The Lewis structures of the given molecules and ions are to be represented. Concept Introduction: In Lewis dot symbol, for each element , dots are mentioned around the symbol of an atom. In Lewis dot symbol, valence electrons are represented by dots. Dots are placed above and below as well as to the left and right of symbol. Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol. In writing symbol pairing is not done until absolutely necessary. For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation. For second period non metals, the number of unpaired dots is the number of bonds the atom can form. Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol. Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule. The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

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Answer 1

Question

Chapter 8, Problem 42QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept Introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Expert Solution & Answer

Answer to Problem 42QP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 1

(b)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 2

(c)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 3

(d)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 4

(e)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 5

(f)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 6

Explanation of Solution

a) $NCl3$

The electronic configuration of nitrogen and chlorine in $NCl3$ is as:

$N=1s22s22p3Cl=1s22s22p63s23p5$

The nitrogen atom contains three valence electrons in its $2p$ subshell and the chlorine atom contains five valence electrons in its $3p$ subshell. Therefore, nitrogen has a tendency to donate three electrons and chlorine has a tendency to accept one electron, to complete their outermost shell. Therefore, the Lewis structure of $NCl3$ contains three $N−Cl$ bonds and one lone pair on the nitrogen atom.

The Lewis structure of $NCl3$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 7

b) $OCS$

The electronic configuration of oxygen, carbon, and sulfur in $OCS$ is as:

$O=1s22s22p4C=1s22s22p2S=1s22s22p63s23p4$

Oxygen and sulfur atoms contain two valence electrons in their $2p$ and $3p$ subshells, respectively and the carbon atom contains four valence electrons in its $2s$ and $2p$ subshells. Therefore, carbon has a tendency to donate four electrons, and oxygen and sulfur have a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $OCS$ contains one oxygen-carbon double bond, one carbon-sulfur double bond, and two lone pairs on oxygen and sulfur atoms.

The Lewis structure of $OCS$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 8

c) $H2O2$

The electronic configuration of oxygen and hydrogen in $H2O2$ is as:

$O=1s22s22p4H=1s1$

The oxygen atom contains four valence electrons in its $2p$ and $3p$ subshell and the hydrogen atom contains one valence electron in its $1s$ subshell. Therefore, hydrogen has a tendency to donate one electron and oxygen has a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $H2O2$ contains two oxygen-hydrogen bonds, one oxygen-oxygen bond, and two lone pairs on oxygen atoms.

The Lewis structure of $H2O2$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 9

d) $CH3COO−$

The electronic configuration of oxygen, carbon, and hydrogen in $CH3COO−$ is as:

$O=1s22s22p4C=1s22s22p2H=1s1$

The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.

The Lewis structure of $CH3COO−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 10

e) $CN−$

The electronic configuration of nitrogen and carbon in $CN−$ is as:

$N=1s22s22p3C=1s22s22p2$

Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.

The Lewis structure of $CN−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 11

f) $CH3CH2NH3+$

The electronic configuration of carbon, nitrogen, and hydrogen in $CH3CH2NH3+$ is as:

$N=1s22s22p3C=1s22s22p2H=1s1$

Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its $2p$ subshell and one lone pair in $2s$ subshell.

The Lewis structure of $CH3CH2NH3+$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 12

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Answer 2

Question

Chapter 8, Problem 42QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept Introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Expert Solution & Answer

Answer to Problem 42QP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 1

(b)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 2

(c)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 3

(d)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 4

(e)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 5

(f)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 6

Explanation of Solution

a) $NCl3$

The electronic configuration of nitrogen and chlorine in $NCl3$ is as:

$N=1s22s22p3Cl=1s22s22p63s23p5$

The nitrogen atom contains three valence electrons in its $2p$ subshell and the chlorine atom contains five valence electrons in its $3p$ subshell. Therefore, nitrogen has a tendency to donate three electrons and chlorine has a tendency to accept one electron, to complete their outermost shell. Therefore, the Lewis structure of $NCl3$ contains three $N−Cl$ bonds and one lone pair on the nitrogen atom.

