Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 60

Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Fluorine has _____ valence electrons and makes______ bond(s) in compounds. (b) Oxygen has _____ valence electrons and makes ______ bond(s) in compounds. (c) Nitrogen has _____ valence electrons and makes ______ bond(s) in compounds. (d) Carbon has _____ valence electrons and makes ______ bond(s) in compounds.

Short Answer

Expert verified
(a) Fluorine has \(7\) valence electrons and makes \(1\) bond(s) in compounds. (b) Oxygen has \(6\) valence electrons and makes \(2\) bond(s) in compounds. (c) Nitrogen has \(5\) valence electrons and makes \(3\) bond(s) in compounds. (d) Carbon has \(4\) valence electrons and makes \(4\) bond(s) in compounds.

Step by step solution

01

Valence Electrons for Fluorine (F)

Fluorine is in Group 17 of the periodic table and has an atomic number of 9. Its electronic configuration is 1s²2s²2p⁵. Since the electrons in the last energy level contribute to the chemical bonding, Fluorine has 7 valence electrons.
02

Fluorine (F) Bonding Behavior

To achieve a full outer electron shell (the octet rule), Fluorine needs to gain one electron. Thus, it forms one single bond with other elements. (a) Fluorine has 7 valence electrons and makes 1 bond(s) in compounds.
03

Valence Electrons for Oxygen (O)

Oxygen is in Group 16 of the periodic table and has an atomic number of 8. Its electronic configuration is 1s²2s²2p⁴. Just like in the case of Fluorine, the last energy level electrons contribute to the chemical bonding. Oxygen has 6 valence electrons.
04

Oxygen (O) Bonding Behavior

To achieve a full outer electron shell, Oxygen needs to gain two electrons. Thus, it can either form two single bonds or one double bond with other elements. (b) Oxygen has 6 valence electrons and makes 2 bond(s) in compounds.
05

Valence Electrons for Nitrogen (N)

Nitrogen is in Group 15 of the periodic table and has an atomic number of 7. Its electronic configuration is 1s²2s²2p³. The last energy level electrons contribute to the chemical bonding. Nitrogen has 5 valence electrons.
06

Nitrogen (N) Bonding Behavior

To achieve a full outer electron shell, Nitrogen needs to gain three electrons. So, it can either form three single bonds, one single bond and one double bond, or one triple bond with other elements. (c) Nitrogen has 5 valence electrons and makes 3 bond(s) in compounds.
07

Valence Electrons for Carbon (C)

Carbon is in Group 14 of the periodic table and has an atomic number of 6. Its electronic configuration is 1s²2s²2p². The last energy level electrons contribute to the chemical bonding. Carbon has 4 valence electrons.
08

Carbon (C) Bonding Behavior

To achieve a full outer electron shell, Carbon needs to gain four electrons. So, it can either form four single bonds, two single bonds and one double bond, or two double bonds with other elements. (d) Carbon has 4 valence electrons and makes 4 bond(s) in compounds.

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

For each of the following molecules or ions of sulfur and oxygen, write a single Lewis structure that obeys the octet rule, and calculate the oxidation numbers and formal charges on all the atoms: (a) SO \(_{2},(\mathbf{b}) \mathrm{SO}_{3},(\mathrm{c}) \mathrm{SO}_{3}^{2-}\) (d) Arrange these molecules/ions in order of increasing \(S-O\) bond length.

What is the Lewis symbol for each of the following atoms or ions? (a) \(\mathrm{K},\) (b) As, (c) \(\mathrm{Sn}^{2+},(\mathbf{d}) \mathrm{N}^{3-}\) .

Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround these atoms: (a) \(\mathrm{PH}_{3},\) (b) AlH_ \(_{3},(\mathbf{c}) \mathrm{N}_{3}^{-}\) (d) \(\mathrm{CH}_{2} \mathrm{Cl}_{2},(\mathbf{e}) \mathrm{SnF}_{6}^{2-}\)

Formic acid has the chemical formula HCOOH. It is a colorless liquid that has a density of 1.220 \(\mathrm{g} / \mathrm{mL}\) . (a) The carbon atom in formic acid is bound to one \(\mathrm{H}\) and both \(\mathrm{O}^{\prime}\) 's. Draw the Lewis structure for formic acid, showing resonance if present. (b) Formic acid can react with NaOH in aqueous solution to produce the formate ion, HCOO- . Write the balanced chemical equation for this reaction. (c) Draw the Lewis structure of the formate ion, showing resonance if present. (d) How many milliliters of a 0.100 M solution of NaOH would it take to completely react with 0.785 \(\mathrm{mL}\) of formic acid?

A carbene is a compound that has a carbon bonded to two atoms and a lone pair remaining on the carbon. Many carbenes are very reactive. (a) Draw the Lewis structure for the simplest carbene, \(\mathrm{H}_{2} \mathrm{C}\) . ( b) Predict the length of the carbon-carbon bond you would expect if two \(\mathrm{H}_{2} \mathrm{C}\) molecules reacted with each other by a combination reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

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