Problem 125 - Chemistry An Atoms First Approach

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, \(\mathbf{B}_{2}\) 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

Answer

The bond order in Li₂ is 1 and in B₂ is 0. To make B₂ isoelectronic with Li₂, 4 electrons must be removed from B₂. To ionize 1.5 kg of B₂ to the desired isoelectronic species, 1.86 x 10⁶ kJ of energy is needed.

Step 1: 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.

Step 2: 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 $$\text{Bond Order} = \frac{1}{2} [4-4] = 0$$

Step 3: 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.

Step 4: 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.

Key Concepts

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom. These are the electrons involved in forming chemical bonds. Understanding valence electrons is crucial as they determine how an atom will interact with others.

In molecular compounds like \( \text{Li}_2 \) and \( \text{B}_2 \), we calculate the total number of valence electrons to predict molecule behavior. For lithium ( ext{Li}), which is in Group 1 of the periodic table, each atom has 1 valence electron. Therefore, two lithium atoms contribute two valence electrons in a \( \text{Li}_2 \) molecule.

• Li: 1 valence electron per atom, so, \( \text{Li}_2 \) has 2 electrons keen to bond.

For boron ( ext{B}), located in Group 13, each atom has 3 valence electrons. Thus, in a \( \text{B}_2 \) molecule, two boron atoms together have 6 valence electrons.

• B: 3 valence electrons per atom, so, \( \text{B}_2 \) has 6 electrons ready for bonding.

This basic understanding helps us proceed to calculate bond order and examine isoelectronic properties.

Isoelectronic Species

Isoelectronic species are molecules or ions that have the same number of electrons. Being isoelectronic means that despite the potential difference in elements, the total electron count is identical.

For instance, the goal can be to make \( \text{B}_2 \) isoelectronic with \( \text{Li}_2 \). \( \text{Li}_2 \) has 2 valence electrons while \( \text{B}_2 \) has 6. To make these two isoelectronic, we must remove 4 electrons from \( \text{B}_2 \). The maintenance of an equal number of electrons can lead to similar chemical properties.

• This is achieved by removing electrons from \( \text{B}_2 \).
• Removing 4 electrons from \( \text{B}_2 \) makes it isoelectronic with \( \text{Li}_2 \).

The process involves removing sufficient electrons (in this case, 4 electrons) such that the total electron count is equal across both species. This concept is fundamental in chemistry when examining similar chemical behaviors across different species.

Molecular Orbitals

Molecular orbitals are critical structures that describe the behavior of electrons in a molecule. They result from the combination of atomic orbitals when atoms interact to form a bond.

In the molecular orbital theory, electrons fill molecular orbitals following specific rules. For \( \text{Li}_2 \), the electrons occupy both the bonding \( \sigma_{1s} \) and antibonding \( \sigma^*_{1s} \) molecular orbitals. This configuration provides a bond order of 1, signifying a stable single bond:

• The bond order calculation is \[\text{Bond Order} = \frac{1}{2} [2-0] = 1 \]

For \( \text{B}_2 \), the situation is slightly more complex, with more electrons distributed among \( \sigma_{1s}, \sigma_{1s}^*, \sigma_{2s}, \sigma_{2s}^* \), and the \( \pi_{2p} \) bonding orbitals. The distribution leads to a different bond order of zero:

• The lack of net bonding in \( \text{B}_2 \) means \[\text{Bond Order} = \frac{1}{2} [4-4] = 0 \]

Understanding the filling and energy levels of molecular orbitals is vital for predicting molecule stability and reactions.