The resonance structure of the PO 2 F 2 − ion should be drawn. Concept Introduction: Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance. All the possible resonance structures are imaginary whereas the resonance hybrid is real. These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei. To find : The resonance structure of PO 2 F 2 − ion

Question

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Answer 1

Question

Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the $PO2F2 −$ ion should be drawn.

Concept Introduction:

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of $PO2F2 −$ ion

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 1

Explanation of Solution

Resonance structure of $PO2F2 −$ ion is drawn below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 2

In the case of $PO2F2 −$ , the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 4 possible resonance structures. In all the 4 resonance structures the position, over whole charge and chemical framework remains intact.

Conclusion

The resonance structures of the $PO2F2 −$ ion were drawn.

Interpretation Introduction

Interpretation: The formal charges of the $PO2F2 −$ ion should be drawn.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

Formal charge of an atom can be determined by the given formula.

$Formal charge (F C) = (no.of valence electron in atom)−12(no.of bonding electrons)−(no.of non-bonding electrons)$

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 3

Explanation of Solution

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 4

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 8Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×8)= +1$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 5

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 6

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 7

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

Conclusion

The formal charge of the $PO2F2 −$ ion was found.

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Answer 2

Question

Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the $PO2F2 −$ ion should be drawn.

Concept Introduction:

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of $PO2F2 −$ ion

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 1

Explanation of Solution

Resonance structure of $PO2F2 −$ ion is drawn below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 2

In the case of $PO2F2 −$ , the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 4 possible resonance structures. In all the 4 resonance structures the position, over whole charge and chemical framework remains intact.

Conclusion

The resonance structures of the $PO2F2 −$ ion were drawn.

Interpretation Introduction

Interpretation: The formal charges of the $PO2F2 −$ ion should be drawn.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

Formal charge of an atom can be determined by the given formula.

$Formal charge (F C) = (no.of valence electron in atom)−12(no.of bonding electrons)−(no.of non-bonding electrons)$

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 3

Explanation of Solution

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 4

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 8Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×8)= +1$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 5

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 6

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 7

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

Conclusion

The formal charge of the $PO2F2 −$ ion was found.

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Answer 3

Question

Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the $PO2F2 −$ ion should be drawn.

Concept Introduction:

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of $PO2F2 −$ ion

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 1

Explanation of Solution

Resonance structure of $PO2F2 −$ ion is drawn below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 2

In the case of $PO2F2 −$ , the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 4 possible resonance structures. In all the 4 resonance structures the position, over whole charge and chemical framework remains intact.

Conclusion

The resonance structures of the $PO2F2 −$ ion were drawn.

Interpretation Introduction

Interpretation: The formal charges of the $PO2F2 −$ ion should be drawn.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

Formal charge of an atom can be determined by the given formula.

$Formal charge (F C) = (no.of valence electron in atom)−12(no.of bonding electrons)−(no.of non-bonding electrons)$

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 3

Explanation of Solution

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 4

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 8Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×8)= +1$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 5

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 6

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 7

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

Conclusion

The formal charge of the $PO2F2 −$ ion was found.

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Answer 4

Question

Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the $PO2F2 −$ ion should be drawn.

Concept Introduction:

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of $PO2F2 −$ ion

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 1

Explanation of Solution

Resonance structure of $PO2F2 −$ ion is drawn below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 2

In the case of $PO2F2 −$ , the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 4 possible resonance structures. In all the 4 resonance structures the position, over whole charge and chemical framework remains intact.

Conclusion

The resonance structures of the $PO2F2 −$ ion were drawn.

Interpretation Introduction

Interpretation: The formal charges of the $PO2F2 −$ ion should be drawn.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

Formal charge of an atom can be determined by the given formula.

$Formal charge (F C) = (no.of valence electron in atom)−12(no.of bonding electrons)−(no.of non-bonding electrons)$

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 3

Explanation of Solution

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 4

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 8Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×8)= +1$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 5

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 6

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 7

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

Conclusion

The formal charge of the $PO2F2 −$ ion was found.

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Answer 5

Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the $PO2F2 −$ ion should be drawn.

Concept Introduction:

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of $PO2F2 −$ ion

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 1

Explanation of Solution

Resonance structure of $PO2F2 −$ ion is drawn below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 2

In the case of $PO2F2 −$ , the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 4 possible resonance structures. In all the 4 resonance structures the position, over whole charge and chemical framework remains intact.