The Lewis structure of $NCl3$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 7

b) $OCS$

The electronic configuration of oxygen, carbon, and sulfur in $OCS$ is as:

$O=1s22s22p4C=1s22s22p2S=1s22s22p63s23p4$

Oxygen and sulfur atoms contain two valence electrons in their $2p$ and $3p$ subshells, respectively and the carbon atom contains four valence electrons in its $2s$ and $2p$ subshells. Therefore, carbon has a tendency to donate four electrons, and oxygen and sulfur have a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $OCS$ contains one oxygen-carbon double bond, one carbon-sulfur double bond, and two lone pairs on oxygen and sulfur atoms.

The Lewis structure of $OCS$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 8

c) $H2O2$

The electronic configuration of oxygen and hydrogen in $H2O2$ is as:

$O=1s22s22p4H=1s1$

The oxygen atom contains four valence electrons in its $2p$ and $3p$ subshell and the hydrogen atom contains one valence electron in its $1s$ subshell. Therefore, hydrogen has a tendency to donate one electron and oxygen has a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $H2O2$ contains two oxygen-hydrogen bonds, one oxygen-oxygen bond, and two lone pairs on oxygen atoms.

The Lewis structure of $H2O2$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 9

d) $CH3COO−$

The electronic configuration of oxygen, carbon, and hydrogen in $CH3COO−$ is as:

$O=1s22s22p4C=1s22s22p2H=1s1$

The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.

The Lewis structure of $CH3COO−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 10

e) $CN−$

The electronic configuration of nitrogen and carbon in $CN−$ is as:

$N=1s22s22p3C=1s22s22p2$

Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.

The Lewis structure of $CN−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 11

f) $CH3CH2NH3+$

The electronic configuration of carbon, nitrogen, and hydrogen in $CH3CH2NH3+$ is as:

$N=1s22s22p3C=1s22s22p2H=1s1$

Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its $2p$ subshell and one lone pair in $2s$ subshell.

The Lewis structure of $CH3CH2NH3+$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 12

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Answer 3

Question

Chapter 8, Problem 42QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept Introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Expert Solution & Answer

Answer to Problem 42QP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 1

(b)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 2

(c)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 3

(d)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 4

(e)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 5

(f)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 6

Explanation of Solution

a) $NCl3$

The electronic configuration of nitrogen and chlorine in $NCl3$ is as:

$N=1s22s22p3Cl=1s22s22p63s23p5$

The nitrogen atom contains three valence electrons in its $2p$ subshell and the chlorine atom contains five valence electrons in its $3p$ subshell. Therefore, nitrogen has a tendency to donate three electrons and chlorine has a tendency to accept one electron, to complete their outermost shell. Therefore, the Lewis structure of $NCl3$ contains three $N−Cl$ bonds and one lone pair on the nitrogen atom.

The Lewis structure of $NCl3$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 7

b) $OCS$

The electronic configuration of oxygen, carbon, and sulfur in $OCS$ is as:

$O=1s22s22p4C=1s22s22p2S=1s22s22p63s23p4$

Oxygen and sulfur atoms contain two valence electrons in their $2p$ and $3p$ subshells, respectively and the carbon atom contains four valence electrons in its $2s$ and $2p$ subshells. Therefore, carbon has a tendency to donate four electrons, and oxygen and sulfur have a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $OCS$ contains one oxygen-carbon double bond, one carbon-sulfur double bond, and two lone pairs on oxygen and sulfur atoms.

The Lewis structure of $OCS$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 8

c) $H2O2$

The electronic configuration of oxygen and hydrogen in $H2O2$ is as:

$O=1s22s22p4H=1s1$

The oxygen atom contains four valence electrons in its $2p$ and $3p$ subshell and the hydrogen atom contains one valence electron in its $1s$ subshell. Therefore, hydrogen has a tendency to donate one electron and oxygen has a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $H2O2$ contains two oxygen-hydrogen bonds, one oxygen-oxygen bond, and two lone pairs on oxygen atoms.

The Lewis structure of $H2O2$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 9

d) $CH3COO−$

The electronic configuration of oxygen, carbon, and hydrogen in $CH3COO−$ is as:

$O=1s22s22p4C=1s22s22p2H=1s1$

The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.

