/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 Sodium azide is a shock-sensitiv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 (蟽) and 2 pi bonds (蟺).

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 脳 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 鈮� 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 (蟽) bonds and pi (蟺) 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 蟽 bond) and 1 triple bond (1 蟽 bond + 2 蟺 bonds). Therefore, the central nitrogen atom in the azide ion makes 2 sigma bonds (蟽) and 2 pi-bonds (蟺).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Formal Charge
When considering the Lewis structure of a molecule, one of the key elements is the formal charge. The formal charge helps determine the most stable Lewis structure by indicating the charge each atom would have if all the atoms in the molecule were sharing electrons equally. To calculate the formal charge, use the formula:
  • Formal charge = (Valence electrons of the atom) 鈥� (Non-bonding electrons) 鈥� (1/2 * Bonding electrons)
In the case of the azide ion \( ext{N}_3^- \), each nitrogen atom strives for a formal charge of zero to achieve stability.
  • The first nitrogen atom in the azide structure is attached to one triple bond, incorporating no lone pairs, and it maintains a formal charge of zero.
  • The central nitrogen, also achieving a formal charge of zero, participates in one single and one triple bond with the other nitrogen atoms.
  • The third nitrogen with a single bond also aims for a formal charge of zero by having appropriate lone pairs.
Minimizing the formal charges is crucial for determining the most favorable structure, which in this case, results in a linear molecular geometry.
Hybridization
Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals. These new orbitals are designed to accommodate electron pairs in bond formation, essentially explaining molecular geometry.
For the central nitrogen atom in the azide ion, which is represented as \(N\equiv N-N\), the hybridization is sp.
  • This sp hybridization arises because the central nitrogen atom has two regions of electron density around it鈥攐ne associated with the triple bond and one with the single bond.
  • To accommodate these two bonding regions, the central nitrogen atom uses one s orbital and one p orbital to form two sp hybrid orbitals.
Sp hybridization results in a linear structure as the angle between the hybrid orbitals is 180 degrees. This linearity aligns with the actual structure of the azide ion, demonstrating how hybridization helps predict molecular shape.
Sigma and Pi Bonds
When examining the covalent bonds within a molecule, they are typically characterized as either sigma \((\sigma)\) or pi \((\pi)\) bonds, based on their formation.
  • Sigma bonds occur when orbitals overlap head-to-head. These bonds are strong and allow free rotation of bonded atoms. Every single bond contains one sigma bond.
  • Pi bonds, by contrast, are formed when orbitals overlap side-to-side. These are weaker than sigma bonds and restrict rotation due to their electron cloud placement above and below the bond axis.
In the azide ion \( (\text{N}_3^-) \):
  • The central nitrogen atom engages in a single bond with the terminal nitrogen, which includes one sigma bond.
  • In addition to the single bond, it participates in a triple bond with the other terminal nitrogen atom, consisting of one sigma bond and two pi bonds.
  • Overall, the central nitrogen atom forms two sigma bonds and two pi bonds, illustrating how its bonding arrangement supports the ion's stability and linear structure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(O\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) 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 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.

(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}{ }^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}^{2-} ?\) For those ions that are paramagnetic, determine the number of unpaired electrons.

Methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO},\) was made infamous in 1984 when an accidental leakage of this compound from a storage tank in Bhopal, India, resulted in the deaths of about 3800 people and severe and lasting injury to many thousands more. (a) Draw a Lewis structure for methyl isocyanate. (b) Draw a ball-and-stick model of the structure, including estimates of all the bond angles in the compound. (c) Predict all the bond distances in the molecule. (d) Do you predict that the molecule will have a dipole moment? Explain.

How many electron domains are surrounding the central atoms which adopt the following geometries? (a) linear (b) trigonal planar (c) trigonal pyramidal (d) trigonal bipyramidal.

Ethyl propanoate, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3},\) gives a fruity pineapple-like smell. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\sigma\) and how many \(\pi\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.