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Chapter 9: Problem 77

(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

Expert verified
(a) Bond order is the number of chemical bonds between a pair of atoms, and as it increases, the bond length decreases, and the bond energy increases. There is an inverse relationship between bond order and bond length, while both bond order and bond energy are directly proportional. (b) Using molecular orbital theory, neither Be鈧� nor Be鈧傗伜 are expected to exist. Be鈧� has a bond order of 0, indicating no bond and making it unstable. Be鈧傗伜 has a weak bond with a bond order of 0.5, not strong enough to overcome the electrostatic repulsion between nuclei, making it non-existent as well.

Step by step solution

01

Part (a): Relationship among Bond Order, Bond Length, and Bond Energy

1. Bond Order: Bond order is the number of chemical bonds between a pair of atoms in a molecule. It is calculated as the difference between the number of bonding electrons and the number of antibonding electrons, divided by 2. A higher bond order implies a stronger bond. 2. Bond Length: Bond length is the distance between the nuclei of two bonded atoms in a molecule. As the bond order increases, the bond becomes stronger, and the two nuclei are pulled closer to each other. Therefore, there is an inverse relationship between bond order and bond length - as bond order increases, bond length decreases. 3. Bond Energy: Bond energy is the energy required to break a chemical bond and form neutral atoms. A higher bond order corresponds to a stronger bond, which means more energy will be required to break it. So, bond energy and bond order are directly proportional. In summary, as bond order increases, the bond length decreases, and the bond energy increases.
02

Part (b): Predicting the Existence of Be鈧� and Be鈧傗伜 using Molecular Orbital Theory

1. Be鈧� Electronic Configuration: Be has an electronic configuration of [He] 2s虏. When two Be atoms come together to form a Be鈧� molecule, the 2s orbitals of both Be atoms will combine to form bonding (蟽) and antibonding (蟽*) molecular orbitals. Thus, the molecular orbital configuration for Be鈧� is (蟽鈧俿)虏(蟽鈧俿*)虏. 2. Be鈧� Bond Order Calculation: The bond order for Be鈧� can be calculated as (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2 = (2 - 2) / 2 = 0 3. Existence of Be鈧�: Since the bond order for Be鈧� is 0, it means that there's essentially no bond, and hence Be鈧� is not stable and not expected to exist. 4. Be鈧傗伜 Electronic Configuration: For Be鈧傗伜, one electron is removed from the antibonding orbital. So, the molecular orbital configuration becomes (蟽鈧俿)虏(蟽鈧俿*)鹿. 5. Be鈧傗伜 Bond Order Calculation: The bond order for Be鈧傗伜 is (2 - 1) / 2 = 0.5 6. Existence of Be鈧傗伜: In this case, the bond order for Be鈧傗伜 is 0.5, meaning that there's a weak bond. However, this weak bond is not enough to overcome the electrostatic repulsion between the two Be nuclei. Therefore, Be鈧傗伜 is also not expected to exist. In conclusion, according to molecular orbital theory, neither Be鈧� nor Be鈧傗伜 are expected to exist due to their bond orders.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bond Order
Bond order refers to the number of chemical bonds between a pair of atoms. It can be defined by subtracting the number of antibonding electrons from the number of bonding electrons, then dividing by two. Imagine it like this: the more pairs of electrons holding two atoms together, the stronger that bond is.

Higher bond orders imply stronger, more stable bonds. Why? Because there are more electron pairs pulling the atoms closer. This is very useful when predicting the stability of a molecule or its chemical properties.

It's like getting a bigger band to keep two beach balls together. A single band might do the trick, but two or three bands will keep them glued much more firmly. As bond order increases, it directly impacts both bond length and bond energy, which we'll cover next.
Bond Length
Bond length is simply the distance between the nuclei of two bonded atoms. A useful analogy is a seesaw game 鈥� higher bond order causes stronger attraction, pulling the nuclei closer for a shorter bond length.

Generally, as the bond order increases, the atoms are held more tightly together, reducing the bond length. This can be seen experimentally: a triple bond, which has a bond order of 3, is shorter than a double bond of the same type of atoms, which itself is shorter than a single bond.

In summary, understanding bond length helps chemists predict how molecules will interact physically with each other, and can be crucial in understanding reaction dynamics and physical properties of substances.
Bond Energy
Bond energy represents the energy required to break a chemical bond and separate atoms in a molecule. It is intimately related to bond order, as higher bond orders generally mean greater bond energy.

Think of it as the effort needed to pull apart glued pieces of paper: the stronger the glue (or the more glue you use), the harder it is to separate. Similarly, higher bond orders result in higher bond energies, making bonds harder to break.

From a practical standpoint, understanding bond energy is vital for any chemical reaction, as it directly affects reaction conditions, such as temperature and catalysts, required to break bonds and form new ones.

In essence, bond energy gives insight into the strength and stability of chemical bonds, crucial for everything from cooking to constructing materials. Understanding how bond energy works helps chemists design and predict reactions efficiently.

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Most popular questions from this chapter

Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\) - that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state? Explain.

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

Consider a molecule with formula \(\mathrm{AX}_{3}\). Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{3}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle increases from \(100^{\circ}\) to \(120^{\circ} ?\)

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher- energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \AA\). (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5,\) is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-}\) ? (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-}\). (e) It is often observed that \(\sigma\) bonds that involve an \(s p\) hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?
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