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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
In H₂O, the bonding can be described using the localized electron model as follows: the molecule has a bent molecular geometry with an H-O-H bond angle of approximately 104.5° and the central oxygen atom exhibits sp³ hybridization. The O-H bonds are σ bonds formed due to the head-on overlap of orbitals.

Step by step solution

01

Determine the Lewis structure

First, we need to find out the total number of valence electrons in the H₂O molecule. Oxygen has 6 valence electrons, while hydrogen has 1 valence electron each. Thus, we have a total of 8 valence electrons (6 from oxygen and 2 from the two hydrogen atoms). Now, we will place the oxygen atom in the center and two hydrogen atoms on either side of it. Distributing the valence electrons in the molecule, we give 2 electrons to each H-O bond (totaling 4 electrons) and the remaining 4 electrons are distributed as lone pairs on the oxygen atom. The Lewis structure is as follows: O / \ H - H
02

Determine the molecular geometry and bond angles

The molecular geometry of a molecule can be identified using the Valence Shell Electron Pair Repulsion (VSEPR) theory. H₂O has 2 bonding electron pairs (between O and H atoms) and 2 lone electron pairs on the central oxygen atom. This corresponds to an AX₂E₂ (A = central atom, X = atoms bonded to central atom, E = lone electron pairs) geometry in the VSEPR model with molecular geometry being bent (angular). The bond angle (H-O-H) in H₂O is approximately 104.5°. This is less than the ideal bond angle of 109.5° in a tetrahedral geometry due to the repulsion between the lone electron pairs on the central oxygen atom, which compresses the H-O-H bond angle.
03

Determine the hybridization of the central atom

To identify the hybridization of the central oxygen atom, we need to look at the number of electron domains (regions of electron density) around the atom. In the case of H₂O, there are 4 electron domains (2 bonding pairs and 2 lone pairs). Hence, the oxygen atom exhibits sp³ hybridization.
04

Identify the type of chemical bonds (σ and π bonds)

In H₂O, there are two O-H bonds. Each of these bonds is formed by the overlap of orbitals, one from the oxygen atom and one from the hydrogen atom. The sigma (σ) bond is formed by the axial head-on overlap of orbitals, which is the case for both O-H bonds in H₂O. Since there is no sideways overlap of orbitals to form additional, parallel bonds (π bonds), both O-H bonds are solely classified as σ bonds. In conclusion, using the localized electron model, we can describe the bonding in H₂O as follows: H₂O has a bent molecular geometry, with an H-O-H angle of 104.5°, and the central oxygen atom exhibits sp³ hybridization. The O-H bonds are σ bonds formed due to the head-on overlap of orbitals.

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Most popular questions from this chapter

The transport of \(\mathrm{O}_{2}\) in the blood is carried out by hemoglobin. Carbon monoxide can interfere with oxygen transport because hemoglobin has a stronger affinity for \(\mathrm{CO}\) than for \(\mathrm{O}_{2}\). If \(\mathrm{CO}\) is present, normal uptake of \(\mathrm{O}_{2}\) is prevented, depriving the body of needed oxygen. Using the molecular orbital model, write the electron configurations for \(\mathrm{CO}\) and \(\mathrm{for} \mathrm{O}_{2}\). From your configurations, give two property differences between \(\mathrm{CO}\) and \(\mathrm{O}_{2}\).

Compare and contrast bonding molecular orbitals with antibonding molecular orbitals.

In the hybrid orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi\).

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity. a. \(\mathrm{CF}_{4}\) e. \(\mathrm{BeH}_{2}\) i. \(\mathrm{KrF}_{4}\) b. \(\mathrm{NF}_{3}\) f. \(\mathrm{TeF}_{4}\) j. \(\mathrm{SeF}_{6}\) c. \(\mathrm{OF}_{2}\) g. \(\mathrm{AsF}_{5}\) k. \(\mathrm{IF}_{5}\) d. \(\mathrm{BF}_{3}\) h. \(\mathrm{KrF}_{2}\) 1\. \(\mathrm{IF}_{3}\)
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