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In addition to those in Table \(14.3,\) other less stable nitrogen oxides exist. Draw a Lewis structure for each of the following: (a) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with an \(\mathrm{N}-\mathrm{N}\) bond (b) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with no \(\mathrm{N}-\mathrm{N}\) bond (c) \(\mathrm{N}_{2} \mathrm{O}_{3}\) with no \(\mathrm{N}-\mathrm{N}\) bond (d) \(\mathrm{NO}^{+}\) and \(\mathrm{NO}_{3}^{-}\), products of the ionization of liquid \(\mathrm{N}_{2} \mathrm{O}_{4}\)

Short Answer

Expert verified
Lewis structures involve placing atoms and electrons to satisfy the octet rule for given molecular formulas and charges.

Step by step solution

01

Draw the Lewis Structure of \(\mathrm{N}_{2}\mathrm{O}_{2}\) with an \(\mathrm{N}-\mathrm{N}\) bond

Start by connecting the two nitrogen atoms with a single bond. Each nitrogen atom will also have one oxygen atom bonded to it. Assign electrons to complete the octets of oxygen atoms first, then place leftover electrons on nitrogen. Add any lone pairs as necessary to complete the Lewis structure.
02

Draw the Lewis Structure of \(\mathrm{N}_{2}\mathrm{O}_{2}\) without an \(\mathrm{N}-\mathrm{N}\) bond

Connect each nitrogen atom to an oxygen atom as done before, but do not connect the nitrogen atoms with an \(\mathrm{N}-\mathrm{N}\) bond. Arrange the electrons to satisfy the octet rule for all atoms involved by making double bonds if necessary.
03

Draw the Lewis Structure of \(\mathrm{N}_{2}\mathrm{O}_{3}\) with no \(\mathrm{N}-\mathrm{N}\) bond

Start by placing the two nitrogen atoms separately. Connect one oxygen atom between them, then attach the remaining oxygen atoms appropriately while ensuring the valence electrons are correct and each atom achieves an octet where possible.
04

Draw the Lewis Structure of \(\mathrm{NO}^{+}\) and \(\mathrm{NO}_{3}^{-}\)

For \(\mathrm{NO}^{+}\), connect the nitrogen and oxygen atoms, and account for the positive charge by removing one electron. For \(\mathrm{NO}_{3}^{-}\), connect the nitrogen atom to three oxygen atoms and add one extra electron to account for the negative charge. Arrange electrons to ensure octets are satisfied.

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Key Concepts

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

Nitrogen Oxides
Nitrogen oxides are a group of compounds made up of nitrogen and oxygen atoms. These compounds vary in their stability and the number of oxygen and nitrogen atoms they include. Some well-known nitrogen oxides include nitrogen monoxide (NO), nitrogen dioxide (NO2), and dinitrogen tetroxide (N2O4). In these exercises, we explore different configurations, including less stable variants like N2O2 and N2O3. These variations can have different bonds and structures, which are important when drawing their Lewis structures. Understanding these differences helps in predicting their chemical behavior and reactivity. Keep in mind these structures often play a significant role in environmental chemistry and industrial processes.
Molecular Bonding
Molecular bonding pertains to how atoms connect in a molecule using different types of bonds. Bonds can be single, double, or triple, representing one, two, or three pairs of shared electrons respectively. Let's take N2O2 as an example: when nitrogen atoms bond to each other (N-N bond) or separately to oxygen atoms, it changes the molecular arrangement. This kind of bonding helps us understand how atoms form compounds and how those compounds will react with others. For instance:
  • Single bonds (like in N-N) share one pair of electrons.
  • Double bonds share two pairs of electrons, which may be seen in NO2 structures.
  • Knowing these bonds helps draw their Lewis structures correctly.
Octet Rule
The octet rule is a fundamental guideline in chemistry stating that atoms tend to form bonds until they are surrounded by eight electrons, achieving a stable electron arrangement similar to noble gases. When drawing Lewis structures, the octet rule ensures each atom achieves this stable configuration. For instance:
  • Nitrogen, with five valence electrons, will typically form three bonds to reach an octet.
  • Oxygen, with six valence electrons, will often form two bonds.
In our exercise, each nitrogen and oxygen atom should be examined to ensure they fulfill the octet rule where possible. However, exceptions exist, especially in molecules with odd numbers of electrons or those involving elements from period three or beyond on the periodic table.
Electron Pairs
Electron pairs play a crucial role in determining the shape and reactivity of a molecule. In Lewis structures, electron pairs fall into two categories:
  • Bonding pairs: Shared between atoms to create a bond.
  • Lone pairs: Not shared and remain on a single atom.
For instance, in the \(NO_3^{-}\) ion:
  • Each oxygen must have three lone pairs, and the nitrogen must have one lone pair and three bonding pairs.
Recognizing where lone and bonding pairs are located helps predict molecular geometry and reactivity. Different configurations of electron pairs can create various molecule shapes, influencing how molecules interact with others.
Chemical Ionization
Chemical ionization involves the gain or loss of electrons, leading to the formation of ions. These ions can greatly influence a molecule's properties. For example, in the provided exercise:
  • The \(NO^{+}\) ion forms by removing one electron, making it positively charged.
  • The \(NO_3^{-}\) ion forms by adding one electron, giving it a negative charge.
Ionization alters how molecules interact, especially in solutions where ions can conduct electricity or participate in specific chemical reactions. Understanding ionization is crucial for predicting behavior and reactions in different chemical environments.

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Most popular questions from this chapter

The main reason alkali metal dihalides \(\left(\mathrm{MX}_{2}\right)\) do \(\mathrm{not}\) form is the high \(\mathrm{IE}_{2}\) of the metal. (a) Why is IE \(_{2}\) so high for alkali metals? (b) The IE \(_{2}\) for Cs is \(2255 \mathrm{~kJ} / \mathrm{mol}\), low enough for \(\mathrm{CsF}_{2}\) to form exothermically \(\left(\Delta H_{i}^{\circ}=-125 \mathrm{~kJ} / \mathrm{mol}\right) .\) This compound cannot be synthesized, however, because CsF forms with a much greater release of heat \(\left(\Delta H_{\mathrm{f}}^{\circ}=-530 \mathrm{~kJ} / \mathrm{mol}\right) .\) Thus, the breakdown of \(\mathrm{CsF}_{2}\) to CsF happens readily. Write the equation for this breakdown, and calculate the enthalpy of reaction per mole of CsF.

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