Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078131615
Author: Martin Silberberg Dr.
Publisher: McGraw-Hill Education
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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,

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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:

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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,

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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:

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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,

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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:

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, Chapter 10.1, Problem 10.5BFP , additional homework tip  6

For structure I:

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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:

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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:

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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:

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change, 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.

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Chapter 10 Solutions

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change

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 - 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...
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