Question

Place the species ${\mathit{B}}_{\mathbf{2}}^{\mathbf{+}}$,${\mathit{B}}_{\mathbf{2}}$, and ${\mathit{B}}_{\mathbf{2}}^{\mathbf{-}}$in order of increasing bond length and increasing bond energy.

Short answer

According to molecular orbital theory, two atomic orbitals of different or the same atoms combine to form a molecular orbital. The configuration of molecular orbital consists of bonding and nonbonding orbitals, the difference between them divided by 2 will give the bond order which have direct relation with bond energy and inverse relation with bond length.

Step-by-step solution

Molecular orbital configuration

As one boron atom has 5 valence electrons, so the molecule of boron will have a total of 10 electrons. Therefore, the molecular orbital configuration for $B_2$will be:

$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2$

In $B_2^{+}$, there are total of 9 electrons as it consists of positive charge so the molecular orbital configuration will be

$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^1$

In $B_2^{-}$, there are a total of 11 electrons as it consists of negative charge so the molecular orbital configuration will be

$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^1$

Calculation of bond order

Bond order for any molecule can be calculated by dividing the difference of bonding and antibonding electrons by 2. Thus, $B_2^{+}$, $B_2$, and $B_2^{-}$the bond order will be:

$$\begin{gathered} Bondorderof=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{6-4}{2} \\ =1 \end{gathered}$$

$$\begin{gathered} Bondorderof=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{5-4}{2} \\ =0.5 \end{gathered}$$

$$\begin{gathered} Bondorderof=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{7-4}{2} \\ =1.5 \end{gathered}$$

Step 3: Order of bond length and bond energy

Now, the relation between bond order, bond length and bond energy is given as:

$\text{Bond order}\propto \text{Bond energy}\propto \frac{1}{\text{Bond length}}$

So, molecule with higher bond order will have lower bond length and higher bond energy. Therefore, increasing order of bond length will be $B_2^{-}⩽{\text{B}}_2⩽B_2^{+}$and increasing order of bond energy will be$B_2^{+}⩽{\text{B}}_2⩽B_2^{-}.$