Chapter 9, Problem 9.67P

(a)

Interpretation Introduction

Interpretation:

The members of the following set are to be ranked in the decreasing order of ionic character. Also, the bond polarity with partial charges is to be indicated.

${\text{PCl}}_3,{\text{PBr}}_3,{\text{PF}}_3$

Concept introduction:

Electronegativity is the tendency of an atom to attract the shared electrons in the bond towards itself. The more electronegative atom will more attract the bonding electrons towards itself than the less electronegative atom. Therefore the electrons will spend more time with the more electronegative atom than an electropositive atom. The electronegative atom will acquire the partial negative charge and the electropositive atom will acquire a partial positive charge.

Here, B is the electronegative atom and A is the electropositive atom.

Bond polarity and ionic character can be estimated by $\Delta \text{EN}$. $\Delta \text{EN}$ is the electronegativity difference between the atoms bonded to each other. The formula to calculate $\Delta \text{EN}$ in AB bond is as follows:

$\Delta \text{EN}=\left(\text{electronegativity of B}\right)-\left(\text{electronegativity of A}\right)$

Here, B is the electronegative atom and A is the electropositive atom.

Answer to Problem 9.67P

The decreasing order of bond polarity and ionic character is as follows:

Explanation of Solution

The formula to calculate $\Delta \text{EN}$ in $\text{P}-\text{Cl}$ bond of ${\text{PCl}}_3$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{P}-\text{Cl}}=\left(\text{electronegativity of Cl}\right)-\left(\text{electronegativity of P}\right)$ (1)

Substitute $2.1$ for electronegativity of $\text{P}$ and $3.0$ for electronegativity of $\text{Cl}$ in the equation (1).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{P}-\text{Cl}}=\left(3.0\right)-\left(2.1\right)\\ =0.9\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{P}-\text{Br}$ bond of ${\text{PBr}}_3$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{P}-\text{Br}}=\left(\text{electronegativity of Br}\right)-\left(\text{electronegativity of P}\right)$ (2)

Substitute $2.8$ for electronegativity of $\text{Br}$ and $2.1$ for electronegativity of $\text{P}$ in the equation (2).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{P}-\text{Br}}=\left(2.8\right)-\left(2.1\right)\\ =0.7\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{P}-\text{F}$ bond of ${\text{PF}}_3$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{P}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of P}\right)$ (3)

Substitute $4.0$ for electronegativity of $\text{F}$ and $2.1$ for electronegativity of $\text{P}$ in the equation (3).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{P}-\text{F}}=\left(4.0\right)-\left(2.1\right)\\ =1.9\end{array}$

Bond polarity and ionic character are directly related to the electronegativity difference. The decreasing order of bond polarity and ionic character is as follows:

Conclusion

$\Delta \text{EN}$ is the electronegativity difference between the atoms bonded to each other. Bond polarity and ionic character are directly related to the electronegativity difference.

(b)

Interpretation Introduction

Interpretation:

The members of the following set are to be ranked in the decreasing order of ionic character. Also, the bond polarity with partial charges is to be indicated.

${\text{BF}}_3,{\text{NF}}_3,{\text{CF}}_3$

Concept introduction:

Electronegativity is the tendency of an atom to attract the shared electrons in the bond towards itself. The more electronegative atom will more attract the bonding electrons towards itself than the less electronegative atom. Therefore the electrons will spend more time with the more electronegative atom than an electropositive atom. The electronegative atom will acquire the partial negative charge and the electropositive atom will acquire a partial positive charge.

Here, B is the electronegative atom and A is the electropositive atom.

Bond polarity and ionic character can be estimated by $\Delta \text{EN}$. $\Delta \text{EN}$ is the electronegativity difference between the atoms bonded to each other. The formula to calculate $\Delta \text{EN}$ in AB bond is as follows:

$\Delta \text{EN}=\left(\text{electronegativity of B}\right)-\left(\text{electronegativity of A}\right)$

Here, B is the electronegative atom and A is the electropositive atom.

