Chapter 6, Problem 6.62QP

Interpretation Introduction

Interpretation: The resonance structure of the ${\text{PO}}_{\text{2}}{\text{F}}^{\text{2}-}$ 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 ${\text{PO}}_{\text{2}}{\text{F}}^{\text{2}-}$ ion

Answer to Problem 6.62QP

Explanation of Solution

Resonance structure of ${\text{PO}}_{\text{2}}{\text{F}}^{\text{2}-}$ ion is drawn below.

In the case of ${\text{PO}}_{\text{2}}{\text{F}}^{\text{2}-}$, 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 ${\text{PO}}_{\text{2}}{\text{F}}^{\text{2}-}$ ion were drawn.

Interpretation Introduction

Interpretation: The formal charges of the ${\text{PO}}_{\text{2}}{\text{F}}^{\text{2}-}$ 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.

$\text{Formal}\text{charge}\left(\text{F}\text{C}\right)\text{=}\left(\begin{array}{l}\text{no}\text{.of}\text{valence}\\ \text{electron}\text{in}\text{atom}\end{array}\right)-\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{no}\text{.of}\text{bonding}\\ \text{electrons}\end{array}\right)-\left(\begin{array}{l}\text{no}\text{.of}\text{non-bonding}\\ \text{electrons}\end{array}\right)$

Answer to Problem 6.62QP

Explanation of Solution

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

The formal charge of the given resonance structure is calculated,

• Phosphorous atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}8\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×8}\right)\\ \text{=}+1\end{array}$

• Oxygen atom (a)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Oxygen atom (b)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Oxygen atom (c)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Fluorine atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=7}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}7-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}0\end{array}$

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

The formal charge of the given resonance structure is calculated,

• Phosphorous atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}10\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×10}\right)\\ \text{=}0\end{array}$

• Oxygen atom (a)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-4\\ \text{=}0\end{array}$

• Oxygen atom (b)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Oxygen atom (c)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Fluorine atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=7}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}7-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}0\end{array}$

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

The formal charge of the given resonance structure is calculated,

• Phosphorous atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}10\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×10}\right)\\ \text{=}0\end{array}$

• Oxygen atom (a)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Oxygen atom (b)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-4\\ \text{=}0\end{array}$

• Oxygen atom (c)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Fluorine atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=7}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}7-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}0\end{array}$

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

The formal charge of the given resonance structure is calculated,

• Phosphorous atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}10\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}0\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}5-\left(\frac{\text{1}}{\text{2}}\text{×10}\right)\\ \text{=}0\end{array}$

• Oxygen atom (a)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Oxygen atom (b)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}-\text{1}\end{array}$

• Oxygen atom (c)

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}4\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}4\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×4}\right)-4\\ \text{=}0\end{array}$

• Fluorine atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=7}\\ \text{Number}\text{of}\text{bonding}\text{electron=}2\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}6\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}7-\left(\frac{\text{1}}{\text{2}}\text{×2}\right)-6\\ \text{=}0\end{array}$

Conclusion

The formal charge of the ${\text{PO}}_{\text{2}}{\text{F}}^{\text{2}-}$ ion was found.