MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 10.1, Problem 10.5BFP

(a)

Interpretation Introduction

Interpretation:

Lewis structure with the lowest formal charges of BeH2 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the molecule are as follows:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  Formalcharge=(numberofvalenceelectrons)((numberofnon-bondingelectrons)+(12)(numberofbondingelectrons))        (1)

(a)

Expert Solution
Check Mark

Answer to Problem 10.5BFP

Lewis structure with lowest formal charges for BeH2 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  1

The formal charge on Be is 0.

The formal charge on each H is 0.

Explanation of Solution

The total number of valence electrons of BeH2 is calculated as,

  Total valence electrons(TVE)=[(Total number of Be atom)(Valence electrons of Be)+(Total number of H atom)(Valence electrons of H)]        (1)

Substitute 3 for the total number of Be atom, 2 for valence electrons of Be, and 2 for the number of H atom, and 1 for valence electrons of H in equation (1).

  Total valence electrons(TVE)=[(1)(2)+(2)(1)]=4

BeH2 has 4 valence electrons. Centre Be atom form two single bonds with each H atom. All electrons are used to form 2 single bonds. This leaves beryllium atom 4 electrons short to complete its octet. Thus, BeH2 is an exception to the octet rule.

The Lewis structure of BeH2 is as follows:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  2

Substitute 2 for the number of valence electrons, 0 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on Be atom.

  Formalcharge=(2)((0)+(12)(4))=0

Substitute 1 for the number of valence electrons, 0 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each hydrogen atom.

  Formalcharge=(7)((6)+(12)(2))=0.

Conclusion

BeH2 has only one Lewis structure with the zero formal charges on Be and H atom.

(b)

Interpretation Introduction

Interpretation:

Lewis structure with the lowest formal charges of I3 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the molecule are as follows:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  Formalcharge=(numberofvalenceelectrons)((numberofnon-bondingelectrons)+(12)(numberofbondingelectrons))        (1)

(b)

Expert Solution
Check Mark

Answer to Problem 10.5BFP

The Lewis structure of I3 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  3

The formal charge on central I is 1.

The formal charge on the terminal I is 0.

Explanation of Solution

The total number of valence electrons of I3 is calculated as,

  Total valence electrons(TVE)=[(Total number of I atom)(Valence electrons of I)+(negative charge)]        (1)

Substitute 3 for the total number of I atom, 7 for valence electrons of I, and 1 for the negative charge in equation (1).

  Total valence electrons(TVE)=[(3)(7)+(1)]=22

I3 has 22 valence electrons. One iodine serves as the central atom and forms two single bonds with other two I atoms. Out of 22 electrons, 4 are used for the formation of 2 single bonds. Remaining 18 electrons are distributed as lone pairs of iodine atoms to complete the octet of each atom.

The Lewis structure of I3 is as follows:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  4

Substitute 7 for the number of valence electrons, 6 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on the central I atom.

  Formalcharge=(7)((6)+(12)(4))=1

Substitute 7 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each terminal I atom.

  Formalcharge=(7)((6)+(12)(2))=0.

Conclusion

Only one Lewis structure is possible for I3 and the central iodine has 1 formal charge.

(c)

Interpretation Introduction

Interpretation:

Lewis structure with the lowest formal charges of XeO3 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the molecule are as follows:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  Formalcharge=(numberofvalenceelectrons)((numberofnon-bondingelectrons)+(12)(numberofbondingelectrons))        (1)

(c)

Expert Solution
Check Mark

Answer to Problem 10.5BFP

Lewis structure of XeO3 with lowest formal charge is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  5

The formal charge on Xe is 0.

The formal charge on each O is 0.

Explanation of Solution

The Lewis structures of XeO3 are written as follows:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  6

For structure I:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  7

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 6 for the number of bonding electrons in equation (1) to calculate the formal charge on Xe atom.

  Formalcharge=(8)((2)+(12)(6))=+3

Substitute 6 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each single bonded oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

For structure II:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  8

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 8 for the number of bonding electrons in equation (1) to calculate the formal charge on Xe atom.

