Chemical Bonding — Molecular Geometry and Hybridization of Atomic Orbitals: Using Molecular Orbital Theory and Bond Order to Determine Ion Charge

Question

What charge would be needed on F2 to generate an ion with a bond order of 2?

 

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The charge would be 2+.

Refer to Section 5.5: Molecular Orbital Theory (1).

Strategy Map

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Table 1: Strategy Map
Strategy Map Steps
1. Fill out a molecular orbital energy diagram.

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Recall molecular orbital energy diagrams.

Refer to Section 5.5.3: Bonding Diatomic Molecules (1).
2. Calculate the bond order of F2.

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Recall the bond order formula.

Refer to Section 5.5.3: Bonding Diatomic Molecules (1).
3. Calculate the number of electrons it would require for a bond order of 2.
4. Identify the electron difference and if you would have to add or subtract electrons to F2 to fix it.

Solution

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Bond order (BO) formula:

[latex]\begin{gathered} \mathrm{BO}=\frac{\text{(Number of bonding electrons)}-\text{(Number of antibonding electrons)}}{2} \end{gathered}[/latex]

The typical bond order for F2:

[latex]\begin{aligned} \mathrm{BO}&=\frac{\text{Bonding electrons}-\text{Antibonding electrons}}{2}\\ \mathrm{BO}&=\frac{8-6}{2}\\ \mathrm{BO}&=1 \end{aligned}[/latex]

Electrons required for bond order of 2:

[latex]\begin{aligned} \mathrm{BO}&=\frac{\text{Bonding electrons }-\text{Antibonding electrons}}{2}\\ 2&=\frac{\mathrm{X}}{2}\\ \mathrm{X}&=4 \end{aligned}[/latex]

Answer: Two electrons would have to be removed, therefore the charge on the F2 molecule would be 2+.

Guided Solution

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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.

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Table 2: Guided Solution
Guided Solution Ideas
This question is both a theory and calculation type problem where you will use the bond order equation to find the charge a polyatomic ion would require with a bond order of 2.

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Refer to Section 5.5: Molecular Orbital Theory (1).
Try filling out a molecular orbital energy diagram for F2:

A blank molecular orbital diagram is shown with lines where orbital electrons should be filled in. In the center it has a single line or pairs of lines and on each side there are additional lines. At the bottom of the diagram is a horizontal line that is connected to the right and left by upward-facing lines to two more horizontal lines. Those two lines are connected by upward-facing lines to another line in the center of the diagram but farther up from the first. Above this structure is a horizontal line split in two that is connected to the right and left by upward-facing lines to two sets of three horizontal lines and those two lines are connected by upward-facing lines to another line in the center of the diagram, but further up from the first. In between the horizontal lines of this structure is another horizontal line that is above the first line but below the second and connected by lines to the side horizontal lines.

This shows how many bonding and antibonding electrons are in this polyatomic ion, which you will use for your calculation.

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Filled out diagram:

A diagram is shown that represents the filled in molecular orbital digram for F2. At the bottom center of the diagram is a horizontal line that has two vertical half arrows drawn on it, one facing up and one facing down. This line is connected to the right and left by upward-facing, dotted lines to two more horizontal lines, each labeled, “2 s,” and with two vertical half arrows drawn on them, one facing up and one facing down. These two lines are connected by upward-facing dotted lines to another line in the center of the diagram, but farther up from the first with two vertical half-arrow drawn on it, one facing up and one facing down. Moving further up the center of the diagram is a horizontal line below two horizontal lines, lying side-by-side. Both the bottom and top lines are connected to the right and left by upward-facing, dotted lines to three more horizontal lines, each labeled, “2 p,” on either side. These sets of lines each hold three upward-facing and two downward-facing half-arrows. They are connected by upward-facing lines to a pair of double lines and then a single line in the center of the diagram, but farther up from the lower lines. The The lower of these two central, horizontal lines each contain an upward-facing half-arrow and a downward facing half arrow. The upper single line is empty.

 
Recall the bond order (BO) equation.

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[latex]\begin{gathered} \mathrm{BO}= \frac{\text{(Number of bonding electrons)}-\text{(Number of antibonding electrons)}}{2} \end{gathered}[/latex]
Since we are adding a charge to the polyatomic ion, we know that we are manipulating the number of electrons it has. This means you will have to identify the difference in electrons for the polyatomic ions normal bond order and if it were to have a bond order of 2.

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Use the bond order equation to calculate the number of electrons it would require having a bond order of 2:

[latex]\begin{equation} 2=\frac{\mathrm{X}}{2} \end{equation}[/latex]

Subtract the number you get for “x” from the number of electrons F2 would usually have.
Will the charge be positive or negative?

