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Chapter 9: Problem 16

Write out the ground-state molecular-orbital electron configurations for \(\mathrm{Na}_{2}\) through \(\mathrm{Ar}_{2}\). Would you predict a stable \(\mathrm{Mg}_{2}\) molecule?

Short Answer

Expert verified
The ground state configuration shows Mg鈧� is not stable due to filled bonding and antibonding orbitals, cancelling out net bonding.

Step by step solution

01

Identify Atomic Orbitals

First, recall the atomic orbitals and their energies for each element from sodium (Na) to argon (Ar). Elements in this range have their valence electrons in the 3s and 3p orbitals.
02

Construct Molecular Orbitals

For diatomic molecules, combine atomic orbitals to form molecular orbitals. For each pair of atomic orbitals, form bonding and antibonding molecular orbitals. Sequence them as follows: 1. Sigma (蟽) from 3s 2. Sigma star (蟽*) from 3s 3. Pi (蟺) from 3p (degenerate) 4. Sigma (蟽) from 3p 5. Pi star (蟺*) from 3p (degenerate) 6. Sigma star (蟽*) from 3p.
03

Fill Molecular Orbitals for Molecular Na

The valence electron configuration for two Na atoms is 3s鹿 for each, totaling two electrons. Fill the 蟽(3s) orbital with these two electrons, resulting in the configuration: \[\sigma(3s)^2\].
04

Determine Na鈧� Stability

Na鈧� has all electrons in bonding orbitals with no electrons in antibonding orbitals, indicating it is relatively stable.
05

Repeat for Mg to Ar

Repeat the process for - Mg: 3s虏 each 鈫� \[\sigma(3s)^2, \sigma^*(3s)^2\] - No net bonding electrons, Mg鈧� is not stable.- Al: 3s虏3p鹿 each 鈫� \[\sigma(3s)^2, \sigma^*(3s)^2, \pi(3p_x)^2\] - Stable.- Si: 3s虏3p虏 each 鈫� \[\sigma(3s)^2, \sigma^*(3s)^2, \pi(3p_x)^2, \pi(3p_y)^2\] - Less stable.- P: 3s虏3p鲁 each 鈫� \[\sigma(3s)^2, \sigma^*(3s)^2, \pi(3p_x)^2, \pi(3p_y)^2, \sigma(3p_z)^2\] - Stable.- S: 3s虏3p鈦� each 鈫� \[\sigma(3s)^2, \sigma^*(3s)^2, \pi(3p_x)^2, \pi(3p_y)^2, \sigma(3p_z)^2, \pi^*(3p_x)^2\] - Less stable.- Cl: 3s虏3p鈦� each 鈫� \[\sigma(3s)^2, \sigma^*(3s)^2, \pi(3p_x)^2, \pi(3p_y)^2, \sigma(3p_z)^2, \pi^*(3p_x)^2, \pi^*(3p_y)^2\] - Relatively stable.- Ar: 3s虏3p鈦� each 鈫� \[\sigma(3s)^2, \sigma^*(3s)^2, \pi(3p_x)^2, \pi(3p_y)^2, \sigma(3p_z)^2, \pi^*(3p_x)^2, \pi^*(3p_y)^2, \sigma^*(3p_z)^2\] - No net bonding electrons, not stable.

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Key Concepts

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

Electron Configuration in Molecules
Electron configuration is key to understanding molecular orbital theory. This theory allows chemists to determine the stability and properties of molecules. When atoms come near each other, their atomic orbitals overlap, forming molecular orbitals. Each molecular orbital is a mix of atomic orbitals from the interacting atoms.

For diatomic molecules like those between sodium (\(\text{Na}_2\)) and argon (\(\text{Ar}_2\)), electrons occupy these molecular orbitals following three simple rules:
  • Electrons fill the lowest energy orbitals first.
  • Each orbital can hold a maximum of two electrons with opposite spins (Pauli exclusion principle).
  • Electrons prefer to remain unpaired when they are of the same energy (Hund鈥檚 rule).

