Chapter 10: Problem 52
Arrange these species in order of increasing stability: \(\mathrm{Li}_{2}, \mathrm{Li}_{2}^{+}, \mathrm{Li}_{2}^{-}\). Justify your choice with a molecular orbital energy level diagram.
Short Answer
Expert verified
Order: \(\mathrm{Li}_2, \mathrm{Li}_2^-, \mathrm{Li}_2^+\).
Step by step solution
01
Draw the Molecular Orbital Diagram for Li_2
For the molecule \(\mathrm{Li}_2\), construct its molecular orbital (MO) diagram. Lithium has an electronic configuration of \(1s^2 \ 2s^1\). In \(\mathrm{Li}_2\), the \(2s\) orbitals from each lithium combine to form bonding and antibonding molecular orbitals, \(\sigma_{2s}\) and \(\sigma^*_{2s}\) respectively.
02
Fill MO Diagram for Li_2
In \(\mathrm{Li}_2\), there are two electrons from each lithium's \(2s\) orbitals that combine, adding a total of four electrons. Two electrons fill the lower energy \(\sigma_{2s}\) orbital, and the next two electrons completely fill the \(\sigma_{2s}^{*}\) antibonding orbital.
03
Calculate Bond Order for Li_2
Bond order \(\text{BO}\) is given by the formula: \(\text{BO} = \frac{N_{bonding} - N_{antibonding}}{2}\). For \(\mathrm{Li}_2\):\[ N_{bonding} = 2 \, \text{and} \, N_{antibonding} = 2 \]Thus,\[ \text{BO} = \frac{2 - 2}{2} = 0\]This indicates that \(\mathrm{Li}_2\) is not stable.
04
Draw the MO Diagram for Li_2^+
For \(\mathrm{Li}_2^+\), remove one electron from the \(\sigma_{2s}^{*}\) antibonding orbital. This results in \(2\) electrons in \(\sigma_{2s}\) and \(1\) in \(\sigma_{2s}^{*}\).
05
Calculate Bond Order for Li_2^+
Plug values into the bond order formula for \(\mathrm{Li}_2^+\):\[ N_{bonding} = 2 \, \text{and} \, N_{antibonding} = 1 \]Thus,\[ \text{BO} = \frac{2 - 1}{2} = 0.5\]This bond order greater than zero indicates moderate stability.
06
Draw the MO Diagram for Li_2^-
For \(\mathrm{Li}_2^-\), add an extra electron to the \(\sigma_{2s}^{*}\) antibonding orbital. This results in \(2\) electrons in both \(\sigma_{2s}\) and \(\sigma_{2s}^{*}\) orbitals.
07
Calculate Bond Order for Li_2^-
Calculate bond order for \(\mathrm{Li}_2^-\) using:\[ N_{bonding} = 2 \, \text{and} \, N_{antibonding} = 2 \]Thus,\[ \text{BO} = \frac{2 - 2}{2} = 0\]This indicates minimal stability.
08
Compare Bond Orders and Rank Stability
The bond orders we calculated indicate stability. Since bond order correlates with stability:- \(\mathrm{Li}_2\) and \(\mathrm{Li}_2^-\) both have a bond order of \(0\), indicating least stability.- \(\mathrm{Li}_2^+\) has a bond order of \(0.5\), indicating the greatest stability.Thus, arrange in order of increasing stability: \(\mathrm{Li}_2\), \(\mathrm{Li}_2^-\), \(\mathrm{Li}_2^+\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bond Order
Bond order is a measure of the stability and strength of a chemical bond in a molecule. It is calculated using the formula: \[ \text{Bond Order} (\text{BO}) = \frac{N_{\text{bonding}} - N_{\text{antibonding}}}{2} \]Here, \(N_{\text{bonding}}\) refers to the number of electrons in bonding molecular orbitals, while \(N_{\text{antibonding}}\) refers to the number in antibonding molecular orbitals.
The higher the bond order, the greater the bond strength and stability. Conversely, a bond order of zero indicates that the molecule is generally unstable or doesn't exist under normal circumstances.
For instance, in the case of the Li\(^+\) ion, the removal of an electron from an antibonding orbital increases its bond order, contributing to its stability compared to neutral Li\(_2\).
The higher the bond order, the greater the bond strength and stability. Conversely, a bond order of zero indicates that the molecule is generally unstable or doesn't exist under normal circumstances.
For instance, in the case of the Li\(^+\) ion, the removal of an electron from an antibonding orbital increases its bond order, contributing to its stability compared to neutral Li\(_2\).
Li2 Molecule
The Li\(_2\) molecule is a diatomic lithium molecule, known for being quite unique due to its place in the periodic table. As lithium is an alkali metal with a single valence electron in its \(2s\) orbital,
Therefore, there is no effective net bonding that holds the Li\(_2\) molecule together consistently.
- The electronic configuration of lithium is \(1s^2 \, 2s^1\).
- When two lithium atoms combine to form Li\(_2\), their \(2s\) orbitals overlap to form molecular orbitals.
- This results in the formation of a bonding molecular orbital \(\sigma_{2s}\) and an antibonding molecular orbital \(\sigma^*_{2s}\).
Therefore, there is no effective net bonding that holds the Li\(_2\) molecule together consistently.
Stability of Molecular Ions
The stability of molecular ions, such as Li\(_2^+\) and Li\(_2^-\), can be understood through bond order and molecular orbital theory. The three species Li\(_2\), Li\(_2^+\), and Li\(_2^-\) each exhibit different stability features due to their bond orders:
- **Li\(_2\)** has a bond order of zero, indicating weak or nonexistent bonding stability.
- **Li\(_2^+\)** has a bond order of 0.5, as removal of one electron reduces the number of antibonding electrons. This ion is relatively more stable than its neutral counterpart.
- **Li\(_2^-\)** also results in a bond order of zero by adding an electron to the antibonding orbital, yielding poor stability.