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Recall that when you add electrons to an atom, it gains a negative charge, and when you remove electrons, its gains a positive charge.
Table 3: Complete Solution
Complete Solution
Bond order (BO) formula:

[latex]\begin{gathered} \mathrm{BO}=\\ \frac{\text{(Number of bonding electrons)}-\text{(Number of antibonding electrons)}}{2} \end{gathered}[/latex]
The typical bond order for F2:

Molecular orbital diagram:

A diagram is shown that represents the filled in molecular orbital digram for F2. At the bottom center of the diagram is a horizontal line that has two vertical half arrows drawn on it, one facing up and one facing down. This line is connected to the right and left by upward-facing, dotted lines to two more horizontal lines, each labeled, “2 s,” and with two vertical half arrows drawn on them, one facing up and one facing down. These two lines are connected by upward-facing dotted lines to another line in the center of the diagram, but farther up from the first with two vertical half-arrow drawn on it, one facing up and one facing down. Moving further up the center of the diagram is a horizontal line below two horizontal lines, lying side-by-side. Both the bottom and top lines are connected to the right and left by upward-facing, dotted lines to three more horizontal lines, each labeled, “2 p,” on either side. These sets of lines each hold three upward-facing and two downward-facing half-arrows. They are connected by upward-facing lines to a pair of double lines and then a single line in the center of the diagram, but farther up from the lower lines. The The lower of these two central, horizontal lines each contain an upward-facing half-arrow and a downward facing half arrow. The upper single line is empty.

 

 

There are 8 bonding electrons and 6 antibonding electrons in F2:

[latex]\begin{aligned} \mathrm{BO}&=\frac{\text{Bonding electrons}-\text{Antibonding electrons}}{2} \\ \mathrm{BO}&=\frac{8-6}{2}\\ \mathrm{BO}&=1 \end{aligned}[/latex]

The bond order for a ground state F2 polyatomic ion would be 4.
Electrons required for bond order of 2:

Since we know we want the bond order to be 2, but we will not know the number of electrons that would be, we can use the same formula to isolate and solve for the number of electrons.

[latex]\begin{aligned} \mathrm{BO}&=\frac{\text{Bonding electrons}-\text{Antibonding electrons }}{2}\\ 2&=\frac{\mathrm{x}}{2}\\ \mathrm{X}&=4 \end{aligned}[/latex]

If we remove 2 of the antibonding electrons from F2, its new bond order would be 2.

[latex]\begin{gathered} \mathrm{BO}=\frac{8-4}{2}=\frac{4}{2}=2 \end{gathered}[/latex]

Answer: Since 2 electrons are being removed, F2 would gain a positive charge of 2 making the polyatomic ion F22+.

 

Check Your Work

For a bond order of 2, 2 electrons have been removed from F2, leaving it with a charge of positive 2.

Does your answer make chemical sense?

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The bond order is the number of bonds in a molecule or polyatomic ion. The higher the bond order, the stronger the bond. It makes sense for the bond order to increase if two electrons are removed from F2, the single bonded atom would have to become double bonded to fulfill the octet rule. A double bond is much stronger than a single bond, this can be mathematically seen when looking at the table of bond energies, it requires significantly more energy to break a double bond than a single bond.

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All figures are by Ashlynn Jensen, from LibreTexts PASS Chemistry Book CHEM 1500 (3) by Blackstock et al., and are used/updated under a CC BY-NC 4.0 license.

References

1. OpenStax. 5.5: Molecular Orbital Theory. In CHEM 1500: Chemical Bonding and Organic Chemistry; LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/CHEM_1500%3A_Chemical_Bonding_and_Organic_Chemistry/05%3A_Chemical_Bonding_II-_Molecular_Geometry_and_Hybridization_of_Atomic_Orbitals/5.05%3A_Molecular_Orbital_Theory.

2. Blackstock, L.; Brewer, S.; Jensen, A. PASS Chemistry Book CHEM 1500; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1500.

3. Blackstock, L.; Brewer, S.; Jensen, A. 5.4: Question 5.E.78 PASS – Using Molecular Orbital Theory and Bond Order to Determine Ion Charge. In PASS Chemistry Book CHEM 1500; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1500/05%3A_Chemical_Bonding_II_-_Molecular_Geometry_and_Hybridization_of_Atomic_Orbitals/5.04%3A_Question_5.E.78_PASS_-_using_molecular_orbital_theory_and_bond_order_to_determine_ion_charge.

4. OpenStax. 8.E: Advanced Theories of Covalent Bonding (Exercises). In Chemistry 1e (OpenSTAX); LibreTexts, 2022. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.E%3A_Advance_Theories_of_Covalent_Bonding_(Exercises).

5. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 8 Exercises. In Chemistry; OpenStax, 2015. https://openstax.org/books/chemistry/pages/8-exercises.

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