Answer to Problem 9.67P

The decreasing order of bond polarity and ionic character is as follows:

Explanation of Solution

The formula to calculate $\Delta \text{EN}$ in $\text{B}-\text{F}$ bond of ${\text{BF}}_3$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{B}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of B}\right)$ (4)

Substitute $2.0$ for electronegativity of $\text{B}$ and $4.0$ for electronegativity of $\text{F}$ in the equation (4).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{B}-\text{F}}=\left(4.0\right)-\left(2.0\right)\\ =2.0\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{N}-\text{F}$ bond of ${\text{NF}}_3$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{N}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of N}\right)$ (5)

Substitute $4.0$ for electronegativity of $\text{F}$ and $3.0$ for electronegativity of $\text{N}$ in the equation (5).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{N}-\text{F}}=\left(4.0\right)-\left(3.0\right)\\ =1.0\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{C}-\text{F}$ bond of ${\text{CF}}_3$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{C}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of C}\right)$ (6)

Substitute $4.0$ for electronegativity of $\text{F}$ and $2.1$ for electronegativity of $\text{P}$ in the equation (6).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{C}-\text{F}}=\left(4.0\right)-\left(2.5\right)\\ =1.5\end{array}$

Bond polarity and ionic character are directly related to the electronegativity difference. The decreasing order of bond polarity and ionic character is as follows:

Conclusion

$\Delta \text{EN}$ is the electronegativity difference between the atoms bonded to each other. Bond polarity and ionic character are directly related to the electronegativity difference.

(c)

Interpretation Introduction

Interpretation:

The members of the following set are to be ranked in the decreasing order of ionic character. Also, the bond polarity with partial charges is to be indicated.

${\text{SeF}}_4,{\text{TeF}}_4,{\text{BrF}}_3$

Concept introduction:

Electronegativity is the tendency of an atom to attract the shared electrons in the bond towards itself. The more electronegative atom will more attract the bonding electrons towards itself than the less electronegative atom. Therefore the electrons will spend more time with the more electronegative atom than an electropositive atom. The electronegative atom will acquire the partial negative charge and the electropositive atom will acquire a partial positive charge.

Here, B is the electronegative atom and A is the electropositive atom.

Bond polarity and ionic character can be estimated by $\Delta \text{EN}$. $\Delta \text{EN}$ is the electronegativity difference between the atoms bonded to each other. The formula to calculate $\Delta \text{EN}$ in AB bond is as follows:

$\Delta \text{EN}=\left(\text{electronegativity of B}\right)-\left(\text{electronegativity of A}\right)$

Here, B is the electronegative atom and A is the electropositive atom.

Answer to Problem 9.67P

The decreasing order of bond polarity and ionic character is as follows:

Explanation of Solution

The formula to calculate $\Delta \text{EN}$ in $\text{Se}-\text{F}$ bond of ${\text{SeF}}_4$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{Se}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of Se}\right)$ (7)

Substitute $2.4$ for electronegativity of $\text{Se}$ and $4.0$ for electronegativity of $\text{F}$ in the equation (7).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{Se}-\text{F}}=\left(4.0\right)-\left(2.4\right)\\ =1.6\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{Te}-\text{F}$ bond of ${\text{TeF}}_4$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{Te}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of Te}\right)$ (8)

Substitute $4.0$ for electronegativity of $\text{F}$ and $2.1$ for electronegativity of $\text{Te}$ in the equation (8).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{Te}-\text{F}}=\left(4.0\right)-\left(2.1\right)\\ =1.9\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{Br}-\text{F}$ bond of ${\text{BrF}}_3$ is as follows:

${\left(\Delta \text{EN}\right)}_{\text{Br}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of Br}\right)$ (9)

Substitute $4.0$ for electronegativity of $\text{F}$ and $2.8$ for electronegativity of $\text{Br}$ in the equation (9).

$\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{Br}-\text{F}}=\left(4.0\right)-\left(2.8\right)\\ =1.2\end{array}$

Bond polarity and ionic character are directly related to the electronegativity difference. The decreasing order of bond polarity and ionic character is as follows:

Conclusion

$\Delta \text{EN}$ is the electronegativity difference between the atoms bonded to each other. Bond polarity and ionic character are directly related to the electronegativity difference.