Question

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?

Short answer

In summary, the bond orders are 1 for \(Li_{2}\) and 2 for \(B_{2}\). To make \(B_{2}\) isoelectronic with \(Li_{2}\), 4 electrons need to be removed. Finally, the amount of energy needed to ionize 1.5 kg of \(B_{2}\) to the desired isoelectronic species is approximately 448,106 kJ.

Step-by-step solution

Calculate the bond order of Li₂ and B₂

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.

Determine the number of electrons to be removed from 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.

Calculate the energy needed to ionize 1.5 kg of B₂

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.

Key concepts

Bond Order in Molecular Orbital Theory

Bond order is a fundamental concept in Molecular Orbital Theory that helps determine the stability of a molecule. The formula to calculate bond order is: \[ \text{Bond order} = \frac{(\text{Number of electrons in bonding molecular orbitals} - \text{Number of electrons in antibonding molecular orbitals})}{2} \] Higher bond orders indicate stronger bonds, as they imply more electrons are contributing to bond formation.
For example, in the case of \(\text{Li}_2\), the 1s electrons fill the bonding and antibonding orbitals, leaving only the 2s electrons to form a bond.
This results in a bond order of 1. For \(\text{B}_2\), with a total of 10 electrons, both the 2s and 2p electrons contribute to bonding, resulting in a bond order of 2.

• Bonds with higher bond orders are generally more stable and have shorter bond lengths.
• A bond order of 0 would suggest no bond, meaning the molecule doesn't exist in a stable form.

Isoelectronic Species

Isoelectronic species are atoms, molecules, or ions that have the same number of electrons.
This concept is crucial for predicting and explaining the stability and reactivity of chemical species.
To understand this, consider the example in the problem where \(\text{B}_2\) needs to become isoelectronic with \(\text{Li}_2\).
\(\text{Li}_2\) has 6 electrons, thus \(\text{B}_2\), originally with 10 electrons, must lose 4 electrons to match \(\text{Li}_2\). Once \(\text{B}_2\) loses these electrons, it becomes \(\text{B}_2^{4+}\), now having the same electronic structure as \(\text{Li}_2\).

• Isoelectronic species often exhibit similar chemical and physical properties.
• They may have different nuclear compositions but the same electron configuration.

Ionization Energy and Its Calculation

Ionization energy refers to the energy required to remove electrons from a molecule or atom.
It is an essential factor in determining the reactivity and stability of elements and compounds.
In the scenario given, we wish to ionize \(\text{B}_2\) to remove 4 electrons, requiring knowledge of \(\Delta H\), the enthalpy change for this reaction.
Given \(\Delta H = 6455\ \text{kJ/mol}\), we start by computing the moles in 1.5 kg of \(\text{B}_2\):
First, convert the mass to grams and then determine the moles using its molar mass (21.62 g/mol).
Calculating the energy involves multiplying the number of moles by \(\Delta H\):
\[ 69.38\ \text{mol} \times 6455\ \text{kJ/mol} \approx 448,106\ \text{kJ} \] This tells us the total energy needed to achieve the desired isoelectronic species. Understanding ionization energy helps predict how easily an atom or molecule can lose electrons.

• Higher ionization energies mean more energy is required, indicating a stable electron configuration.
• It also generally increases across a periodic table period due to increased nuclear charge.