Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

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 Vaia!

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.

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

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{H}_{2}+, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2-}\) b. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)

Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

Use the localized electron model to describe the bonding in \(\mathrm{C}_{2} \mathrm{H}_{2}\) (exists as \(\left.\mathrm{HCCH}\right)\).

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

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: \(\mathrm{M}-\mathrm{C}=\mathrm{O}\) a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. \(9.42\) and \(9.43\).) Antibonding MOs place more. electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free