Question

Place the species \(\mathrm{B}_{2}^{+}, \mathrm{B}_{2},\) and \(\mathrm{B}_{2}^{-}\) in order of increasing bond length and increasing bond energy.

Short answer

The order of increasing bond length for the species \(\mathrm{B}_{2}^{+}, \mathrm{B}_{2},\) and \(\mathrm{B}_{2}^{-}\) is: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\). The order of increasing bond energy for these species is also: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\).

Step-by-step solution

Calculate the number of valence electrons for each species

Count the valence electrons of each species to determine their electron configurations. - \(\mathrm{B}_{2}^{+}\): Each boron atom contributes 3 valence electrons, and there is a charge of +1, so we have a total of 3 + 3 - 1 = 5 valence electrons. - \(\mathrm{B}_{2}\): Each boron atom contributes 3 valence electrons, so we have a total of 3 + 3 = 6 valence electrons. - \(\mathrm{B}_{2}^{-}\): Each boron atom contributes 3 valence electrons, and there is a charge of -1, so we have a total of 3 + 3 + 1 = 7 valence electrons.

Calculate bond order for each species

Utilize the formula for bond order which is: \[Bond \, Order = \frac{1}{2} \times (number \, of \, bonding \, electrons - number \, of \, antibonding \, electrons)\] - For \(\mathrm{B}_2^+\), there are 5 valence electrons. According to molecular orbital theory, the electron configuration is \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^1\). Therefore, the bond order is \(\frac{1}{2}(3-2)=0.5\). - For \(\mathrm{B}_2\), there are 6 valence electrons. The electron configuration is \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\). Therefore, the bond order is \(\frac{1}{2}(4-2)=1\). - For \(\mathrm{B}_2^-\), there are 7 valence electrons. The electron configuration is \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma_{2p_x}^1\). Therefore, the bond order is \(\frac{1}{2}(5-2)=1.5\).

Arrange species in order of bond length

As mentioned, higher bond order corresponds to shorter bond length. Arrange the species based on their bond orders found in step 2: - \(\mathrm{B}_{2}^{+}\): Bond order = 0.5 (Longest bond length) - \(\mathrm{B}_{2}\): Bond order = 1 (Medium bond length) - \(\mathrm{B}_{2}^{-}\): Bond order = 1.5 (Shortest bond length) So the order of increasing bond length is: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\).

Arrange species in order of bond energy

Higher bond order also corresponds to greater bond energy. Using the bond orders calculated in step 2, arrange the species based on their bond energies: - \(\mathrm{B}_{2}^{+}\): Bond order = 0.5 (Lowest bond energy) - \(\mathrm{B}_{2}\): Bond order = 1 (Medium bond energy) - \(\mathrm{B}_{2}^{-}\): Bond order = 1.5 (Highest bond energy) So the order of increasing bond energy is: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\).