Conclusion

The resonance structures of the $PO2F2 −$ ion were drawn.

Interpretation Introduction

Interpretation: The formal charges of the $PO2F2 −$ ion should be drawn.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

Formal charge of an atom can be determined by the given formula.

$Formal charge (F C) = (no.of valence electron in atom)−12(no.of bonding electrons)−(no.of non-bonding electrons)$

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 3

Explanation of Solution

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 4

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 8Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×8)= +1$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 5

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 6

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 7

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

Conclusion

The formal charge of the $PO2F2 −$ ion was found.

Answer 6

Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the $PO2F2 −$ ion should be drawn.

Concept Introduction:

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of $PO2F2 −$ ion

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 1

Explanation of Solution

Resonance structure of $PO2F2 −$ ion is drawn below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 2

In the case of $PO2F2 −$ , the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 4 possible resonance structures. In all the 4 resonance structures the position, over whole charge and chemical framework remains intact.

Conclusion

The resonance structures of the $PO2F2 −$ ion were drawn.

Answer 7

Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the $PO2F2 −$ ion should be drawn.

Concept Introduction:

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

To find: The resonance structure of $PO2F2 −$ ion

Answer 8

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 1

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Resonance structure of $PO2F2 −$ ion is drawn below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 2

In the case of $PO2F2 −$ , the chemical bonding of a molecule cannot be represented using a single Lewis structure. The chemical bonding are described by delocalization of electrons forming 4 possible resonance structures. In all the 4 resonance structures the position, over whole charge and chemical framework remains intact.

Answer 13

Conclusion

The resonance structures of the $PO2F2 −$ ion were drawn.

Answer 14

Interpretation Introduction

Interpretation: The formal charges of the $PO2F2 −$ ion should be drawn.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

Formal charge of an atom can be determined by the given formula.

$Formal charge (F C) = (no.of valence electron in atom)−12(no.of bonding electrons)−(no.of non-bonding electrons)$

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 3

Explanation of Solution

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 4

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 8Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×8)= +1$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 5

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 6

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 7

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

Conclusion

The formal charge of the $PO2F2 −$ ion was found.

Answer 15

Interpretation Introduction

Interpretation: The formal charges of the $PO2F2 −$ ion should be drawn.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

Formal charge of an atom can be determined by the given formula.

$Formal charge (F C) = (no.of valence electron in atom)−12(no.of bonding electrons)−(no.of non-bonding electrons)$

Answer 16

Expert Solution

Answer to Problem 6.62QP

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 3

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 4

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 8Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×8)= +1$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 5

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 6

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

The formal charge of the given resonance structure is given below.

Chemistry Atoms First, Second Edition, Chapter 6, Problem 6.62QP , additional homework tip 7

The formal charge of the given resonance structure is calculated,

Phosphorous atom

$Number of valence electron=5Number of bonding electron= 10Number of non-bonding electron = 0$

Substituting these values to the equation,

$FC = 5− (12×10)= 0$

Oxygen atom (a)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (b)

$Number of valence electron=6Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 6− (12×2)−6= −1$

Oxygen atom (c)

$Number of valence electron=6Number of bonding electron= 4Number of non-bonding electron = 4$

Substituting these values to the equation,

$FC = 6− (12×4)−4= 0$

Fluorine atom

$Number of valence electron=7Number of bonding electron= 2Number of non-bonding electron = 6$

Substituting these values to the equation,

$FC = 7−(12×2) −6= 0$

Answer 21

Conclusion

The formal charge of the $PO2F2 −$ ion was found.

Answer 22

Ch. 6.2 - Classify the following bonds as nonpolar, polar,...Ch. 6.2 - Classify the following bonds as nonpolar, polar,...Ch. 6.2 - Prob. 1PPB Ch. 6.2 - Electrostatic potential maps are shown for HCl and...Ch. 6.2 - Prob. 6.2WE Ch. 6.2 - Prob. 2PPA Ch. 6.2 - Prob. 2PPB Ch. 6.2 - Prob. 2PPC Ch. 6.2 - Prob. 6.3WE Ch. 6.2 - Prob. 3PPA

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