Problem 65 - Chemistry: The Central Science

Question

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}_{2}\) or \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\) be expected to exist? Explain.

Step-by-Step Solution

Answer

(a) Bond order, bond length, and bond energy are related as follows: As bond order increases, bond length decreases, and bond energy increases. (b) According to molecular orbital theory, \(\mathrm{Be}_{2}\) is not expected to exist due to its bond order of 0, while \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\) is expected to exist but with a weak bond due to its bond order of 0.5.

Step 1: Relationship among bond order, bond length, and bond energy

Bond order is the measure of the number of bonds shared between two atoms and is calculated as the difference between the number of bonding electrons and the number of antibonding electrons, divided by 2. Bond length represents the distance between the nuclei of two bonded atoms, and bond energy refers to the energy required to break a bond. 1. As bond order increases, it means more electrons are being shared in the bond, resulting in a stronger and more stable bond. 2. A higher bond order leads to shorter bond lengths since the bonding electrons hold the nuclei closer together. 3. A higher bond order also correlates with higher bond energy because a stronger bond requires more energy to break. In summary: - Bond order ↑, bond length ↓ - Bond order ↑, bond energy ↑

Step 2: Molecular Orbital Theory for \(\mathrm{Be}_{2}\) and \(\mathrm{Be}_{2}{ }^{+}\)

To predict if \(\mathrm{Be}_{2}\) or \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\) are expected to exist, we will use molecular orbital (MO) theory to calculate their bond orders. Step 1: Determine the electron configuration of Be Beryllium has 4 electrons, so its electron configuration is \(1s^{2}2s^{2}\). Step 2: Construct the MO diagram for \(\mathrm{Be}_{2}\) In \(\mathrm{Be}_{2}\), there are 8 electrons. We fill the molecular orbitals in this order: \(\sigma_{1s},\sigma_{1s}^*,\sigma_{2s}, \sigma_{2s}^*, \pi_{2p}\), and \(\pi_{2p}^*\), with each of the first four orbitals accommodating 2 electrons. So, for \(\mathrm{Be}_{2}\), the MO diagram has 4 bonding and 4 antibonding electrons. Step 3: Calculate bond order for \(\mathrm{Be}_{2}\) Bond order: \(\frac{1}{2} \times \text{(bonding electrons - antibonding electrons)}\) For \(\mathrm{Be}_{2}\): \(\frac{1}{2}(4-4)=0\) Since the bond order for \(\mathrm{Be}_{2}\) is 0, this molecule is not expected to exist. Step 4: Calculate bond order for \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\) In \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\), there are 7 electrons. We remove one electron from the highest energy MO, which is \(\sigma_{2s}^*\). For \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\): \(\frac{1}{2}(4-3)=0.5\) Since the bond order for \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\) is 0.5, this ion is expected to exist, although it will be weakly bonded. In conclusion, according to molecular orbital theory, \(\mathrm{Be}_{2}\) is not expected to exist due to its bond order of 0, while \(\mathrm{Be}_{2}{\underline{\phantom{xx}}}^{+}\) is expected to exist but with a weak bond due to its bond order of 0.5.