Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 79

Write Lewis structures that obey the octet rule (duet rule for 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. \(\mathrm{F}_{2}\) \(\mathbf{b} . \mathbf{O}_{2}\) c. CO d. \(\overline{\mathrm{CH}_{4}}\) \(\mathbf{e} . \mathrm{NH}_{3}\) \(\mathbf{f .} \quad \mathbf{H}_{2} \mathbf{O}\) g. IIF

Short Answer

Expert verified
Lewis structures: a. F2: F - F b. O2: O = O c. CO: C ≡ O d. CH4: H | C - H | H e. NH3: H | N - H - H f. H2O: H - O - H g. IIF: I - F

Step by step solution

01

a. \(\mathrm{F}_{2}\)

1. Count the valence electrons: Fluorine has 7 valence electrons, so the total for F2 is 14. 2. Create a single bond between the fluorine atoms, accounting for 2 electrons. 3. Distribute the remaining 12 electrons in pairs to the outside atoms (fluorine) to satisfy the octet rule for each atom. Lewis structure for F2: F - F
02

b. \(\mathrm{O}_{2}\)

1. Count the valence electrons: Oxygen has 6 valence electrons, so the total for O2 is 12. 2. Create a double bond between the oxygen atoms, accounting for 4 electrons. 3. Distribute the remaining 8 electrons in pairs to the outside atoms (oxygen) to satisfy the octet rule for each atom. Lewis structure for O2: O = O
03

c. CO

1. Count the valence electrons: Carbon has 4 and oxygen has 6, so the total is 10. 2. Create a triple bond between the carbon and oxygen atoms, accounting for 6 electrons. 3. Distribute the remaining 4 electrons in pairs to the outside atom (oxygen) to satisfy the octet rule for each atom. Lewis structure for CO: C ≡ O
04

d. \(\mathrm{CH}_{4}\)

1. Count the valence electrons: Carbon has 4 and each hydrogen has 1, so the total is 8. 2. Create a single bond between the central atom (carbon) and the hydrogen atoms, accounting for 4 electrons. 3. Distribute the remaining 4 electrons in pairs to the hydrogen atoms to satisfy the duet rule for each hydrogen atom. Lewis structure for CH4: H | C - H | H
05

e. \(\mathrm{NH}_{3}\)

1. Count the valence electrons: Nitrogen has 5 and each hydrogen has 1, so the total is 8. 2. Create a single bond between the central atom (nitrogen) and the hydrogen atoms, accounting for 3 electrons. 3. Distribute the remaining 5 electrons: 3 as a lone pair on the nitrogen atom, and 2 as a pair to one hydrogen atom to satisfy the duet rule for each hydrogen atom. Lewis structure for NH3: H | N - H - H
06

f. \(\mathrm{H}_{2}\mathrm{O}\)

1. Count the valence electrons: Oxygen has 6 and each hydrogen has 1, so the total is 8. 2. Create a single bond between the central atom (oxygen) and the hydrogen atoms, accounting for 2 electrons. 3. Distribute the remaining 6 electrons in pairs to the outside hydrogen atoms to satisfy the duet rule for each hydrogen atom. Lewis structure for H2O: H - O - H
07

g. IIF

1. Count the valence electrons: Iodine has 7 and fluorine has 7, so the total is 14. 2. Create a single bond between the central atom (iodine) and the fluorine atoms, accounting for 2 electrons. 3. Distribute the remaining 12 electrons in pairs to the outside atoms (iodine and fluorine) to satisfy the octet rule for each atom. Lewis structure for IIF: I - F

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following reaction: $$ \mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB} \quad \Delta E=-285 \mathrm{kJ} $$ The bond energy for \(A_{2}\) is one-half the amount of the AB bond energy. The bond energy of \(\mathbf{B}_{2}=432 \mathrm{kJ} / \mathrm{mol} .\) What is the bond energy of \(\mathrm{A}_{2} ?\)

Name each of the following compounds. Assume the acids are dissolved in water. a. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) b. \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) c. \(\mathrm{Co}_{2} \mathrm{S}_{3}\) d. ICl e. \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) f. \(\mathrm{KClO}_{3}\) g. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(\mathbf{h} . \mathrm{Sr}_{3} \mathrm{N}_{2}\) i. \(\quad \mathrm{Al}_{2}\left(\mathrm{SO}_{3}\right)_{3}\) J. \(\mathrm{SnO}_{2}\) \(\mathbf{k} . \mathrm{Na}_{2} \mathrm{CrO}_{4}\) I. \(\mathrm{HClO}\)

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)

Consider the following: $$\mathrm{Li}(s)+\frac{1}{2} \mathrm{I}_{2}(s) \longrightarrow \mathrm{LiI}(s) \quad \Delta E=-272 \mathrm{kJ} / \mathrm{mol}$$ LiI(s) has a lattice energy of -753 kJ/mol. The ionization energy of \(\mathrm{Li}(g)\) is \(520 . \mathrm{kJ} / \mathrm{mol},\) the bond energy of \(\mathrm{I}_{2}(g)\) is \(151 \mathrm{kJ} /\) mol, and the electron affinity of \(\mathrm{I}(g)\) is \(-295 \mathrm{kJ} / \mathrm{mol} .\) Use these data to determine the energy of sublimation of Li(s).

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. Al and Cl b. Na and O c. Sr and F d. Ca and Se
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free