  Formalcharge=(8)((2)+(12)(8))=+2

Substitute 6 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each single bonded oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

Substitute 6 for the number of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on the double bonded oxygen atom.

  Formalcharge=(6)((4)+(12)(4))=0

For structure III:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  9

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 10 for the number of bonding electrons in equation (1) to calculate the formal charge on Xe atom.

  Formalcharge=(8)((2)+(12)(10))=+1

Substitute 6 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on the single bonded oxygen atom.

  Formalcharge=(6)((6)+(12)(2))=1

Substitute 6 for the number of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on each double bonded oxygen atom.

  Formalcharge=(6)((4)+(12)(4))=0

For structure IV:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10.1, Problem 10.5BFP , additional homework tip  10

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 12 for the number of bonding electrons in equation (1) to calculate the formal charge on Xe atom.

  Formalcharge=(8)((2)+(12)(12))=0

Substitute 6 for the number of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on each double bonded oxygen atom.

  Formalcharge=(6)((4)+(12)(4))=0.

Conclusion

Four Lewis structures are possible for XeO3. Among 4 structures, only one structure has zero formal charge on xenon atom.

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!

Chapter 10 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 10.2 - Prob. 10.6AFPCh. 10.2 - Prob. 10.6BFPCh. 10.2 - Prob. 10.7AFPCh. 10.2 - Prob. 10.7BFPCh. 10.2 - Prob. 10.8AFPCh. 10.2 - Prob. 10.8BFPCh. 10.3 - Prob. 10.9AFPCh. 10.3 - Prob. 10.9BFPCh. 10.3 - Prob. B10.1PCh. 10.3 - Prob. B10.2PCh. 10 - Prob. 10.1PCh. 10 - When is a resonance hybrid needed to adequately...Ch. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Draw a Lewis structure for (a) SiF4; (b) SeCl2;...Ch. 10 - Draw a Lewis structure for (a) ; (b) C2F4; (c)...Ch. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.11PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - Molten beryllium chloride reacts with chloride ion...Ch. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Phosgene is a colorless, highly toxic gas that was...Ch. 10 - If you know the formula of a molecule or ion, what...Ch. 10 - In what situation is the name of the molecular...Ch. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Consider the following molecular shapes. (a) Which...Ch. 10 - Use wedge-bond perspective drawings (if necessary)...Ch. 10 - Prob. 10.33PCh. 10 - Determine the electron-group arrangement,...Ch. 10 - Determine the electron-group arrangement,...Ch. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Determine the shape, ideal bond angle(s), and the...Ch. 10 - Prob. 10.41PCh. 10 - Determine the shape around each central atom in...Ch. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Arrange the following ACln species in order of...Ch. 10 - State an ideal value for each of the bond angles...Ch. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - How can a molecule with polar covalent bonds not...Ch. 10 - Prob. 10.54PCh. 10 - Consider the molecules SCl2, F2, CS2, CF4, and...Ch. 10 - Consider the molecules BF3, PF3, BrF3, SF4, and...Ch. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - When SO3 gains two electrons, forms. (a) Which...Ch. 10 - The actual bond angle in NO2 is 134.3°, and in it...Ch. 10 - Prob. 10.69PCh. 10 - Propylene oxide is used to make many products,...Ch. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - A gaseous compound has a composition by mass of...Ch. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Pure HN3 (atom sequence HNNN) is explosive. In...Ch. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Oxalic acid (H2C2O4) is found in toxic...Ch. 10 - Prob. 10.87PCh. 10 - Hydrazine (N2H4) is used as a rocket fuel because...Ch. 10 - Prob. 10.89PCh. 10 - Prob. 10.90PCh. 10 - Prob. 10.91PCh. 10 - Consider the following molecular shapes: Match...Ch. 10 - Prob. 10.93PCh. 10 - Prob. 10.94PCh. 10 - Prob. 10.95PCh. 10 - Phosphorus pentachloride, a key industrial...
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
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
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY