5.4: Question 5.E.78 PASS - using molecular orbital theory and bond order to determine ion charge
- Page ID
- 452260
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What charge would be needed on F2 to generate an ion with a bond order of 2?
- Answer
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The charge would be positive 2, or 2+.
See LibreText 5.5 Molecular Orbital Theory
- Strategy Map
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Step Hint 1. Fill out a molecular orbital bond energy diagram. Recall molecular orbital energy diagrams (see LibreText section 5.5.1) 2. Calculate the bond order of F2. Recall bond order calculations (see LibreText section 5.5.2) 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 formula:
\(\text{bond order} = \frac{\text { (number of bonding electrons) }- \text { (number of antibonding electrons) }}{2}\)
The typical bond order for F2:
\(\begin{gathered}
\text{bond order} =\frac{\text { Bonding electrons }- \text { antibonding electrons }}{2} \\
\text{bond order} =\frac{8-6}{2} \\
\text{bond order} =1
\end{gathered}\)Electrons required for bond order of 2:
\(\begin{gathered}
\text{bond order} =\frac{\text { Bonding electrons }- \text { antibonding electrons }}{2} \\
2=\frac{x}{2} \\
X=4
\end{gathered}\)Two electrons would have to be removed, therefore the charge on the molecule would be 2+.
- Guided Solution
-
Download Guided Solution as a pdf
Guided Solution Hint This 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 having a bond order of 2.
What charge would be needed on F2 to generate an ion with a bond order of 2?
See LibreText 5.5 Molecular Orbital Theory Try filling out a molecular orbital energy diagram for F2.
This shows how many bonding and antibonding electrons are in this polyatomic ion which you will use for your calculation.
Check your molecular orbital energy diagram for F2 by clicking here
Recall the bond order equation. \(\text{bond order} =\frac{\text { (number of bonding electrons) }- \text { (number of antibonding electrons) }}{2}\) 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 two. Use the bond order equation to calculate the number of electrons it would require having a bond order of two:
\(2=\frac{x}{2}\)
Subtract the number you get for “x” from the number of electrons F2 would usually have.
Will the charge be positive or negative?
Recall that when you add electrons to an atom it gains a negative charge and when you remove electrons its gains a positive charge.
Complete Solution:
Bond Order formula:
\(\text{bond order} = \frac{\text { (number of bonding electrons) }- \text { (number of antibonding electrons) }}{2}\)
The typical bond order for F2:
molecular orbital energy diagram
There are eight bonding electrons and six antibonding electrons in F2.
\(\begin{gathered}
\text{bond order} =\frac{\text { Bonding electrons }- \text { antibonding electrons }}{2} \\
\text{bond order} =\frac{8-6}{2} \\
\text{bond order} =1
\end{gathered}\)The bond order for a ground state F2 polyatomic ion would be four.
Electrons required for bond order of 2:
Since we know we want the bond order to be two, but we wont know the number of electrons that would be, we can use the same formula to isolate and solve for the number of electrons.
\(\begin{gathered}
\text{bond order} =\frac{\text { Bonding electrons }- \text { antibonding electrons }}{2} \\
2=\frac{x}{2} \\
X=4
\end{gathered}\)If we remove two of the antibonding electrons from F2 its new bond order would be two.
\(\text{bond order} =\frac{8-4}{2}=\frac{4}{2}=2\)
Since two 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, two electrons have been removed from F2 leaving it with a charge of positive 2.
Why does this answer make chemical sense?
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.
(question source from page titled 8.E: Advanced Theories of Covalent Bonding (Exercises) https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.E%3A_Advanced_Theories_of_Covalent_Bonding_(Exercises), shared under a CC BY 4.0 license, authored, remixed, and/or curated by OpenStax, original source https://openstax.org/books/chemistry/pages/8-exercises, Access for free at https://openstax.org/books/chemistry/pages/1-introduction)