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Chapter 4: Problem 39

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

Short Answer

Expert verified
In the water molecule, H₂O, the oxygen atom serves as the central atom and forms two single covalent bonds with hydrogen atoms, sharing a pair of electrons in each bond. The molecule exhibits a bent or angular molecular geometry due to the influence of the two lone pairs of electrons on the oxygen atom, which also results in a tetrahedral electron-pair geometry according to the localized electron model.

Step by step solution

01

1. Determine the valence electrons

Each atom in the molecule has a certain number of valence electrons that can be involved in bonding. For hydrogen atoms, there is one valence electron; for oxygen, there are six valence electrons.
02

2. Calculate the total number of valence electrons

Combine the valence electrons from all atoms in the molecule. For a water molecule, this is \( 2 \times (\text{1 valence electron for hydrogen}) + \text{6 valence electrons for oxygen} = 2 + 6 = 8 \text{ valence electrons} \)
03

3. Place the central atom

Identify the central atom, which is the atom with the highest bonding capacity or the lowest electronegativity. In the case of the water molecule, the central atom is oxygen since it has more available bonding sites than hydrogen.
04

4. Distribute the electrons around the atoms

Arrange the remaining atoms and valence electrons around the central atom to form bonds and satisfy the octet rule. Each bond requires 2 electrons. In the case of water, connect the oxygen atom with the two hydrogen atoms using covalent bonds (single bonds), which account for \(2 \times 2 = 4\) valence electrons. The remaining \(8 - 4 = 4\) valence electrons should be placed on the oxygen atom as two lone pairs, satisfying the octet rule for oxygen.
05

5. Determine the electron-pair geometry

Based on the distribution of electrons around the central atom (oxygen) and the orientation of the bonded atoms (hydrogen), determine the electron-pair geometry of the molecule. In the case of water, the electron-pair geometry is tetrahedral since there are two bonding electron pairs and two lone pairs.
06

6. Determine the molecular geometry

Ignoring the lone pairs, consider the positions of the atoms only. In the case of water, the molecular geometry is bent or angular since the hydrogen atoms are connected to the central oxygen atom in an angle.
07

7. Describe the bonding using the localized electron model

In the water molecule, the bonding can be described using the localized electron model as follows: the oxygen atom forms two single covalent bonds with each of the hydrogen atoms, sharing two bonding electron pairs between them. This forms a bent molecular geometry, with two lone pairs of electrons remaining on the oxygen atom, causing the molecule to have a tetrahedral electron-pair geometry.

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Most popular questions from this chapter

Predict the molecular structure (including bond angles) for each of the following. (See Exercises 25 and \(26 .\) ) a. ICls b. \(\mathrm{XeCl}_{4}\) c. \(\mathrm{SeCl}_{6}\)

Two different compounds have the formula \(\mathrm{XeF}_{2} \mathrm{Cl}_{2}\). Write Lewis structures for these two compounds, and describe how measurement of dipole moments might be used to distinguish between them.

The compound \(\mathrm{NF}_{3}\) is quite stable, but \(\mathrm{NCl}_{3}\) is very unstable (NCl \(_{3}\) was first synthesized in 1811 by P. L. Dulong, who lost three fingers and an eye studying its properties). The compounds \(\mathrm{NBr}_{3}\) and \(\mathrm{NI}_{3}\) are unknown, although the explosive compound \(\mathrm{NI}_{3} \cdot \mathrm{NH}_{3}\) is known. Account for the instability of these halides of nitrogen.

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following. a. \(\mathrm{XeO}_{3}\) b. \(\mathrm{XeO}_{4}\) c. \(\mathrm{XeOF}_{4}\) d. \(\mathrm{XeOF}_{2}\) e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{kJ} / \mathrm{mol},\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen 1s character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of OH. Explain.
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