Chapter 11, Problem 11.37P

(a)

Interpretation Introduction

Interpretation:

The order of decreasing bond energy of ${\text{B}}_2{}^{+}$, ${\text{B}}_2$ and ${\text{B}}_2{}^{-}$ on the basis of the molecular orbital diagram is to be ranked.

Concept introduction:

A molecular orbital diagram is a tool that is used to describe the chemical bonding formed between different molecules. It is used to predict the bond strength and the electronic transitions that a molecule can undergo.

Answer to Problem 11.37P

The order of decreasing bond energy of ${\text{B}}_2{}^{+}$, ${\text{B}}_2$ and ${\text{B}}_2{}^{-}$ is ${\text{B}}_2{}^{-}>{\text{B}}_2>{\text{B}}_2{}^{+}$.

Explanation of Solution

The molecular orbital diagram of ${\text{B}}_2{}^{-}$ is as follows:

The formula to calculate the bond order is as follows:

$\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\text{bonding}\text{orbitals}-\\ \text{number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{orbitals}\end{array}\right)$ (1)

Substitute 7 for the number of electrons in bonding orbitals and 4 for the number of electrons in antibonding orbitals in equation(1).

$\begin{array}{c}\text{Bond}\text{order}\text{of}{\text{B}}_2{}^{-}=\frac{1}{2}\left(7-4\right)\\ =1.5\end{array}$

The molecular orbital diagram of ${\text{B}}_2$ is as follows:

Substitute 6 for the number of electrons in bonding orbitals and 4 for the number of electrons in antibonding orbitals in equation(1).

$\begin{array}{c}\text{Bond}\text{order}\text{of}{\text{B}}_2=\frac{1}{2}\left(6-4\right)\\ =1\end{array}$

The molecular orbital diagram of ${\text{B}}_2{}^{+}$ is as follows:

Substitute 5 for the number of electrons in bonding orbitals and 4 for the number of electrons in antibonding orbitals in equation(1).

$\begin{array}{c}\text{Bond}\text{order}\text{of}{\text{B}}_2{}^{+}=\frac{1}{2}\left(5-4\right)\\ =0.5\end{array}$

Bond energy is directly proportional to the bond order. The highest bond order of ${\text{B}}_2{}^{-}$ indicates that its bond is very strong and therefore bond energy of ${\text{B}}_2{}^{-}$ will be the highest, followed by ${\text{B}}_2$ and ${\text{B}}_2{}^{+}$ has the least bond energy.

So the order of decreasing bond energy is ${\text{B}}_2{}^{-}>{\text{B}}_2>{\text{B}}_2{}^{+}$.

Conclusion

The order of decreasing bond energy of ${\text{B}}_2{}^{+}$, ${\text{B}}_2$ and ${\text{B}}_2{}^{-}$ is ${\text{B}}_2{}^{-}>{\text{B}}_2>{\text{B}}_2{}^{+}$.

(b)

Interpretation Introduction

Interpretation:

The order of decreasing bond length of ${\text{B}}_2{}^{+}$, ${\text{B}}_2$ and ${\text{B}}_2{}^{-}$ on the basis of the molecular orbital diagram is to be ranked.

Concept introduction:

A molecular orbital diagram is a tool that is used to describe the chemical bonding formed between different molecules. It is used to predict the bond strength and the electronic transitions that a molecule can undergo.

Answer to Problem 11.37P

The order of the decreasing bond length is ${\text{B}}_2{}^{+}>{\text{B}}_2>{\text{B}}_2{}^{-}$.

Explanation of Solution

The molecular orbital diagram of ${\text{B}}_2{}^{-}$ is as follows:

The formula to calculate the bond order is as follows:

$\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\text{bonding}\text{orbitals}-\\ \text{number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{orbitals}\end{array}\right)$ (1)

Substitute 7 for the number of electrons in bonding orbitals and 4 for the number of electrons in antibonding orbitals in equation(1).

$\begin{array}{c}\text{Bond}\text{order}\text{of}{\text{B}}_2{}^{-}=\frac{1}{2}\left(7-4\right)\\ =1.5\end{array}$

The molecular orbital diagram of ${\text{B}}_2$ is as follows:

Substitute 6 for the number of electrons in bonding orbitals and 4 for the number of electrons in antibonding orbitals in equation(1).

$\begin{array}{c}\text{Bond}\text{order}\text{of}{\text{B}}_2=\frac{1}{2}\left(6-4\right)\\ =1\end{array}$

The molecular orbital diagram of ${\text{B}}_2{}^{+}$ is as follows:

Substitute 5 for the number of electrons in bonding orbitals and 4 for the number of electrons in antibonding orbitals in equation(1).

$\begin{array}{c}\text{Bond}\text{order}\text{of}{\text{B}}_2{}^{+}=\frac{1}{2}\left(5-4\right)\\ =0.5\end{array}$

Bond length decreases with the increase in bond energy. The highest bond order of ${\text{B}}_2{}^{-}$ indicates that its bond is very strong and therefore the bond length of ${\text{B}}_2{}^{-}$ will be the smallest, followed by ${\text{B}}_2$ and ${\text{B}}_2{}^{+}$ has the largest bond length.

So the order of the decreasing bond length is ${\text{B}}_2{}^{+}>{\text{B}}_2>{\text{B}}_2{}^{-}$.

Conclusion

The order of the decreasing bond length is ${\text{B}}_2{}^{+}>{\text{B}}_2>{\text{B}}_2{}^{-}$.