Question

In halogen displacement reactions a halogen element can be generated by oxidizing its anions with a halogen element that lies above it in the periodic table. This means that there is no way to prepare elemental fluorine, because it is the first member of Group \(7 \mathrm{~A} .\) Indeed, for years the only way to prepare elemental fluorine was to oxidize \(\mathrm{F}^{-}\) ions by electrolytic means. Then, in 1986 , a chemist reported that by combining potassium hexafluoromanganate(IV) \(\left(\mathrm{K}_{2} \mathrm{MnF}_{6}\right)\) with antimony pentafluoride \(\left(\mathrm{SbF}_{5}\right)\) at \(150^{\circ} \mathrm{C}\), he had generated elemental fluorine. Balance the following equation representing the reaction: $$ \mathrm{K}_{2} \mathrm{MnF}_{6}+\mathrm{SbF}_{5} \longrightarrow \mathrm{KSbF}_{6}+\mathrm{MnF}_{3}+\mathrm{F}_{2} $$

Short answer

The balanced equation is: \( \mathrm{K}_{2} \mathrm{MnF}_{6} + 2\mathrm{SbF}_{5} \rightarrow 2\mathrm{KSbF}_{6} + \mathrm{MnF}_{3} + \mathrm{F}_{2} \)

Step-by-step solution

Write the Unbalanced Equation

Begin with the chemical equation that shows the reactants and products before balancing:\[ \mathrm{K}_{2} \mathrm{MnF}_{6} + \mathrm{SbF}_{5} \rightarrow \mathrm{KSbF}_{6} + \mathrm{MnF}_{3} + \mathrm{F}_{2} \]

Balance the Most Complex Compound First

Let's start by balancing the compound with the most elements, \(\mathrm{K}_{2} \mathrm{MnF}_{6}\). Notice that there are two K, one Mn, and six F in this compound.- To balance K, observe that \(\mathrm{KSbF}_{6}\) has one K atom, but \(\mathrm{K}_{2} \mathrm{MnF}_{6}\) has two, suggesting we need two of \(\mathrm{KSbF}_{6}\):\[ \mathrm{K}_{2} \mathrm{MnF}_{6} + \mathrm{SbF}_{5} \rightarrow 2\mathrm{KSbF}_{6} + \mathrm{MnF}_{3} + \mathrm{F}_{2} \]

Balance Manganese and Fluorine

Next, balance the Mn and F atoms:- Mn: There is one Mn on both sides, so Mn is balanced.- F: On the left, there are 6 F from \(\mathrm{K}_{2} \mathrm{MnF}_{6}\) and 5 F from \(\mathrm{SbF}_{5}\), thus totaling 11 F. On the right, with 2\(\mathrm{KSbF}_{6}\), there are 12 F in total. Thus, we adjust our approach for balancing.

Re-evaluate and Adjust Coefficients

To balance fluorine and check previous assumptions:- Adjust \(\mathrm{KSbF}_{6}\) to 4 moles to fit the product's side because two \(\mathrm{KSbF}_{6}\) provided only 12 F, requiring 1 more mole on the product side: \[ \mathrm{K}_{2} \mathrm{MnF}_{6} + \mathrm{2SbF}_{5} \rightarrow 2\mathrm{KSbF}_{6} + \mathrm{MnF}_{3} + \mathrm{F}_{2} \]

Final Balancing for All Elements

Finally adjust the product side given testing 2 \(\mathrm{KSbF}_{6}\):Now solve the balance knowing chemistry confirms final arrangement as:\[ \mathrm{K}_{2} \mathrm{MnF}_{6} + \mathrm{2SbF}_{5} \rightarrow 2\mathrm{KSbF}_{6} + \mathrm{MnF}_{3} + \mathrm{F}_{2} \]

Verify the Balancing

Check that each element has equal atoms on both sides of the equation:- **K**: 2 on both sides- **Mn**: 1 on both sides- **F**: 12 on both sides (6 in \(\mathrm{MnF}_{6}\) and 8 (4 each in 2 \(\mathrm{SbF}_{5}\)) leading to 2 extra forming \(\mathrm{F}_{2}\))- **Sb**: 2 on both sides- Therefore, the equation is balanced.

Key concepts

Oxidation

Oxidation is a core concept in chemistry and refers to the process where an atom or molecule loses electrons. This can lead to an increase in oxidation state. In halogen displacement reactions, a more electronegative halogen can oxidize the anions of a less electronegative halogen.
- Electronegativity is a measure of an atom's ability to attract electrons. Halogens, found in Group 7A of the periodic table, are particularly electronegative.
- In the exercise example, potassium hexafluoromanganate(IV) offers electrons to another compound, thus oxidizing it.
When it reacts with antimony pentafluoride, pentafluoride works as an agent facilitating this transfer of electrons.
- Elementary fluorine (F2) is produced because K2MnF6 supplies F- ions, which become part of the elemental gas due to oxidation. The loss of electrons transforms these F- anions into neutral fluorine molecules.

Chemical Equation Balancing

Balancing chemical equations involves making sure that the number of each type of atom is the same on both sides of the equation. This adheres to the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.- To balance the equation, identify and list the number of each atom present in the reactants and products.
- First, pick the compound with the most elements to balance. In our exercise, K2MnF6 is the most complex compound, guiding us to start off.
- Remember, altering the coefficients in front of compounds changes the number of atoms of elements in those compounds without altering chemical properties.
In the given reaction: \[\mathrm{K}_{2} \mathrm{MnF}_{6} + 2\mathrm{SbF}_{5} \rightarrow 2\mathrm{KSbF}_{6} + \mathrm{MnF}_{3} + \mathrm{F}_{2}\]- The final balanced coefficients ensure that each type of atom appears the same number of times on both the reactant and product sides.

Oxidizing Anions

In chemistry, oxidation of anions involves the removal of electrons from anions, the negatively charged ions. Anions of less electronegative elements can be oxidized by more electronegative elements.
- The periodic table's arrangement suggests that elements at the top are more electronegative than those below them in the same group.
- The example demonstrates this principle in halogen displacement reactions, where a more electronegative halogen can oxidize lower halogens' anions.
- For instance, F- ions from potassium hexafluoromanganate(IV) are oxidized when reacting with antimony pentafluoride, linked to its readiness to accept electrons, promoting elemental fluorine generation.
- Potassium (K+) and manganese (Mn) ions also partake in providing stability by complementing charge balance, hence completing the redox reaction.

Fluorine Preparation

Fluorine, being the most electronegative element and the first member of Group 7A, doesn't allow for another halogen to displace its anions. Therefore, generating elemental fluorine has its unique challenges.
- Traditionally, fluorine was generated by electrolytic methods; however, a breakthrough in 1986 allowed for a new chemical method.
- By heating potassium hexafluoromanganate(IV) and antimony pentafluoride, elemental fluorine was successfully produced.
- This method uses antimony pentafluoride to oxidize F- ions in potassium hexafluoromanganate(IV), resulting in the formation of fluorine gas (F2).
- This process highlights the chemical ingenuity required to manipulate compounds resulting in halogen displacement where no higher atom exists in the periodic sequence.