Understanding the electron configuration involves knowing the arrangement of electrons in molecular orbitals. This helps predict both chemical bonding and molecular stability, as well as relates directly to the bonding and antibonding nature of these orbitals.
Bonding and Antibonding Orbitals
In molecular orbital theory, atomic orbitals combine to form molecular orbitals, which are classified into two types: bonding and antibonding orbitals. Bonding orbitals arise when atomic orbitals combine constructively, meaning their wave functions add up, resulting in a lower energy orbital. These orbitals promote molecule stability by increasing electron density between the nuclei.

Antibonding orbitals, in contrast, form through a destructive combination of atomic orbitals. These orbitals have higher energy levels, as electrons in them reduce the molecule's stability. The notation for antibonding orbitals typically includes an asterisk (e.g. \(\sigma^*\)).

Each bonding orbital occupied by electrons contributes positively to stability, whereas each antibonding orbital subtracts from it. Therefore, the stability of a molecule is heavily influenced by the presence of bonding and antibonding electrons. For example, Mg鈧� shows no net bonding benefit due to equal occupancy of bonding and antibonding orbitals, proving it's unstable.
Molecular Stability Considerations
The stability of a molecule depends on the electron occupancy of its bonding and antibonding orbitals. Molecules are more stable when there are more electrons in bonding orbitals than in antibonding orbitals. This is known as having a positive bond order.

Bond order is calculated as: \[ \text{Bond Order} = \frac{\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals}}{2} \]

A bond order greater than zero indicates that a molecule is likely to be stable, as seen in \(\text{Na}_2\), where all electrons occupy bonding orbitals. Conversely, a bond order of zero suggests instability. For instance, \(\text{Ar}_2\) has a bond order of zero because equal numbers of bonding and antibonding electrons cancel each other鈥檚 effects.

In summary, predicting the stability of a molecule through molecular orbital theory involves examining the placement of electrons in specific orbitals and computing the bond order. Higher bond orders typically signify a stronger and more stable bond, reflecting the unique balance of forces that hold atoms together in a molecule.

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Most popular questions from this chapter

Use molecular-orbital theory to explain why the dissociation energy of \(\mathrm{N}_{2}\) is greater than that of \(\mathrm{N}_{2}^{+},\) but the dissociation energy of \(\mathrm{O}_{2}^{+}\) is greater than that of \(\mathrm{O}_{2}\).

The dipole moment of \(\mathrm{HCl}\) is \(3.697 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\). The bond length of \(\mathrm{HCl}\) is \(127.5 \mathrm{pm}\). If \(\mathrm{HCl}\) is modeled as two point charges separated by its bond length, then what are the net charges on the \(\mathrm{H}\) and Cl atom?

9-10.

