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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 Hâ‚‚O 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 Hâ‚‚O 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 Hâ‚‚O 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 Hâ‚‚O, 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 Hâ‚‚O 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 (Hâ‚‚O) 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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Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom that are involved in chemical bonding. They play a crucial role in determining how atoms interact with each other. In the case of water (\( \mathrm{H}_{2}\mathrm{O} \)), the total number of valence electrons can be calculated by summing the valence electrons of each individual atom in the molecule.

- Each hydrogen atom has 1 valence electron.- The oxygen atom has 6 valence electrons.
For two hydrogen atoms and one oxygen atom, the total number of valence electrons is \( (2 \times 1) + (1 \times 6) = 8 \) valence electrons. These electrons are used in chemical bonding and in forming lone pairs around atoms. Understanding valence electrons is fundamental to writing the Lewis structure for a molecule.
Central Atom
The central atom in a molecule is typically the atom with the highest ability to attract electrons, known as electronegativity, or having the most bonds with other atoms. In water (\( \mathrm{H}_{2}\mathrm{O} \)), oxygen becomes the central atom due to its higher electronegativity compared to hydrogen.

A skeleton structure is first drawn with oxygen at the center and hydrogen atoms bonded to either side, forming a rough "H-O-H" shape.
Positioning in a molecule is vital because it guides the bonding and arrangement of electrons, thus predicting molecular shape and reactivity.
Lone Pairs
Lone pairs consist of valence electrons that are not involved in bonding and remain with a single atom. They are very important in shaping the molecule and determining its polarity and properties.

For water (\( \mathrm{H}_{2}\mathrm{O} \)), we start with 8 valence electrons. Two are used for the \( O-H \) bonds, leaving 6 electrons.
These remaining electrons are arranged as three lone pairs on the oxygen atom. - These lone pairs make the oxygen electron-rich.- Each pair contributes to the bent shape of the molecule.With the presence of these lone pairs, the shape of water is angular rather than linear, influencing many properties like its high boiling point and solvent abilities.
Lewis Structure
The Lewis structure of a molecule visualizes its valence electrons, bonds, and lone pairs. It's a simple way to represent how electrons are distributed around atoms.

For \( \mathrm{H}_{2}\mathrm{O} \), the Lewis structure displays oxygen as the central atom:
- Two single bonds link oxygen to each hydrogen atom, using 2 electrons per bond.- Three pairs of lone electrons (\( 6 \) electrons) are depicted around the oxygen.This structure visually shows how all 8 valence electrons are accounted for:
- Bonding electrons in \( O-H \) bonds- Lone pairs on the oxygen atom
Understanding the Lewis structure helps predict molecular geometry, bond angles, and even reactivity, making it a foundational tool in chemistry for students and professionals alike.

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Most popular questions from this chapter

Describe the bonding in \(\mathrm{NO}^{+}, \mathrm{NO}^{-},\) and \(\mathrm{NO}\) using both the localized electron and molecular orbital models. Account for any discrepancies between the two models.

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

The diatomic molecule OH exists in the gas phase. OH plays an important part in combustion reactions and is a reactive oxidizing agent in polluted air. 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 the MOs result from the overlap of a \(p_{z}\) orbital from oxygen and the 1\(s\) orbital of hydrogen (the O-H bond lies along the z axis). a. Draw pictures of the sigma bonding and antibonding molecular orbitals in OH. b. Which of the two MOs has the greater hydrogen 1\(s\) character? c. Can the 2\(p_{x}\) orbital of oxygen form MOs with the 1\(s\) orbital of hydrogen? Explain. d. Knowing that only the 2\(p\) orbitals of oxygen interact significantly with the 1\(s\) orbital of hydrogen, complete the MO energy-level diagram for OH. Place the correct number of electrons in the energy levels. e. Estimate the bond order for OH. f. Predict whether the bond order of \(\mathrm{OH}^{+}\) is greater than, less than, or the same as that of OH. Explain.

Draw the Lewis structures for \(\mathrm{SeO}_{2}, \mathrm{PCl}_{3}, \mathrm{NNO}, \mathrm{COS},\) and \(\mathrm{PF}_{3} .\) Which of the compounds are polar? Which of the compounds exhibit at least one bond angle that is approximately \(120^{\circ}\) Which of the compounds exhibit \(s p^{3}\) hybridization by the central atom? Which of the compounds have a linear molecular structure?

In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. $$ \text {a} C N \text { or } N O \qquad \text { b. } O_{2}^{2+} \text { or } N_{2}^{2+} $$

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