The Lewis structure of $CH3COO−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 10

e) $CN−$

The electronic configuration of nitrogen and carbon in $CN−$ is as:

$N=1s22s22p3C=1s22s22p2$

Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.

The Lewis structure of $CN−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 11

f) $CH3CH2NH3+$

The electronic configuration of carbon, nitrogen, and hydrogen in $CH3CH2NH3+$ is as:

$N=1s22s22p3C=1s22s22p2H=1s1$

Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its $2p$ subshell and one lone pair in $2s$ subshell.

The Lewis structure of $CH3CH2NH3+$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 12

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Answer 4

Question

Chapter 8, Problem 42QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept Introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Expert Solution & Answer

Answer to Problem 42QP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 1

(b)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 2

(c)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 3

(d)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 4

(e)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 5

(f)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 6

Explanation of Solution

a) $NCl3$

The electronic configuration of nitrogen and chlorine in $NCl3$ is as:

$N=1s22s22p3Cl=1s22s22p63s23p5$

The nitrogen atom contains three valence electrons in its $2p$ subshell and the chlorine atom contains five valence electrons in its $3p$ subshell. Therefore, nitrogen has a tendency to donate three electrons and chlorine has a tendency to accept one electron, to complete their outermost shell. Therefore, the Lewis structure of $NCl3$ contains three $N−Cl$ bonds and one lone pair on the nitrogen atom.

The Lewis structure of $NCl3$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 7

b) $OCS$

The electronic configuration of oxygen, carbon, and sulfur in $OCS$ is as:

$O=1s22s22p4C=1s22s22p2S=1s22s22p63s23p4$

Oxygen and sulfur atoms contain two valence electrons in their $2p$ and $3p$ subshells, respectively and the carbon atom contains four valence electrons in its $2s$ and $2p$ subshells. Therefore, carbon has a tendency to donate four electrons, and oxygen and sulfur have a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $OCS$ contains one oxygen-carbon double bond, one carbon-sulfur double bond, and two lone pairs on oxygen and sulfur atoms.

The Lewis structure of $OCS$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 8

c) $H2O2$

The electronic configuration of oxygen and hydrogen in $H2O2$ is as:

$O=1s22s22p4H=1s1$

The oxygen atom contains four valence electrons in its $2p$ and $3p$ subshell and the hydrogen atom contains one valence electron in its $1s$ subshell. Therefore, hydrogen has a tendency to donate one electron and oxygen has a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $H2O2$ contains two oxygen-hydrogen bonds, one oxygen-oxygen bond, and two lone pairs on oxygen atoms.

The Lewis structure of $H2O2$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 9

d) $CH3COO−$

The electronic configuration of oxygen, carbon, and hydrogen in $CH3COO−$ is as:

$O=1s22s22p4C=1s22s22p2H=1s1$

The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.

The Lewis structure of $CH3COO−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 10

e) $CN−$

The electronic configuration of nitrogen and carbon in $CN−$ is as:

$N=1s22s22p3C=1s22s22p2$

Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.

The Lewis structure of $CN−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 11

f) $CH3CH2NH3+$

The electronic configuration of carbon, nitrogen, and hydrogen in $CH3CH2NH3+$ is as:

$N=1s22s22p3C=1s22s22p2H=1s1$

Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its $2p$ subshell and one lone pair in $2s$ subshell.

The Lewis structure of $CH3CH2NH3+$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 12

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!

Answer 5

Chapter 8, Problem 42QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept Introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Expert Solution & Answer

Answer to Problem 42QP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 1

(b)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 2

(c)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 3

(d)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 4

(e)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 5

(f)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 6

Explanation of Solution

a) $NCl3$

The electronic configuration of nitrogen and chlorine in $NCl3$ is as:

$N=1s22s22p3Cl=1s22s22p63s23p5$

The nitrogen atom contains three valence electrons in its $2p$ subshell and the chlorine atom contains five valence electrons in its $3p$ subshell. Therefore, nitrogen has a tendency to donate three electrons and chlorine has a tendency to accept one electron, to complete their outermost shell. Therefore, the Lewis structure of $NCl3$ contains three $N−Cl$ bonds and one lone pair on the nitrogen atom.

