orbitals. Pi bonds, on the other hand, occur when parallel orbital overlap happens and always accompany a sigma bond in multiple bonds.

For the central nitrogen in the azide ion, there are two sigma bonds and two pi bonds. One sigma bond originates from the end-to-end overlap with each of the neighboring nitrogen atoms. The triple bond with one of the terminal nitrogen atoms consists of one sigma and two pi bonds. This configuration is vital for understanding the chemical reactivity and physical properties of the azide ion.
Nitrogen Hybridization
The concept of hybridization explains how atomic orbitals mix to form new hybrid orbitals, affecting both molecular geometry and bonding. For nitrogen, an element in the second period of the periodic table, sp hybridization occurs when one s orbital and one p orbital blend together. This results in two sp-hybrid orbitals.

In the azide ion's central nitrogen atom, sp hybridization is anticipated due to the combination of one single bond and one triple bond, leading to two regions of electron density鈥攃onsistent with the linear geometry of the molecule. Understanding hybridization helps in predicting the electron distribution in a molecule, which in turn helps in anticipating molecular reactivity.

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Most popular questions from this chapter

Describe the bond angles to be found in each of the following molecular structures: (a) trigonal planar, (b) tetrahedral, (c) octahedral, (d) linear.

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 p}-\pi_{2 p}^{\star}\) ,\(\sigma_{25}-\sigma_{25}^{*},\) and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(N_{2} .\) (b) Is \(N_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(N_{2}\) molecule has the lowest energy peak at 170 nm. To what orbital transition does this corre- spond? (a) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of kJ/mol. (e) Is the N-N bondin the first excited state stronger or weaker compared to that in the ground state?

Which of the following statements about hybrid orbitals is or are true? (i) After an atom undergoes sp hybridization, there is one unhybridized \(p\) orbital on the atom, (ii) Under \(s p^{2}\) hybridization, the large lobes point to the vertices of an equilateral triangle, and (iii) The angle between the large lobes of \(s p^{3}\) hybrids is \(109.5^{\circ} .\)

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_ \(_{4} )\) according to the following unbalanced reaction: \begin{equation}\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) \end{equation} The O atom and the four \(\mathrm{F}\) atoms in OSF \(_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of OSF_ in which the formal charges of all atoms are zero.(c) Use average bond enthalpies (Table 8.3 ) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(I_{2} ?\) Explain your answer.
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