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, \(\mathbf{B}_{2}\) 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 b) as follows: $$\mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta E=6455 \mathrm{kJ} / \mathrm{mol}$$ How much energy is needed to ionize \(1.5 \mathrm{kg} \mathrm{B}_{2}\) to the desired isoelectronic species?

Step by step solution

01

Find the valence electrons

For each molecule, we need to find the total number of valence electrons. For Li₂: Each Li atom has 1 valence electron, so 2 Li atoms contribute 2 valence electrons in total. For B₂: Each B atom has 3 valence electrons, so 2 B atoms contribute 6 valence electrons in total.

02

Determine the bond order

The bond order can be found using the formula: $$\text{Bond Order} = \frac{1}{2} [\text{electrons in bonding MOs} - \text{electrons in antibonding MOs}]$$ For Li₂: The valence electrons occupy the σ₁s bonding and σ₁s* antibonding molecular orbitals. So, the bond order is $$\text{Bond Order} = \frac{1}{2} [2-0] = 1$$ For B₂: The valence electrons occupy the σ₁s, σ₁s*, σ₂s, σ₂s* as two electrons each, and there are two electrons shared by the π₂p bonding orbitals. Thus, the bond order is $$\text{Bond Order} = \frac{1}{2} [4-4] = 0$$

03

Determine the difference in electrons

To make B₂ isoelectronic with Li₂, we need to find the difference in electrons between these two molecules. The number of valence electrons in Li₂ is 2 (from step 1). The number of valence electrons in B₂ is 6 (from step 1). So, the number of electrons to be removed from B₂ is \(6 - 2 = 4\) electrons.

04

Calculate the total energy required to ionize the given mass of B₂.

We are given the ionization energy as \(\Delta E = 6455\, kJ/mol\). We need to find the energy required to ionize 1.5 kg of B₂. First, we need to find the number of moles in 1.5 kg of B₂. To do this, we will use the molar mass of B₂, which is approximately 20.8 g/mol (as the molar mass of B is 10.4 g/mol). Now, convert the mass to moles: $$\text{moles}=\frac{1.5 \times 10^3 \, g}{20.8 \, g/ mol} = 72 \, mol$$ Finally, we can calculate the energy required to ionize 1.5 kg of B₂: $$E = \Delta E \times N_\text{moles} \times N_\text{electrons} = 6455 \, kJ/mol \times 72 \, mol \times 4 = 1.86 \times 10^6 \, kJ$$ So, the energy needed to ionize 1.5 kg of B₂ to the desired isoelectronic species is \(1.86 \times 10^6\) kJ.

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