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In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. a. CN or NO b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}^{2+}\)

Short Answer

Expert verified
In terms of the molecular orbital model, CN is more likely to gain an electron compared to NO, due to its lower energy unoccupied σ\(2p_z\) orbital. On the other hand, \(\mathrm{N}_{2}^{2+}\) is more likely to gain an electron compared to \(\mathrm{O}_{2}^{2+}\), given its higher effective nuclear charge resulting in a stronger attraction between the newly added electron and the nucleus.

Step by step solution

01

Identify the molecular orbitals for each species

We'll use the abbreviation HOMO for the highest occupied molecular orbital and LUMO for the lowest unoccupied molecular orbital. a. CN or NO: - CN: HOMO is π\(2p_y\) and LUMO is σ\(2p_z\) - NO: HOMO is σ\(2p_z\) and LUMO is π\(2p_y\) b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}^{2+}\): - \(\mathrm{O}_{2}^{2+}\): HOMO is π\(2p_y\) and LUMO is π\(2p_z\) - \(\mathrm{N}_{2}^{2+}\): HOMO is π\(2p_y\) and LUMO is π\(2p_z\)
02

Compare the energy levels

We need to compare the energy levels of the HOMO and LUMO for each species in the pair and determine which one has a lower energy unoccupied orbital. a. CN or NO: - CN: LUMO is σ\(2p_z\) - NO: LUMO is π\(2p_y\) The energy level of the σ\(2p_z\) orbital is lower than the π\(2p_y\) orbital. Therefore, CN has a lower energy unoccupied orbital compared to NO. b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}^{2+}\): - \(\mathrm{O}_{2}^{2+}\): LUMO is π\(2p_z\) - \(\mathrm{N}_{2}^{2+}\): LUMO is π\(2p_z\) Both species have the same LUMO, so we need to look at other factors, such as atomic size and effective nuclear charge, to determine which species has a lower energy unoccupied orbital. \(\mathrm{N}_{2}^{2+}\) has higher effective nuclear charge, which leads to a stronger attraction between the newly added electron and the nucleus. Therefore, \(\mathrm{N}_{2}^{2+}\) is more likely to gain an electron.
03

Conclusions

Based on the comparison of the HOMO and LUMO energy levels for each species in the pair, we can conclude that: a. CN is more likely to gain an electron compared to NO. b. \(\mathrm{N}_{2}^{2+}\) is more likely to gain an electron compared to \(\mathrm{O}_{2}^{2+}\).

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Key Concepts

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

The Molecular Orbitals
Molecular orbitals represent the quantum states of electrons in molecules that extend over the entire molecule. They are vital for understanding how atoms bond to form molecules. In the molecular orbital model, we use the principles of quantum mechanics to describe the behavior of electrons within a molecule. Unlike atomic orbitals, which are centered around a single atom, molecular orbitals arise from the combination or overlap of atomic orbitals belonging to the atoms in a molecule.

This results in the creation of two types of orbitals - bonding orbitals, which have lower energy and ant-bonding orbitals with higher energy. Electrons will fill the lower energy molecular orbitals first, following the principles of the Aufbau principle, Hund's rule, and Pauli's exclusion principle, just as they do with atomic orbitals. The understanding of molecular orbitals allows chemists to explain the bond order, magnetism, and other properties of molecules.
Electron Gain Likelihood
In the context of molecular orbitals, the electron gain likelihood refers to how likely a molecule is to accept an additional electron. This is an important factor in chemical reactions and stability. When assessing the likelihood of electron gain, the focus is on the energy of the molecule's lowest unoccupied molecular orbital (LUMO).

A lower-energy LUMO means that less energy is required for an electron to be added to the molecule, increasing the likelihood of electron gain. Factors that can affect the energy of the LUMO include the type of atomic orbitals that combine to form the molecular orbitals, the electronegativity of the atoms involved, and the overall molecular geometry.
HOMO and LUMO Comparison
Comparing the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) is crucial for predicting the reactivity of molecules. The HOMO represents the highest energy level occupied by an electron within a molecule under normal conditions, while the LUMO represents the next available energy level that can accept an electron.

The energy gap between these orbitals is known as the HOMO-LUMO gap. A smaller gap often signifies higher chemical reactivity and a greater likelihood of a molecule to participate in reactions, as it suggests that less energy is required to remove an electron from the HOMO or add an electron to the LUMO. By understanding the relative energy levels of the HOMO and LUMO, chemists can better predict which molecular species are likely to gain electrons. In the exercise, CN and NO molecules were compared. CN, with a lower energy LUMO, is more likely to gain an electron than NO. Similarly, \(\mathrm{N}_{2}^{2+}\) has a lower energy LUMO compared to \(\mathrm{O}_{2}^{2+}\), due in part to a higher effective nuclear charge, making it more electron-accepting.

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Most popular questions from this chapter

Use the localized electron model to describe the bonding in \(\mathrm{CCl}_{4}\).

A variety of chlorine oxide fluorides and related cations and anions are known. They tend to be powerful oxidizing and fluorinating agents. \(\mathrm{FClO}_{3}\) is the most stable of this group of compounds and has been studied as an oxidizing component in rocket propellants. Draw a Lewis structure for \(\mathrm{F}_{3} \mathrm{ClO}\) \(\mathrm{F}_{2} \mathrm{ClO}_{2}^{+},\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\). What is the molecular structure for each species, and what is the expected hybridization of the central chlorine atom in each compound or ion?

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{kJ} / \mathrm{mol},\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen 1s character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of OH. Explain.

Compare and contrast the MO model with the local electron model. When is each useful?

The three most stable oxides of carbon are carbon monoxide \((\mathrm{CO}),\) carbon dioxide \(\left(\mathrm{CO}_{2}\right),\) and carbon suboxide \(\left(\mathrm{C}_{3} \mathrm{O}_{2}\right) .\) The space-filling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).

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