MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 10, Problem 10.76P

(a)

Interpretation Introduction

Interpretation:

A Lewis structure for C3H4 is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

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.

Step 5: Convert the lone pair into bond pair.

(a)

Expert Solution
Check Mark

Answer to Problem 10.76P

The Lewis structure for C3H4 is as follows:

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  1

The carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Explanation of Solution

The total number of valence electrons of C3H4 is calculated as,

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

Substitute 3 for the total number of atom, 4 for valence electrons of , 1 for the total valence electrons of H, and 4 for the total number of H atom in equation (1).

  Total valence electrons=[(3)(4)+(4)(1)]=16.

In C3H4, the three atom use 6 electrons to form single bond with each other and form three-membered ring and the two carbon atom in the ring also form one double bond with each other. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

The Lewis structure of C3H4 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  2

The single bond has the bond order 1 and double bond has the bond order 2. Therefore, the carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.

(b)

Interpretation Introduction

Interpretation:

A Lewis structure for C3H6 is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as octet rule.

To draw the Lewis structure of the molecule there are following steps:

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.

Step 5: Convert the lone pair into bond pair.

(b)

Expert Solution
Check Mark

Answer to Problem 10.76P

The Lewis structure of C3H6 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  3

There is only single bond in C3H6 so the carbon-carbon bond order is 1.

Explanation of Solution

The total number of valence electrons of C3H6 is calculated as,

  Total valence electrons=[[(Total number of C atom)(Valence electrons of C)]+[(Total number of H atom)(Valence electrons of H)]]        (2)

Substitute 3 for the total number of atom, 4 for valence electrons of , 1 for the total valence electrons of H, and 6 for total number of H atom in equation (2).

  Total valence electrons=[(3)(4)+(6)(1)]=18.

In C3H6, the three atom use 6 electrons to form single bond with each other and form three-membered ring. The remaining electrons are used by the carbon atoms to form bond with hydrogen atom. Each carbon atom forms two carbon-hydrogen bonds respectively to complete the valence electrons.

The Lewis structure of C3H6 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  4

The single bond has the bond order 1 and double bond has the bond order 2. Therefore, there is only single bond in C3H6 so the carbon-carbon bond order is 1.

Conclusion

The single bond has the bond order 1, double bond has the bond order 2 and triple bond has the bond order 3.

(c)

Interpretation Introduction

Interpretation:

A Lewis structure for C4H6 is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

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.

Step 5: Convert the lone pair into bond pair.

(c)

Expert Solution
Check Mark

Answer to Problem 10.76P

The Lewis structure of C4H6 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  5

The carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Explanation of Solution

The total number of valence electrons of C4H6 is calculated as,

  Total valence electrons=[[(Total number of C atom)(Valence electrons of C)]+[(Total number of H atom)(Valence electrons of H)]]        (3)

Substitute 4 for the total number of atom, 4 for valence electrons of , 1 for the total valence electrons of H, and 6 for total number of H atom in equation (3).

  Total valence electrons=[(4)(4)+(6)(1)]=22.

In C4H6, the four atom use 8 electrons to form single bond with each other and form a four-membered ring and the two carbon atom in the ring also form one double bond with each other. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

The Lewis structure of C4H6 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  6

The single bond has the bond order 1 and double bond has the bond order 2. Therefore, the carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.

(d)

Interpretation Introduction

Interpretation:

A Lewis structure for C4H4 is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

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.

Step 5: Convert the lone pair into bond pair.

(d)

Expert Solution
Check Mark

Answer to Problem 10.76P

The Lewis structure of C4H4 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  7

The average bond order in C4H4 is 1.5.

Explanation of Solution

The total number of valence electrons of C4H4 is calculated as,

  Total valence electrons=[[(Total number of C atom)(Valence electrons of C)]+[(Total number of H atom)(Valence electrons of H)]]        (4)

Substitute 4 for the total number of atom, 4 for valence electrons of , 1 for the total valence electrons of H, and 4 for total number of H atom in equation (4).

  Total valence electrons=[(4)(4)+(4)(1)]=20

In C4H4, the four atom use 8 electrons to form single bond with each other and form four-membered ring and the four atoms in the ring also forms two alternate double bond with other carbon other in the ring. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

There is alternate single and double present in the ring so the molecule will show resonance.

The Lewis structure of C4H4 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  8

There are 4 single bonds and 2 double bonds in C4H4.

The formula to calculate the average bond order in C4H4 is,

  Average bond order=Number of shared electron pairNumber of bonded atoms        (5)

Substitute 6 for the number of shared electron pair and 4 for the number of bonded atoms in equation (5).

  Average bond order=64=1.5.

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.

(e)

Interpretation Introduction

Interpretation:

A Lewis structure for C6H6 is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

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.

Step 5: Convert the lone pair into bond pair.

(e)

Expert Solution
Check Mark

Answer to Problem 10.76P

The Lewis structure of C6H6 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  9

The average bond order in C6H6 is 1.5.

Explanation of Solution

The total number of valence electrons of C6H6 is calculated as,

  Total valence electrons=[[(Total number of C atom)(Valence electrons of C)]+[(Total number of H atom)(Valence electrons of H)]]        (6)

Substitute 6 for the total number of atom, 4 for valence electrons of , 1 for the total valence electrons of H, and 6 for total number of H atom in equation (6).

  Total valence electrons=[(6)(4)+(6)(1)]=30

In C6H6, the four atom use 12 electrons to form single bond with each other and form a six-membered ring and the six atoms in the ring also forms three alternate double bonds with other carbon other in the ring. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

There is alternate single and double present in the ring so the molecule will show resonance.

The Lewis structure of C6H6 is,

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 10, Problem 10.76P , additional homework tip  10

There are 6 single bonds and 3 double bonds in C6H6.

The formula to calculate the average bond order in C6H6 is,

  Average bond order=Number of shared electron pairNumber of bonded atoms        (7)

Substitute 9 for the number of shared electron pair and 6 for the number of bonded atoms in equation (7).

  Average bond order=96=1.5.

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.

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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...
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