The Lewis structure of $NCl3$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 7

b) $OCS$

The electronic configuration of oxygen, carbon, and sulfur in $OCS$ is as:

$O=1s22s22p4C=1s22s22p2S=1s22s22p63s23p4$

Oxygen and sulfur atoms contain two valence electrons in their $2p$ and $3p$ subshells, respectively and the carbon atom contains four valence electrons in its $2s$ and $2p$ subshells. Therefore, carbon has a tendency to donate four electrons, and oxygen and sulfur have a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $OCS$ contains one oxygen-carbon double bond, one carbon-sulfur double bond, and two lone pairs on oxygen and sulfur atoms.

The Lewis structure of $OCS$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 8

c) $H2O2$

The electronic configuration of oxygen and hydrogen in $H2O2$ is as:

$O=1s22s22p4H=1s1$

The oxygen atom contains four valence electrons in its $2p$ and $3p$ subshell and the hydrogen atom contains one valence electron in its $1s$ subshell. Therefore, hydrogen has a tendency to donate one electron and oxygen has a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $H2O2$ contains two oxygen-hydrogen bonds, one oxygen-oxygen bond, and two lone pairs on oxygen atoms.

The Lewis structure of $H2O2$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 9

d) $CH3COO−$

The electronic configuration of oxygen, carbon, and hydrogen in $CH3COO−$ is as:

$O=1s22s22p4C=1s22s22p2H=1s1$

The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.

The Lewis structure of $CH3COO−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 10

e) $CN−$

The electronic configuration of nitrogen and carbon in $CN−$ is as:

$N=1s22s22p3C=1s22s22p2$

Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.

The Lewis structure of $CN−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 11

f) $CH3CH2NH3+$

The electronic configuration of carbon, nitrogen, and hydrogen in $CH3CH2NH3+$ is as:

$N=1s22s22p3C=1s22s22p2H=1s1$

Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its $2p$ subshell and one lone pair in $2s$ subshell.

The Lewis structure of $CH3CH2NH3+$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 12

Answer 6

Chapter 8, Problem 42QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept Introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Expert Solution & Answer

Answer to Problem 42QP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 1

(b)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 2

(c)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 3

(d)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 4

(e)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 5

(f)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 6

Explanation of Solution

a) $NCl3$

The electronic configuration of nitrogen and chlorine in $NCl3$ is as:

$N=1s22s22p3Cl=1s22s22p63s23p5$

The nitrogen atom contains three valence electrons in its $2p$ subshell and the chlorine atom contains five valence electrons in its $3p$ subshell. Therefore, nitrogen has a tendency to donate three electrons and chlorine has a tendency to accept one electron, to complete their outermost shell. Therefore, the Lewis structure of $NCl3$ contains three $N−Cl$ bonds and one lone pair on the nitrogen atom.

The Lewis structure of $NCl3$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 7

b) $OCS$

The electronic configuration of oxygen, carbon, and sulfur in $OCS$ is as:

$O=1s22s22p4C=1s22s22p2S=1s22s22p63s23p4$

Oxygen and sulfur atoms contain two valence electrons in their $2p$ and $3p$ subshells, respectively and the carbon atom contains four valence electrons in its $2s$ and $2p$ subshells. Therefore, carbon has a tendency to donate four electrons, and oxygen and sulfur have a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $OCS$ contains one oxygen-carbon double bond, one carbon-sulfur double bond, and two lone pairs on oxygen and sulfur atoms.

The Lewis structure of $OCS$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 8

c) $H2O2$

The electronic configuration of oxygen and hydrogen in $H2O2$ is as:

$O=1s22s22p4H=1s1$

The oxygen atom contains four valence electrons in its $2p$ and $3p$ subshell and the hydrogen atom contains one valence electron in its $1s$ subshell. Therefore, hydrogen has a tendency to donate one electron and oxygen has a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $H2O2$ contains two oxygen-hydrogen bonds, one oxygen-oxygen bond, and two lone pairs on oxygen atoms.

The Lewis structure of $H2O2$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 9

d) $CH3COO−$

The electronic configuration of oxygen, carbon, and hydrogen in $CH3COO−$ is as:

$O=1s22s22p4C=1s22s22p2H=1s1$

The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.

