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Chapter 8: Problem 79

Write Lewis structures that obey the octet rule (duet rule for \(\mathrm{H}\) ) for each of the following molecules. Carbon is the central atom in \(\mathrm{CH}_{4}\), nitrogen is the central atom in \(\mathrm{NH}_{3}\), and oxygen is the central atom in \(\mathrm{H}_{2} \mathrm{O}\). a. \(\mathrm{F}_{2}\) e. \(\mathrm{NH}_{3}\) b. \(\mathrm{O}_{2}\) f. \(\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{CO}\) g. HF d. \(\mathrm{CH}_{4}\)

Short Answer

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Step by step solution

01

1. Determine the total number of valence electrons for each molecule.

In order to draw the Lewis structures for the given molecules, we must first determine the total number of valence electrons for each molecule. Valence electrons are the outermost electrons of an element. We can find the number of valence electrons of an element by checking its group number in the periodic table.
02

2. Drawing Lewis structures for the given molecules:

a. \(\mathrm{F}_{2}\): 1. Fluorine is a group 17 element, so each fluorine atom has 7 valence electrons. Since there are 2 fluorine atoms in the molecule, the total number of valence electrons is 14. 2. Place the two fluorine atoms side by side and draw a single bond between them. 3. Now, distribute the remaining electrons to complete the octet rule for each fluorine atom. : F - F : b. \(\mathrm{O}_{2}\): 1. Each oxygen atom has 6 valence electrons (group 16). So, the total number of valence electrons is 12. 2. Place the two oxygen atoms side by side and connect them with a double bond. 3. Distribute the remaining electrons to complete the octet rule for each oxygen atom. :: O=O :: c. \(\mathrm{CO}\): 1. Carbon has 4, and oxygen has 6 valence electrons; hence, the total number of valence electrons is 10. 2. Create a triple bond between the carbon and oxygen atom. 3. Distribute the remaining electrons to complete the octet rule. :: C鈮 : d. \(\mathrm{CH}_{4}\): 1. Carbon has 4, and each hydrogen atom has 1 valence electron, giving a total of 8 valence electrons. 2. Place the carbon atom in the center and connect it with single bonds to the four hydrogen atoms. 3. Distribute the remaining electrons to complete the octet for carbon (the duet rule for hydrogen is already satisfied). H | H-C-H | H e. \(\mathrm{NH}_{3}\): 1. Nitrogen has 5 valence electrons, and each hydrogen atom has 1, giving a total of 8 valence electrons. 2. Connect the central nitrogen atom to each of the three hydrogen atoms through single bonds. 3. Distribute the remaining electrons to complete the octet for nitrogen (the duet rule for hydrogen is already satisfied). H | H-N-H | : f. \(\mathrm{H}_{2}\mathrm{O}\): 1. Oxygen has 6 valence electrons, and each hydrogen atom has 1, giving a total of 8 valence electrons. 2. Connect the central oxygen atom to each of the two hydrogen atoms with single bonds. 3. Distribute the remaining electrons to complete the octet rule for the oxygen atom (the duet rule for hydrogen is already satisfied). H | H-O-H | :: g. \(\mathrm{HF}\): 1. Hydrogen has 1 valence electron, and fluorine has 7 valence electrons, giving a total of 8 valence electrons. 2. Connect the hydrogen and fluorine atoms with a single bond. 3. Distribute the remaining electrons to complete the octet rule for fluorine and the duet rule for hydrogen. H - F

