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Chapter 10: Problem 105

Write one Lewis structure of \(\mathrm{N}_{2} \mathrm{O}_{5}\left(\mathrm{O}_{2} \mathrm{NONO}_{2}\right.\) skeleton structure). What are the bond angles around the central oxygen atom and the two nitrogen atoms? What is the hybridization of each?

Short Answer

Expert verified
The N-O-N bond angles are around 120掳, N-O-O angles are ~109.5掳, and the oxygen hybridization is sp. Nitrogens are sp虏 and sp鲁 hybridized.

Step by step solution

01

Determine Total Valence Electrons

To draw the Lewis structure, we first calculate the total number of valence electrons in \( \text{N}_2\text{O}_5 \). Nitrogen has 5 valence electrons, and oxygen has 6 valence electrons. Thus, we have: \[ 2 \times 5 + 5 \times 6 = 40 \text{ valence electrons}. \]
02

Establish the Skeleton Structure

The structural formula given is \( \text{O}_2 \text{NONO}_2 \). This means we connect the atoms as follows: O-N-O-N-O-O. The two nitrogen atoms are central, each bonded to an oxygen atom on either side.
03

Add Single Bonds

Draw single bonds between each pair of bonded atoms (O-N, N-O, N-O, O-O). Each single bond represents 2 electrons. We initially use 8 electrons for these four bonds.
04

Complete the Octet for Oxygen Atoms

Add lone pairs to the oxygen atoms to complete their octets, keeping in mind the electrons already used in bonding. Each oxygen needs 6 additional electrons to complete its octet, except for those forming double bonds.
05

Use Double Bonds to Satisfy Octets

Change O-N single bonds to double bonds where needed to ensure that all atoms have complete octets. The oxygens not involved in peroxide linkage need a double bond to the nitrogen to satisfy the octet rule.
06

Check Formal Charges

Check formal charges to ensure the most stable structure. The preferable structure is the one with the least formal charges on the atoms. Adjust any necessary bonding with this in consideration.
07

Determine Bond Angles and Hybridization

Around the central nitrogen atoms, the bond angles are about \(109.5^\circ\) and \(120^\circ\) due to their tetrahedral and trigonal planar geometries, respectively. The central oxygen has an angle of \(180^\circ\) due to its linear arrangement caused by its single bond with two nitrogen atoms. The hybridizations are \( \text{sp}^3\) for one nitrogen, \( \text{sp}^2\) for the other, and \( \text{sp} \) for oxygen.

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Key Concepts

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

Valence Electrons
Understanding valence electrons is fundamental to drawing Lewis structures. Valence electrons are the outermost electrons of an atom and determine its bonding behavior.
For example, nitrogen has 5 valence electrons, which makes it capable of sharing electrons to achieve a stable configuration. Oxygen, on the other hand, has 6 valence electrons.
This means when you calculate the valence electrons for a compound, like \( \text{N}_2\text{O}_5 \), you sum up all the valence electrons present:
  • Nitrogen: 2 atoms \( \times \) 5 electrons = 10 electrons
  • Oxygen: 5 atoms \( \times \) 6 electrons = 30 electrons
Total = 40 valence electrons

This total helps in placing these electrons correctly in the Lewis structure, ensuring that each atom achieves a full valence shell, obeying the octet rule.
Hybridization
Hybridization is a key concept when determining the geometry around atoms in a molecule. It involves mixing atomic orbitals to form new orbitals for bonding.
In \( \text{N}_2\text{O}_5 \):
  • One nitrogen atom is \( \text{sp}^3\) hybridized. This means it forms four orbitals: one \( s \) and three \( p \) orbitals. This configuration usually results in a tetrahedral shape, explaining bond angles around \(109.5^\circ\).
  • The second nitrogen is \( \text{sp}^2\) hybridized, mixing one \( s \) and two \( p \) orbitals. This results in a trigonal planar arrangement with bond angles of approximately \(120^\circ\).
  • The central oxygen atom is \( \text{sp} \) hybridized. With one \( s \) and one \( p \) orbital, this linear configuration leads to \(180^\circ\) bond angles.
Hybridization not only ensures atoms are properly linked but also affects molecule geometry and reactivity.
Bond Angles
Bond angles are a result of electron pair repulsion around an atom, influenced by hybridization and the molecular shape.
For \( \text{N}_2\text{O}_5 \), the varied hybridizations lead to different bond angles:
  • Around the tetrahedral nitrogen with \( \text{sp}^3\) hybridization, bond angles are about \(109.5^\circ\). This arrangement minimizes repulsions between all pairs of bonding electrons.
  • The \( \text{sp}^2\) hybridized nitrogen exhibits bond angles near \(120^\circ\). The trigonal planar arrangement arranges bonds evenly on a plane.
  • Finally, the central oxygen, \( \text{sp} \) hybridized, forms a linear \(180^\circ\) bond angle. This direct alignment reduces potential electron pair repulsion to a minimum.
Comprehending bond angles gives insight into a molecule's shape and functional groups, which can influence its physical and chemical properties.

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Most popular questions from this chapter

A Formamide, \(\mathrm{HC}(\mathrm{O}) \mathrm{NH}_{2},\) is prepared at high pressures from carbon monoxide and ammonia, and serves as an industrial solvent (the parentheses around the \(\mathrm{O}\) indicate that it is bonded only to the carbon atom and that the carbon atom is also bonded to the \(\mathrm{H}\) and the \(\mathrm{N}\) atoms). Two resonance forms (one with formal charges) can be written for formamide. Write both resonance structures, and predict the bond angles about the carbon and nitrogen atoms for each resonance form. Are they the same? Describe how the experimental determination of the \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle could be used to indicate which resonance form is more important.

Give the bonded-atom lone-pair arrangement expected for a central atom that has (a) two bonded atoms and one lone pair. (b) three bonded atoms and two lone pairs. (c) four bonded atoms and two lone pairs. (d) five bonded atoms and one lone pair.

Identify two homonuclear diatomic molecules or ions with each of the following molecular orbital electron configurations. Are these species stable? (a) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\left(\pi_{2 p}^{*}\right)^{3}\) (b) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\) (c) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\)

Identify the hybrid orbitals on the oxygen atoms that form the \(\sigma\) bonds in the following species. (a) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (b) \(\mathrm{H}_{3} \mathrm{COH}\) (c) \(\mathrm{Cl}_{2} \mathrm{O}\)

Explain why the molecular shape of HCl provides no information about the hybridization of the chlorine atom.
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