The Lewis structure of $CH3COO−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 10

e) $CN−$

The electronic configuration of nitrogen and carbon in $CN−$ is as:

$N=1s22s22p3C=1s22s22p2$

Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.

The Lewis structure of $CN−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 11

f) $CH3CH2NH3+$

The electronic configuration of carbon, nitrogen, and hydrogen in $CH3CH2NH3+$ is as:

$N=1s22s22p3C=1s22s22p2H=1s1$

Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its $2p$ subshell and one lone pair in $2s$ subshell.

The Lewis structure of $CH3CH2NH3+$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 12

Answer 7

Chapter 8, Problem 42QP

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be represented.

Concept Introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period non metals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Answer 8

Expert Solution & Answer

Answer to Problem 42QP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 1

(b)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 2

(c)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 3

(d)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 4

(e)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 5

(f)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 6

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

a) $NCl3$

The electronic configuration of nitrogen and chlorine in $NCl3$ is as:

$N=1s22s22p3Cl=1s22s22p63s23p5$

The nitrogen atom contains three valence electrons in its $2p$ subshell and the chlorine atom contains five valence electrons in its $3p$ subshell. Therefore, nitrogen has a tendency to donate three electrons and chlorine has a tendency to accept one electron, to complete their outermost shell. Therefore, the Lewis structure of $NCl3$ contains three $N−Cl$ bonds and one lone pair on the nitrogen atom.

The Lewis structure of $NCl3$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 7

b) $OCS$

The electronic configuration of oxygen, carbon, and sulfur in $OCS$ is as:

$O=1s22s22p4C=1s22s22p2S=1s22s22p63s23p4$

Oxygen and sulfur atoms contain two valence electrons in their $2p$ and $3p$ subshells, respectively and the carbon atom contains four valence electrons in its $2s$ and $2p$ subshells. Therefore, carbon has a tendency to donate four electrons, and oxygen and sulfur have a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $OCS$ contains one oxygen-carbon double bond, one carbon-sulfur double bond, and two lone pairs on oxygen and sulfur atoms.

The Lewis structure of $OCS$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 8

c) $H2O2$

The electronic configuration of oxygen and hydrogen in $H2O2$ is as:

$O=1s22s22p4H=1s1$

The oxygen atom contains four valence electrons in its $2p$ and $3p$ subshell and the hydrogen atom contains one valence electron in its $1s$ subshell. Therefore, hydrogen has a tendency to donate one electron and oxygen has a tendency to accept two electrons, to complete their outermost shell. Therefore, the Lewis structure of $H2O2$ contains two oxygen-hydrogen bonds, one oxygen-oxygen bond, and two lone pairs on oxygen atoms.

The Lewis structure of $H2O2$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 9

d) $CH3COO−$

The electronic configuration of oxygen, carbon, and hydrogen in $CH3COO−$ is as:

$O=1s22s22p4C=1s22s22p2H=1s1$

The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.

The Lewis structure of $CH3COO−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 10

e) $CN−$

The electronic configuration of nitrogen and carbon in $CN−$ is as:

$N=1s22s22p3C=1s22s22p2$

Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.

The Lewis structure of $CN−$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 11

f) $CH3CH2NH3+$

The electronic configuration of carbon, nitrogen, and hydrogen in $CH3CH2NH3+$ is as:

$N=1s22s22p3C=1s22s22p2H=1s1$

Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its $2p$ subshell and one lone pair in $2s$ subshell.

The Lewis structure of $CH3CH2NH3+$ is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 42QP , additional homework tip 12

Answer 12

Ch. 8.1 - Practice ProblemATTEMPT Write Lewis dot symbols...Ch. 8.1 - Practice Problem BUILD Indicate the charge on...Ch. 8.1 - Practice ProblemCONCEPTUALIZE For each of the...Ch. 8.1 - 8.1.1 Using only a periodic table, determine the...Ch. 8.1 - 8.1.2 Using only a periodic table, determine the...Ch. 8.1 - To which group does element X belong if its Lewis...Ch. 8.1 - Prob. 4CP Ch. 8.2 - Prob. 1PPA Ch. 8.2 - Practice ProblemBUILD Arrange the compounds NaF,...Ch. 8.2 - Practice ProblemCONCEPTUALIZE Common ions of four...

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