When we built up the molecular orbitals for diatomic molecules, we combined only those orbitals with the same energy because we said that only those with similar energies mix well. This problem is meant to illustrate this idea. Consider two atomic orbitals \(\chi_{\mathrm{A}}\) and \(\chi_{\mathrm{B}}\) Show that a linear combination of these orbitals leads to the secular determinant $$\left|\begin{array}{cc} \alpha_{\mathrm{A}}-E & \beta-E S \\\\\beta-E S & S \alpha_{\mathrm{B}}-E\end{array}\right|=0$$ where $$\begin{aligned}\alpha_{\mathrm{A}} &=\int \chi_{\mathrm{A}} h^{\mathrm{eff}} \chi_{\mathrm{A}} d \tau \\\\\alpha_{\mathrm{B}} &=\int \chi_{\mathrm{B}} h^{\mathrm{eff}} \chi_{\mathrm{B}} d \tau \\ \beta &=\int \chi_{\mathrm{B}} h^{\mathrm{eff}} \chi_{\mathrm{A}} d \tau=\int \chi_{\mathrm{A}} h^{\mathrm{eff}} \chi_{\mathrm{B}} d \tau \\\S &=\int \chi_{\mathrm{A}} \chi_{\mathrm{B}} d\tau\end{aligned}$$ where \(h^{\text {eff }}\) is some effective one-electron Hamiltonian operator for the electron that occupies the molecular orbital \(\phi .\) Show that this secular determinant expands to give $$\left(1-S^{2}\right) E^{2}+\left[2 \beta S-\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}\right] E+\alpha_{\mathrm{A}} \alpha_{\mathrm{B}}-\beta^{2}=0$$ It is usually a satisfactory first approximation to neglect \(S .\) Doing this, show that $$E_{\pm}=\frac{\alpha_{\mathrm{A}}+\alpha_{\mathrm{B}} \pm\left[\left(\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}\right)^{2}+4 \beta^{2}\right]^{1 / 2}}{2}$$ Now if \(\chi_{\mathrm{A}}\) and \(\chi_{\mathrm{B}}\) have the same energy, show that \(\alpha_{\mathrm{A}}=\alpha_{\mathrm{B}}=\alpha\) and that $$E_{\pm}=\alpha \pm \beta$$ giving one level of \(\beta\) units below \(\alpha\) and one level of \(\beta\) units above \(\alpha ;\) that is, one level of \(\beta\) units more stable than the isolated orbital energy and one level of \(\beta\) units less stable. Now investigate the case in which \(\alpha_{\mathrm{A}} \neq \alpha_{\mathrm{B}},\) say \(\alpha_{\mathrm{A}}>\alpha_{\mathrm{B}} .\) Show that $$\begin{aligned}E_{\pm} &=\frac{\alpha_{\mathrm{A}}+\alpha_{\mathrm{B}}}{2} \pm \frac{\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}}{2}\left[1+\frac{4 \beta^{2}}{\left(\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}\right)^{2}}\right]^{1 / 2} \\\&=\frac{\alpha_{\mathrm{A}}+\alpha_{\mathrm{B}}}{2} \pm \frac{\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}}{2}\left[1+\frac{2 \beta^{2}}{\left(\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}\right)^{2}}-\frac{2 \beta^{4}}{\left(\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}\right)^{4}}+\cdots\right] \\\E_{\mathrm{t}}&=\frac{\alpha_{\mathrm{A}}+\alpha_{\mathrm{B}}}{2} \pm \frac{\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}}{2} \pm \frac{\beta^{2}}{\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}}+\cdots\end{aligned}$$ where we have assumed that \(\beta^{2}<\left(\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}\right)^{2}\) and have used the expansion $$(1+x)^{1 / 2}=1+\frac{x}{2}-\frac{x^{2}}{8}+\cdots$$ Show that $$\begin{array}{l} E_{+}=\alpha_{\mathrm{A}}+\frac{\beta^{2}}{\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}}+\cdots \\\ E_{-}=\alpha_{\mathrm{B}}-\frac{\beta^{2}}{\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}}+\cdots \end{array}$$ Using this result, discuss the stabilization-destabilization of \(\alpha_{\mathrm{A}}\) and \(\alpha_{\mathrm{B}}\) versus the case above in which \(\alpha_{\mathrm{A}}=\alpha_{\mathrm{B}} .\) For simplicity, assume that \(\alpha_{\mathrm{A}}-\alpha_{\mathrm{B}}\) is large.

A common light source used in photoelectron spectroscopy is a helium discharge, which generates light at \(58.4 \mathrm{nm}\). A photoelectron spectrometer measures the kinetic energy of the electrons ionized when the molecule absorbs this light. What is the largest electron binding energy that can be measured using this radiation source? Explain how a measurement of the kinetic energy of the ionized electrons can be used to determine the energy of the occupied molecular orbitals of a molecule. Hint: Recall the photoelectron effect discussed in Chapter 1
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