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Chapter 9: Problem 106

Place the following molecules and ions in order from smallest to largest bond order: \(\mathrm{H}_{2}^{+}, \mathrm{B}_{2}, \mathrm{N}_{2}^{+}, \mathrm{F}_{2}^{+},\) and \(\mathrm{Ne}_{2}\) .

Short Answer

Expert verified
The order of the molecules and ions from smallest to largest bond order is: \(\mathrm{Ne}_{2} < \mathrm{H}_{2}^{+} < \mathrm{B}_{2} < \mathrm{N}_{2}^{+} = \mathrm{F}_{2}^{+}\).

Step by step solution

01

Molecular Orbital Diagrams for the molecules and ions

Using Molecular Orbital Theory, we need to create molecular orbital diagrams for each of the given molecules and ions. This will be necessary for determining the bond order of each entity. Here, we'll use the following electron configuration order for the molecule orbitals: σ1s, σ*1s, σ2s, σ*2s, π2p_x/π2p_y, π*2p_x/π*2p_y, and σ2p_z. We are interested only in the electron configurations since electron count is needed for bond order calculation. 1. \(\mathrm{H}_{2}^{+}\): This ion has one hydrogen atom with one electron, and one proton. Its electron configuration is (σ1s)^1. 2. \(\mathrm{B}_{2}\): This molecule has two boron atoms with three electrons each, thus having a total of 6 electrons. Its electron configuration is (σ1s)^2(σ*1s)^2(σ2s)^2. 3. \(\mathrm{N}_{2}^{+}\): This ion has two nitrogen atoms with seven electrons each, minus the one removed to form the ion, giving a total of 13 electrons. Its electron configuration is (σ1s)^2(σ*1s)^2(σ2s)^2(σ*2s)^2(π2p_x)^2(π2p_y)^1. 4. \(\mathrm{F}_{2}^{+}\): This ion has two fluorine atoms with nine electrons each, minus the one removed to form the ion, giving a total of 17 electrons. Its electron configuration is (σ1s)^2(σ*1s)^2(σ2s)^2(σ*2s)^2(π2p_x)^2(π2p_y)^2(σ2p_z)^1(π*2p_x)^1. 5. \(\mathrm{Ne}_{2}\): This molecule has two neon atoms with ten electrons each, thus having a total of 20 electrons. Its electron configuration is (σ1s)^2(σ*1s)^2(σ2s)^2(σ*2s)^2(π2p_x)^2(π2p_y)^2(σ2p_z)^2(π*2p_x)^2(π*2p_y)^2.
02

Calculate bond orders

Now, we'll apply the bond order formula for each molecule and ion using their electron configurations: 1. \(\mathrm{H}_{2}^{+}\): \(\frac{1}{2}(1 - 0)\) = 0.5 2. \(\mathrm{B}_{2}\): \(\frac{1}{2}(6 - 4)\) = 1 3. \(\mathrm{N}_{2}^{+}\): \(\frac{1}{2}(10 - 6)\) = 2 4. \(\mathrm{F}_{2}^{+}\): \(\frac{1}{2}(12 - 8)\) = 2 5. \(\mathrm{Ne}_{2}\): \(\frac{1}{2}(10 - 10)\) = 0
03

Order the bond orders

We have found bond orders for all the given molecules and ions. Now, we'll order them from the smallest bond order to the largest: \(\mathrm{Ne}_{2} < \mathrm{H}_{2}^{+} < \mathrm{B}_{2} < \mathrm{N}_{2}^{+} = \mathrm{F}_{2}^{+}\) So, the order is Ne₂, H₂⁺, B₂, N₂⁺, and F₂⁺.

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Key Concepts

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

Molecular Orbital Theory
Molecular Orbital Theory (MOT) is a fundamental chemical concept that describes the electronic structure of molecules. Unlike the Lewis structure approach, which focuses on electron pairs and their arrangement around atoms, MOT delves into the quantum mechanics realm, considering the wave-like nature of electrons. In MOT, atomic orbitals from each atom in a molecule combine to form molecular orbitals, which are regions in space where electrons are likely to be found.

These molecular orbitals are of two types: bonding orbitals, which lower the energy of the system and are hence stabilizing, and antibonding orbitals, denoted with an asterisk (*) and which raise the energy of the system when occupied, being destabilizing. Electrons occupying these molecular orbitals are what ultimately determine the molecule's bond order, stability, and other chemical properties. Bond order, a key concept, is essentially the difference between the number of bonding and antibonding electrons, divided by two, symbolically represented as \(\frac{{(\text{{number of bonding electrons}}) - (\text{{number of antibonding electrons}})}}{2}\). The higher the bond order, the stronger and more stable the bond between the atoms.
Molecular Orbital Diagrams
Molecular Orbital Diagrams are visual representations of the molecular orbital theory. They provide a schematic for arranging the electrons of a molecule into orbitals. In diagrams, the atomic orbitals are often depicted on the sides, with the resulting molecular orbitals positioned in between. This setup demonstrates how atomic orbitals combine and overlap to form bonding and antibonding molecular orbitals.

The arrangement of these orbitals follows an energy hierarchy, with the lowest energy orbitals filled first, a principle known as Aufbau's principle. Sigma (σ) orbitals are usually the lowest in energy, followed by pi (π) orbitals, with the corresponding antibonding σ* and π* orbitals having a higher energy. Hund's rule and the Pauli exclusion principle are also taken into consideration, ensuring that electrons are distributed in the most stable configuration possible. For example, in the molecular orbital diagram of diatomic nitrogen (N₂), we would observe electrons pairing in the lowest energy orbitals first and then occupying higher orbitals as necessary.
Electron Configuration
Electron Configuration in the context of molecular orbital theory refers to the distribution of electrons among the available orbitals in a molecule. Each electron is assigned to a molecular orbital according to its energy state and as per the rules established by quantum mechanics - the aforementioned Aufbau's principle, Hund's rule, and the Pauli exclusion principle.

An accurate electron configuration is crucial for predicting chemical behavior. For instance, in a molecular orbital diagram, electrons will fill the lowest energy molecular orbitals first. Once a bonding orbital is filled, it can contribute to bond formation between atoms, while electrons in antibonding orbitals can weaken or even prevent a bond. By analyzing the electron configurations, such as those provided for \( \mathrm{H}_{2}^{+} \) and \( \mathrm{N}_{2}^{+} \), we understand their stability and bond strength.

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Most popular questions from this chapter

What are the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom? (a) Three bonding domains and no nonbonding domains, (b) three bonding domains and one nonbonding domain, (c) two bonding domains and two nonbonding domains.

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

In ozone, \(\mathrm{O}_{3}\) , the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

The lactic acid molecule, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COH},\) gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\pi\) and how many \(\sigma\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right),\) ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule?
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