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Chapter 9: Problem 21

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

Short Answer

Expert verified
Using the localized electron model, the bonding in H₂O can be described as follows: The central oxygen atom forms single bonds with two hydrogen atoms, each having a single electron pair in their bonds. The oxygen atom also has three lone pairs of electrons, completing its octet. The Lewis structure for H₂O is: H | O - H with 3 more lone pairs (6 electrons) around O.

Step by step solution

01

Determine the total number of valence electrons

We will first determine the total number of valence electrons in H₂O by adding the valence electrons of each atom in the molecule. Hydrogen has 1 valence electron, and oxygen has 6 valence electrons. Since there are 2 hydrogen atoms and 1 oxygen atom in H₂O, we have a total of: (2 x 1) + (1 x 6) = 2 + 6 = 8 valence electrons
02

Determine the central atom and draw a skeleton structure

Oxygen is the central atom because it has a higher electronegativity than hydrogen. The skeleton structure of H₂O could be represented as O with H atoms bonded on its either side: H - O - H
03

Place remaining electrons as lone pairs on outer atoms

The hydrogen atoms have formed single bonds with the oxygen atom, so their octet is complete (H only requires 2 electrons to complete its valence shell). We used 2 of the 8 valence electrons in the single bonds and are left with 8 - 2 = 6 electrons. Now, place the remaining electrons as lone pairs on the outer atoms which are already complete with their octet.
04

Place remaining lone pairs on the central atom

As the octet of hydrogen atoms is already complete, we will place the remaining 6 electrons as lone pairs on the central oxygen atom to complete its octet. The oxygen atom will have 3 lone pairs and 1 single bond with each of the hydrogen atoms. The Lewis structure could be represented as: H | O - H with 3 more lone pairs (6 electrons) around O.
05

Form double or triple bonds if necessary

Since the octet of the central oxygen atom is complete, there is no need for additional double or triple bonds in this molecule. In conclusion, using the localized electron model, the water molecule (H₂O) has an oxygen atom forming single bonds with two hydrogen atoms, each having a single electron pair in their bonds. The oxygen atom also has three lone pairs of electrons completing its octet.

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Most popular questions from this chapter

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d, \text { or } 6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

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. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

Use the MO model to explain the bonding in BeH. When con- structing the MO energy-level diagram, assume that the Be's 1 s electrons are not involved in bond formation.

Consider the following molecular orbitals formed from the combination of two hydrogen 1s orbitals: a. Which is the bonding molecular orbital and which is the antibonding molecular orbital? Explain how you can tell by looking at their shapes. b. Which of the two molecular orbitals is lower in energy? Why is this true?

Explain the difference between the \(\sigma\) and \(\pi\) MOs for homo- nuclear diatomic molecules. How are bonding and antibonding orbitals different? Why are there two \(\pi\) MOs and one \(\sigma\) MO? Why are the \(\pi\) MOs degenerate?
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