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Chapter 9: Problem 17

Use the localized electron model to describe the bonding in \(\mathrm{H}_{2} \mathrm{O} .\)

Short Answer

Expert verified
The bonding in $\mathrm{H}_{2}\mathrm{O}$ based on the localized electron model consists of a Lewis structure with two O-H single bonds and two lone pairs on the central oxygen atom. This results in a bent molecular geometry, with a bond angle of 104.5 degrees. The central oxygen atom is sp^3 hybridized, indicating tetrahedral electron geometry.

Step by step solution

01

Drawing the Lewis Structure

To draw the Lewis structure of water (H2O), first count the total number of valence electrons. Oxygen has 6 valence electrons and each hydrogen atom has 1 valence electron. Therefore, H2O has a total of 6 + 1 + 1 = 8 valence electrons. The least electronegative atom, oxygen, is placed in the center. We will now distribute the valence electrons as follows: 1. Place a pair of electrons between each bonded atom (oxygen and hydrogen) to form two single bonds. This accounts for 4 electrons. 2. Complete the octet rule for the central atom (oxygen) by placing the remaining 4 electrons, as lone pairs, around oxygen. The resulting Lewis structure of water is: O / \\ H - O - H | |
02

Determine the Molecular Geometry

To determine the molecular geometry of water, we count both the bonding pairs and lone pairs around the central atom (oxygen). In this case, there are 2 bonding pairs (one for each O-H bond) and 2 lone pairs. According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the geometry that minimizes the repulsion between these electron pairs is bent, with an angle of about 104.5 degrees between the H-O-H atoms.
03

Determine the Hybridization of the Central Atom

The hybridization of the central atom (oxygen) in a molecule can be determined by counting the number of electron domains around the atom. In this case, there are 4 electron domains (2 bonding pairs and 2 lone pairs) around oxygen. This corresponds to an sp^3 hybridization, where the 2s and three 2p orbitals of the oxygen atom combine to form four new, equivalent hybrid orbitals. Each hybrid orbital accommodates a single electron pair (either a bonding or lone pair), resulting in tetrahedral electron geometry with a bent molecular shape due to the lone pairs. In conclusion, the localized electron model describes the bonding in H2O as follows: 1. The Lewis structure features two O-H single bonds and two lone pairs on the central oxygen atom. 2. The resulting molecular geometry is bent, with a bond angle of 104.5 degrees. 3. The hybridization of the central oxygen atom is sp^3, according to the number of electron domains present.

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Key Concepts

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

Lewis Structure
The Lewis structure is a visual representation that helps us understand how the atoms in a molecule are bonded. For water ( H_2O ), the first step in drawing the Lewis structure is to determine the number of valence electrons. Oxygen has 6 valence electrons, while each hydrogen contributes 1, resulting in a total of 8 valence electrons for H_2O .

In the structure, oxygen is placed in the center. A pair of electrons is shared between oxygen and each hydrogen atom, forming two O-H single bonds. After forming the bonds, the remaining electrons settle as lone pairs around oxygen. The final structure shows the importance of the octet rule, with oxygen achieving a complete octet.
Molecular Geometry
Molecular geometry describes the shape of a molecule, considering both bonding pairs and lone pairs of electrons. For water ( H_2O ), the central oxygen atom has 2 bonding pairs and 2 lone pairs.

According to molecular geometry principles, these electron domains repel each other, creating a bent shape. This bent geometry is crucial in forming water's unique properties, such as its surface tension and boiling point.

The bond angle in water is slightly less than the tetrahedral angle, measuring about 104.5 degrees. This deviation is due to the lone pairs pushing the hydrogen atoms closer together.
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory explains how the geometry of a molecule is determined by electron pair repulsion. VSEPR theory suggests that electron pairs around a central atom arrange themselves as far apart as possible to minimize repulsion. This principle helps predict molecular shapes.

In the case of H_2O , there are two bonding pairs and two lone pairs around the central oxygen atom. The VSEPR theory tells us that the geometry that reduces repulsion between these pairs is bent, rather than linear or tetrahedral.

By considering the repulsion between lone pairs as being greater than that between bonding pairs, the VSEPR model accurately describes the bent shape and the bond angle of about 104.5 degrees.
Hybridization
Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals suitable for pairing electrons to form chemical bonds. For the oxygen in water ( H_2O ), hybridization is described as sp^3 .

This occurs due to the combination of one 2s and three 2p orbitals from oxygen, creating four equivalent sp^3 hybrid orbitals. Each of these orbitals contains one electron pair.

The sp^3 hybridization aligns with the electron domain geometry seen in water, forming a tetrahedral arrangement around the oxygen. However, because only two of these orbitals are used for bonding (with hydrogen), the molecule has a bent shape.
Valence Electrons
Valence electrons are the outermost electrons of an atom, and they play a key role in bonding. These electrons are involved in forming chemical bonds, which gives us the structure and properties of molecules.

In H_2O , we calculate the total valence electrons by adding those from oxygen (6 electrons) and hydrogen (1 electron per atom), giving a total of 8.

Understanding valence electrons is crucial in drawing Lewis structures and predicting molecular interactions. They help determine how atoms will bond and what the resulting molecular structure will look like, both of which are essential to understanding the chemistry of water and similar molecules.

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Most popular questions from this chapter

Which is the more correct statement: "The methane molecule \(\left(\mathrm{CH}_{4}\right)\) is a tetrahedral molecule because it is \(s p^{3}\) hybridized" of "The methane molecule \(\left(\mathrm{CH}_{4}\right)\) is \(s p^{3}\) hybridized because it is a tetrahedral molecule"? What, if anything, is the difference between these two statements?

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \\ \text { Cyanamide } \end{aligned} $$ Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: Dicyandiamide not shown) a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

Arrange the following from lowest to highest ionization energy: \(\mathrm{O}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

An unusual category of acids known as superacids, which are defined as any acid stronger than \(100 \%\) sulfuric acid, can be prepared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with \(\mathrm{SbF}_{5}\) produces the superacid \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) $$ 2 \mathrm{HF}(l)+\mathrm{SbF}_{5}(l) \longrightarrow\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}(l) $$ a. What are the molecular structures of all species in this reaction? What are the hybridizations of the central atoms in each species? b. What mass of \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) can be prepared when \(2.93 \mathrm{~mL}\) anhydrous \(\mathrm{HF}\) (density \(=0.975 \mathrm{~g} / \mathrm{mL}\) ) and \(10.0 \mathrm{~mL} \mathrm{SbF}_{5}\) (density \(=3.10 \mathrm{~g} / \mathrm{mL}\) ) are allowed to react?

Show how a hydrogen \(1 s\) atomic orbital and a fluorine \(2 p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?
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