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Key Concepts

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

Octet Rule
The octet rule is a fundamental principle in chemistry that helps us understand how atoms form molecules. It states that atoms are most stable when they have eight electrons in their outer shell, mimicking the electron configuration of a noble gas. This rule applies predominantly to the main group elements, where atoms tend to gain, lose, or share electrons to fulfill the octet. An important exception to the octet rule is hydrogen, which is satisfied with only two electrons, known as the duet rule. This is because hydrogen only has one shell, which can hold a maximum of two electrons. The octet rule serves as a guiding principle when constructing Lewis structures, allowing us to predict the arrangement of electrons and thus the stability and behavior of a molecule.
Valence Electrons
Valence electrons play a crucial role in chemical bonding, as they are the electrons in the outermost shell of an atom. These electrons determine how an atom interacts with others to form chemical compounds. The number of valence electrons can be identified in the representative elements using the group number in the periodic table. For instance, elements in Group 1 have one valence electron, while those in Group 17 have seven. When drawing Lewis structures, we first count the total valence electrons for all atoms involved in a molecule. This number guides us in allocating electrons to fulfill the octet rule. By representing valence electrons as dots or lines in a Lewis structure, we can depict the formation of bonds and lone pairs, providing a visual representation of molecular bonding.
Chemical Bonding
Chemical bonding is the force of attraction between atoms or ions, holding them together in a compound. The main types of chemical bonds are covalent, ionic, and metallic. In covalent bonding, which is central to Lewis structures, atoms share pairs of valence electrons. This sharing enables each atom to achieve a full outer shell in accordance with the octet rule.For example, in a water molecule (\(H_2O\)), oxygen forms two covalent bonds with the hydrogen atoms. This allows oxygen to reach an octet of electrons, while hydrogen achieves a duet. Visualizing these bonds in Lewis structures helps in understanding the spatial arrangement of atoms in a molecule and their interactions.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. After constructing the Lewis structure, we can use it to predict the shape of the molecule. The geometrical arrangement impacts the molecule's properties, such as reactivity, polarity, phase of matter, color, magnetism, and biological activity.The VSEPR (Valence Shell Electron Pair Repulsion) theory is often used to determine molecular geometry. It states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion. For instance, in a methane molecule (\(CH_4\)), this results in a tetrahedral shape, while water (\(H_2O\)) has a bent shape due to the two lone pairs on oxygen. Understanding molecular geometry is crucial for predicting molecular behavior and its interactions with other substances.

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Most popular questions from this chapter

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Li}\) and \(\mathrm{N}\) c. \(\mathrm{Rb}\) and \(\mathrm{Cl}\) b. Ga and \(\mathrm{O}\) d. \(\mathrm{Ba}\) and \(\mathrm{S}\)

Write Lewis structures that obey the octet rule for the following species. Assign the formal charge for each central atom. a. \(\mathrm{POCl}_{3}\) e. \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) b. \(\mathrm{SO}_{4}^{2-}\) f. \(\mathrm{XeO}_{4}\) c. \(\mathrm{ClO}_{4}^{-}\) g. \(\mathrm{ClO}_{3}^{-}\) d. \(\mathrm{PO}_{4}^{3-}\) h. \(\mathrm{NO}_{4}^{3-}\)

Use the following data to estimate \(\Delta H\) for the reaction \(\mathrm{S}^{-}(g)+\) \(\mathrm{e}^{-} \rightarrow \mathrm{S}^{2-}(g)\). Include an estimate of uncertainty. $$ \begin{array}{|lcccc|} \hline & & \text { Lattice } & & \Delta H_{\text {sub }} \\ & \Delta \boldsymbol{H}_{\mathrm{t}}^{\circ} & \text { Energy } & \text { I.E. of } \mathbf{M} & \text { of M } \\ \hline \mathrm{Na}_{2} \mathrm{~S} & -365 & -2203 & 495 & 109 \\ \mathrm{~K}_{2} \mathrm{~S} & -381 & -2052 & 419 & 90 \\ \mathrm{Rb}_{2} \mathrm{~S} & -361 & -1949 & 409 & 82 \\ \mathrm{Cs}_{2} \mathrm{~S} & -360 & -1850 & 382 & 78 \\ \hline \end{array} $$ $$ \begin{aligned} \mathrm{S}(s) & \longrightarrow \mathrm{S}(g) & \Delta H &=227 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{S}(g)+\mathrm{e}^{-} & \longrightarrow \mathrm{S}^{-}(g) & \Delta H &=-200 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Assume that all values are known to \(\pm 1 \mathrm{~kJ} / \mathrm{mol}\).

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) consists of a six- membered ring of carbon atoms with one hydrogen bonded to each carbon. Write Lewis structures for benzene, including resonance structures.

The lattice energies of \(\mathrm{FeCl}_{3}, \mathrm{FeCl}_{2}\), and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are (in no particular order \(-2631,-5359\), and \(-14,774 \mathrm{~kJ} / \mathrm{mol}\). Match the appropriate formula to each lattice energy. Explain.
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