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(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 \(\mathrm{Be}_{2}{ }^{+}\) be expected to exist? Explain.

Short Answer

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(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 \(\mathrm{Be}_{2}{ }^{+}\) is expected to exist but with a weak bond due to its bond order of 0.5.

Step by step solution

01

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 ↑
02

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

To predict if \(\mathrm{Be}_{2}\) or \(\mathrm{Be}_{2}{ }^{+}\) 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 \(\mathrm{Be}_{2}{ }^{+}\) In \(\mathrm{Be}_{2}{ }^{+}\), there are 7 electrons. We remove one electron from the highest energy MO, which is \(\sigma_{2s}^*\). For \(\mathrm{Be}_{2}{ }^{+}\): \(\frac{1}{2}(4-3)=0.5\) Since the bond order for \(\mathrm{Be}_{2}{ }^{+}\) 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 \(\mathrm{Be}_{2}{ }^{+}\) is expected to exist but with a weak bond due to its bond order of 0.5.

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Most popular questions from this chapter

Sulfur tetrafluoride (SF \(_{4}\) ) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(S\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electrondomain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

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The molecules \(\mathrm{SiF}_{4}, \mathrm{SF}_{4}\), and \(\mathrm{XeF}_{4}\) have molecular formulas of the type \(\mathrm{AF}_{4}\), but the molecules have different molecular geometries. Predict the shape of each molecule, and explain why the shapes differ.

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