Basic Chemistry Plus Mastering Chemistry With Pearson Etext -- Access Card Package (6th Edition)
Basic Chemistry Plus Mastering Chemistry With Pearson Etext -- Access Card Package (6th Edition)
6th Edition
ISBN: 9780134983783
Author: Karen C. Timberlake
Publisher: PEARSON
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 10, Problem 59UTC

State the number of valence electrons, bonding pairs, and lone pairs in each of the following Lewis structures: (10.1) Chapter 10, Problem 59UTC, State the number of valence electrons, bonding pairs, and lone pairs in each of the following Lewis

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Interpret number of valence electron, bond pair and lone pair in the given Lewis structure.

Concept Introduction:

Valence electrons are the electrons present in the outermost orbital or shell of an atom which participates in the formation of bond with another atom.

The bond formed between two atoms by sharing of electrons is known as covalent bond. A single chemical bond is formed by sharing of 2 electrons while double bond is formed by sharing of 4 and triple bond is formed by sharing of 6 electrons.

Total number of valence electron can be determined by adding all the electrons present in the outermost shell of each atom present in a molecule.

For example, inHxOy

Total number of valence electron = number of H (valence electron of H) + number of O (valence electron of O)

  = x (1) + y (6)

Answer to Problem 59UTC

Total number of valence electron = 2

Bond pair = 1

Lone pair = 0

Explanation of Solution

The given Lewis structure is as follows:

  H:H

Total number of valence electron = number of H (valence electron of H)

Total number of valence electron = 2

Bond pair = 1

Lone pair = 0

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Interpret number of valence electron, bond pair and lone pair in given Lewis structure.

Concept Introduction:

Valence electrons are the electrons present in the outermost orbital or shell of an atom which participates in the formation of bond with another atom.

The bond formed between two atoms by sharing of electrons is known as covalent bond. A single chemical bond is formed by sharing of 2 electrons while double bond is formed by sharing of 4 and triple bond is formed by sharing of 6 electrons.

Total number of valence electron can be determined by adding all the electrons present in the outermost shell of each atom present in a molecule.

For example, inHxOy

Total number of valence electron = number of H (valence electron of H) + number of O (valence electron of O)

  = x (1) + y (6)

Answer to Problem 59UTC

Total number of valence electron = 8

Bond pair = 1

Lone pair = 3

Explanation of Solution

The given Lewis structure is of HBr.

Total number of valence electron = number of H (valence electron of H) + number of Br (valence electron of Br)

Total number of valence electron = 1 (1) + 1 (7) = 8

Bond pair = 1

Lone pair = 3

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Interpret number of valence electron, bond pair and lone pair in Br-Br

Concept Introduction:

Valence electrons are the electrons present in the outermost orbital or shell of an atom which participates in the formation of bond with another atom.

The bond formed between two atoms by sharing of electrons is known as covalent bond. A single chemical bond is formed by sharing of 2 electrons while double bond is formed by sharing of 4 and triple bond is formed by sharing of 6 electrons.

Total number of valence electron can be determined by adding all the electrons present in the outermost shell of each atom present in a molecule.

For example, inHxOy

Total number of valence electron = number of H (valence electron of H) + number of O (valence electron of O)

  = x (1) + y (6)

Answer to Problem 59UTC

Total number of valence electron = 2 (7) = 14

Bond pair = 1

Lone pair = 6

Explanation of Solution

The given Lewis structure is of Br2.

Total number of valence electron = number of Br (valence electron of Br)

Total number of valence electron = 2 (7) = 14

Bond pair = 1

Lone pair = 6

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Calculate delta H for the following Thanks
Calcium citrate; MM=498.46 g/mol, a common preservative, has a Ksp of 1.12 x 10^-15 at 25°C.
175g of water was heated from 15 c to 88 c how many kilojoules were absorbed by the water

Chapter 10 Solutions

Basic Chemistry Plus Mastering Chemistry With Pearson Etext -- Access Card Package (6th Edition)

