As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, B. a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2} ?\) b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part \(\mathrm{b}\) ) as follows: $$ \mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta H=6455 \mathrm{kJ} / \mathrm{mol} $$ How much energy is needed to ionize 1.5 \(\mathrm{kg} \mathrm{B}_{2}\) to the desired isoelectronic species?

To calculate the bond order, we need to use the formula: Bond order = (Number of electrons in bonding molecular orbitals - Number of electrons in antibonding molecular orbitals) / 2 For Li₂: Both lithium atoms have 3 electrons (1s² 2s¹), so Li₂ has 6 electrons in total. The first 4 (two from each lithium atom) are in 1s orbitals (both bonding and antibonding) and the remaining 2 are in the bonding molecular orbital (2sσ). Thus, the bond order for Li₂ is: Bond order = (2 - 0) / 2 = 1 For B₂: Both boron atoms have 5 electrons (1s² 2s² 2p¹), so B₂ has 10 electrons in total. The first 4 are in 1s orbitals (both bonding and antibonding), the next 4 are in 2s orbitals (both bonding and antibonding), and the remaining 2 are in the bonding molecular orbital (2pσ). Thus, the bond order for B₂ is: Bond order = (4 - 0) / 2 = 2 So, the bond orders are 1 for Li₂ and 2 for B₂.

To make B₂ isoelectronic with Li₂, it must have the same number of electrons. Since Li₂ has 6 electrons, we need to remove 4 electrons from B₂ (which has 10 electrons) so that it is isoelectronic with Li₂: B₂ → B₂⁴⁺ + 4 e⁻ The number of electrons to be removed is 4.

We are given that the energy required for the reaction (ΔH) is 6455 kJ/mol. We need to find how many moles are in 1.5 kg of B₂ and then calculate the total energy required for the ionization process. First, find the molar mass of B₂: Molar mass of B₂ = 2 × Atomic weight of B = 2 × 10.81 = 21.62 g/mol Now convert the mass of B₂ from kg to g: 1.5 kg × (1000 g/kg) = 1500 g Now, find the moles of B₂: Moles of B₂ = mass / molar mass = 1500 g / 21.62 g/mol = 69.38 mol Finally, calculate the total energy required for ionization: Energy = moles × ΔH = 69.38 mol × 6455 kJ/mol ≈ 448,106 kJ The amount of energy needed to ionize 1.5 kg of B₂ to the desired isoelectronic species is approximately 448,106 kJ.