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Place the species \(\mathrm{B}_{2}^{+}, \mathrm{B}_{2},\) and \(\mathrm{B}_{2}^{-}\) in order of increasing bond length and increasing bond energy.

Short Answer

Expert verified
The orders for bond length and bond energy are as follows: Bond Length: B鈧 > B鈧傗伜 > B鈧傗伝 Bond Energy: B鈧傗伝 > B鈧傗伜 > B鈧

Step by step solution

01

Recall the molecular orbital diagram for dinitrogen (N鈧)

We can use the molecular orbital diagram of dinitrogen (N鈧) as a guide since boron (B) is in the same period as nitrogen (N) in the periodic table. The molecular orbital diagram for N鈧 can be easily found in textbooks or online resources. The molecular orbitals in order from lowest to highest energy are: 蟽1s, 蟽*1s, 蟽2s, 蟽*2s, 蟺2p, 蟽2p, 蟺*2p, and 蟽*2p.
02

Write electron configurations for B鈧傗伜, B鈧, and B鈧傗伝

Boron has an atomic number of 5, so it has 5 electrons. B鈧 will have a total of 10 electrons, B鈧傗伜 will have 9 electrons, and B鈧傗伝 will have 11 electrons. We can fill the molecular orbitals according to the Aufbau principle: - B鈧傗伜: 蟽1s虏, 蟽*1s虏, 蟽2s虏, 蟽*2s虏, 蟺2p鹿 - B鈧: 蟽1s虏, 蟽*1s虏, 蟽2s虏, 蟽*2s虏, 蟺2p虏 - B鈧傗伝: 蟽1s虏, 蟽*1s虏, 蟽2s虏, 蟽*2s虏, 蟺2p鲁
03

Calculate the bond order for each species

Bond order is calculated as (number of bonding electrons - number of antibonding electrons) / 2. For B鈧傗伜: \(Bond \,order = \frac{9 - 4}{2} = 2.5\) For B鈧: \(Bond \,order = \frac{8 - 4}{2} = 2\) For B鈧傗伝: \(Bond \,order = \frac{11 - 4}{2} = 3.5\)
04

Determine the bond length order

The species with the lowest bond order will have the longest bond length, and the species with the highest bond order will have the shortest bond length. The bond length order is: B鈧 > B鈧傗伜 > B鈧傗伝
05

Determine the bond energy order

The species with the highest bond order will have the highest bond energy, as it is the most stable and requires more energy to break the bond. The bond energy order is: B鈧傗伝 > B鈧傗伜 > B鈧 To summarize, we have the following orders for bond length and bond energy: Bond Length: B鈧 > B鈧傗伜 > B鈧傗伝 Bond Energy: B鈧傗伝 > B鈧傗伜 > B鈧

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Key Concepts

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

Bond Order Calculation
Understanding bond order is crucial as it reveals the strength and stability of a molecule's bonds. Bond order is calculated by subtracting the number of electrons in antibonding molecular orbitals from the number of electrons in bonding molecular orbitals, and then dividing the result by two. Mathematically, it's represented as:
\( Bond \text{ } Order = \frac{{(Number \text{ } of \text{ } bonding \text{ } electrons ) - (Number \text{ } of \text{ } antibonding \text{ } electrons )}}{2} \)
Higher bond orders indicate stronger and shorter bonds, since more electrons are engaged in bond formation. Conversely, a low bond order suggests a weaker and longer bond. In our case with the boron molecules (B鈧傗伜, B鈧, B鈧傗伝), the bond order calculation allows us to determine that B鈧傗伝, with the highest bond order of 3.5, has the strongest and shortest bond, while B鈧, with a bond order of 2, has the weakest and longest bond.
Molecular Orbital Diagram
A molecular orbital diagram visually represents the bonding interaction between atoms in a molecule. It shows the relative energy levels and the occupancy of electrons in various molecular orbitals. When constructing such a diagram, key rules need to be followed including the Pauli exclusion principle, the Aufbau principle (lowest-energy orbitals fill first), and Hund's rule (the most stable arrangement has the maximum number of unpaired electrons).

Molecular orbital diagrams are particularly useful in predicting magnetic properties and the bond order of a molecule. For example, in our reference molecule, B鈧, two electrons occupy each of the 蟽1s, 蟽*1s, 蟽2s, and 蟽*2s orbitals, while the 蟺2p orbitals hold the remaining electrons. This electron configuration impacts the overall stability and properties of the molecule.

Electron Configuration
The electron configuration is the distribution of electrons of an atom or molecule in atomic or molecular orbitals. For molecular systems, the electron configuration is essential in understanding the molecule's chemical behavior. Each electron is placed according to the energy levels of the orbitals, from the lowest to the highest energy. In the molecular orbital theory, we fill the orbitals following the energy layout presented in the molecular orbital diagram.

In our exercise, the species B鈧傗伜, B鈧, and B鈧傗伝 have different numbers of electrons. As a result, their electron configurations are distinct, which affects their respective bond orders. Since B鈧傗伜 has one less electron than B鈧, and B鈧傗伝 has one more, their configurations and thus their chemical properties vary.

Bond Length and Energy Relationship
The relationship between bond length and bond energy is an inverse one. Generally, the shorter the bond length, the greater the bond energy will be. This is due to the stronger electrostatic attraction between atoms when they are closer together, which makes the bond more stable and requires more energy to break.
In the context of the B鈧傗伜, B鈧, and B鈧傗伝 molecules, we utilized the bond order as an indication of this relationship. The species B鈧傗伜, with a bond order of 2.5, represents an intermediate bond length and energy, while B鈧, with the longest bond and the least bond energy due to the lowest bond order of 2. These fundamental concepts allow us to predict and compare the stability of different molecules based on their bond orders, lengths, and bond energies.

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Most popular questions from this chapter

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and \(\mathrm{NO}\) than the hybrid orbital model?

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) can be produced from the reaction of calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) with water. Use both the localized electron and molecular orbital models to describe the bonding in the acetylide anion \(\left(\mathrm{C}_{2}^{2-}\right)\).

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?

Use the localized electron model to describe the bonding in \(\mathrm{H}_{2} \mathrm{CO}\) (carbon is the central atom).

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