Ch. 10.2 - Draw two resonance structures for each of the...Ch. 10.2 - Draw two resonance structures for each of the...Ch. 10.3 - Prob. 13PPCh. 10.3 - Choose the shape (1 to 6) that matches each of the...Ch. 10.3 - Prob. 15PPCh. 10.3 - Prob. 16PPCh. 10.3 - Prob. 17PPCh. 10.3 - Prob. 18PPCh. 10.3 - Use VSEPR theory to predict the shape of each of...Ch. 10.3 - Prob. 20PPCh. 10.3 - Prob. 21PPCh. 10.3 - Draw the Lewis structure and predict the shape for...Ch. 10.4 - Describe the trend in electronegativity as...Ch. 10.4 - Describe the trend in electronegativity as...Ch. 10.4 - Prob. 25PPCh. 10.4 - Which electronegativity difference (a, b, or c)...Ch. 10.4 - Using the periodic table, arrange the atoms in...Ch. 10.4 - Using the periodic table, arrange the atoms in...Ch. 10.4 - Predict whether the bond between each of the...Ch. 10.4 - Predict whether the bond between each of the...Ch. 10.4 - For the bond between each of the following pairs...Ch. 10.4 - For the bond between each of the following pairs...Ch. 10.5 - Why is F2 a nonpolar molecule, but HF is a polar...Ch. 10.5 - Why is CCl4 a nonpolar molecule, but PCl3 is a...Ch. 10.5 - Identify each of the following molecules as polar...Ch. 10.5 - Identify each of the following molecules as polar...Ch. 10.5 - Prob. 37PPCh. 10.5 - Prob. 38PPCh. 10.6 - Prob. 39PPCh. 10.6 - Prob. 40PPCh. 10.6 - Identify the strongest intermolecular forces...Ch. 10.6 - Identify the strongest intermolecular forces...Ch. 10.6 - Prob. 43PPCh. 10.6 - Prob. 44PPCh. 10.7 - Using Figure 10.6, calculate the heat change...Ch. 10.7 - Using Figure 10.6, calculate the heat change...Ch. 10.7 - Prob. 47PPCh. 10.7 - Using Figure 10.6, calculate the heat change...Ch. 10.7 - Using Figure 10.6 and the specific heat of water,...Ch. 10.7 - Using Figure 10.6 and the specific heat of water,...Ch. 10.7 - An ice bag containing 275 g of ice at 0 °C was...Ch. 10.7 - Prob. 52PPCh. 10.7 - Prob. 53PPCh. 10.7 - In the preparation of liquid nitrogen, how many...Ch. 10.7 - Using the electronegativity values in Figure 10.2,...Ch. 10.7 - Prob. 56PPCh. 10.7 - Prob. 57PPCh. 10.7 - a. Draw two resonance structures for bicarbonate...Ch. 10 - State the number of valence electrons, bonding...Ch. 10 - State the number of valence electrons, bonding...Ch. 10 - Prob. 61UTCCh. 10 - Prob. 62UTCCh. 10 - Consider the following bonds: Ca and O, C and O, K...Ch. 10 - Consider the following bonds: F and Cl, Cl and Cl,...Ch. 10 - Identify the major intermolecular forces between...Ch. 10 - Prob. 66UTCCh. 10 - Prob. 67UTCCh. 10 - Prob. 68UTCCh. 10 - Prob. 69UTCCh. 10 - Prob. 70UTCCh. 10 - Prob. 71UTCCh. 10 - Prob. 72UTCCh. 10 - Prob. 73APPCh. 10 - Determine the total number of valence electrons in...Ch. 10 - Draw the Lewis structures for each of the...Ch. 10 - Draw the Lewis structures for each of the...Ch. 10 - Draw resonance structures for each of the...Ch. 10 - Prob. 78APPCh. 10 - Use the periodic table to arrange the following...Ch. 10 - Use the periodic table to arrange the following...Ch. 10 - Select the more polar bond in each of the...Ch. 10 - Select the more polar bond in each of the...Ch. 10 - Show the dipole arrow for each of the following...Ch. 10 - Show the dipole arrow for each of the following...Ch. 10 - Calculate the electronegativity difference and...Ch. 10 - Calculate the electronegativity difference and...Ch. 10 - Prob. 87APPCh. 10 - For each of the following, draw the Lewis...Ch. 10 - For each of the following, draw the Lewis...Ch. 10 - For each of the following, draw the Lewis...Ch. 10 - Prob. 91APPCh. 10 - Predict the shape and polarity of each of the...Ch. 10 - Prob. 93APPCh. 10 - Prob. 94APPCh. 10 - Prob. 95APPCh. 10 - Indicate the major type of intermolecular...Ch. 10 - When it rains or snows, the air temperature seems...Ch. 10 - Prob. 98APPCh. 10 - Using Figure 10.6, calculate the grams of ice that...Ch. 10 - Using Figure 10.6, calculate the grams of ethanol...Ch. 10 - Prob. 101APPCh. 10 - Using Figure 10.6, calculate the grams of benzene...Ch. 10 - Prob. 103CPCh. 10 - Prob. 104CPCh. 10 - Prob. 105CPCh. 10 - Prob. 106CPCh. 10 - Prob. 107CPCh. 10 - The melting point of benzene is 5.5 °C, and its...Ch. 10 - A 45.0-g piece of ice at 0.0 °C is added to a...Ch. 10 - An ice cube at 0 °C with a mass of 115 g is added...Ch. 10 - Prob. 111CPCh. 10 - Prob. 112CPCh. 10 - Prob. 13CICh. 10 - Prob. 14CICh. 10 - Prob. 15CICh. 10 - Ethanol, C2H6O , is obtained from renewable crops...Ch. 10 - Chloral hydrate, a sedative and hypnotic, was the...Ch. 10 - Ethylene glycol, C2H6O2 , used as a coolant and...Ch. 10 - Prob. 19CICh. 10 - Prob. 20CI
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY