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

First, let's determine the total number of electrons in each species: - \(\mathrm{B_{2}^{+}}\): Boron has 5 electrons, but this species has a positive charge, so it has lost an electron. Therefore, it has (2 × 5) - 1 = 9 electrons total. - \(\mathrm{B}_{2}\): This is a neutral molecule with each boron atom having 5 electrons, so it has 2 × 5 = 10 electrons total. - \(\mathrm{B}_{2}^{-}\): This species has a negative charge, indicating that it has gained an electron. So, it has (2 × 5) + 1 = 11 electrons total.

Now, let's fill the molecular orbitals with electrons using the molecular orbital diagram for B₂. Use the Aufbau principle (fill lower energy orbitals first), the Pauli Exclusion Principle (maximum of 2 electrons per orbital), and Hund's rule (maximize the number of parallel spins). The molecular orbitals for B₂ are in the following order (starting from lowest energy): 1. σ(2s) 2. σ*(2s) 3. [σ(2p) and two π(2p) orbitals degenerate in energy, i.e., at the same energy level] 4. [σ(2p)* and two π*(2p) orbitals degenerate in energy] Energies of molecular orbitals decrease with increasing bond order. Bond order determines bond strength and length. Higher bond order correlates with shorter bond length and greater bond strength. Now, fill the molecular orbitals for each molecular species based on their total number of electrons: - \(\mathrm{B_{2}^{+}}\) (9 electrons): σ(2s)², σ*(2s)², σ(2p)², π(2p)³ - \(\mathrm{B}_{2}\) (10 electrons): σ(2s)², σ*(2s)², σ(2p)², π(2p)⁴ - \(\mathrm{B}_{2}^{-}\) (11 electrons): σ(2s)², σ*(2s)², σ(2p)², π(2p)⁴, σ(2p)*¹

Next, we'll calculate the bond order for each species by using the formula: Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals)/2 - \(\mathrm{B_{2}^{+}}\): Bond order = (6-3)/2 = 1.5 - \(\mathrm{B}_{2}\): Bond order = (6-4)/2 = 1 - \(\mathrm{B}_{2}^{-}\): Bond order = (6-5)/2 = 0.5

We know that higher bond order means greater bond energy (strength) and shorter bond length. Based on the bond order values we calculated, we can arrange the species as follows: - Bond length (increasing order): \(\mathrm{B_{2}^{+}}\) < \(\mathrm{B}_{2}\) < \(\mathrm{B}_{2}^{-}\) - Bond energy (increasing order): \(\mathrm{B}_{2}^{-}\) < \(\mathrm{B}_{2}\) < \(\mathrm{B_{2}^{+}}\) So, the species \(\mathrm{B}_{2}+, \mathrm{B}_{2}\), and \(\mathrm{B}_{2}^{-}\) can be ordered by increasing bond length as \(\mathrm{B_{2}^{+}}\), \(\mathrm{B}_{2}\), \(\mathrm{B}_{2}^{-}\) and by increasing bond energy as \(\mathrm{B}_{2}^{-}\), \(\mathrm{B}_{2}\), \(\mathrm{B_{2}^{+}}\).