Question

Fluorine can be prepared by the reaction of hexafluoromanganate(IV) ion, MnF \(_{6}^{2-}\), with antimony pentafluoride to produce manganese(IV) fluoride and \(\mathrm{SbF}_{6}^{-}\) followed by the disproportionation of manganese(IV) fluoride to manganese(III) fluoride and \(\mathrm{F}_{2}(\mathrm{g}) .\) Write chemical equations for these two reactions.

Short answer

Balanced equations: \[ \mathrm{MnF_{6}^{2-}} + 2\mathrm{SbF_{5}} \rightarrow \mathrm{MnF_{4}} + 2\mathrm{SbF_{6}^{-}} \] and \[ 2\mathrm{MnF_{4}} \rightarrow 2\mathrm{MnF_{3}} + \mathrm{F_{2}(\mathrm{g})}\]

Step-by-step solution

Write Reactants and Products for Reaction 1

Reaction 1 involves hexafluoromanganate(IV) ion, or \( \mathrm{MnF_{6}^{2-}} \), and antimony pentafluoride, or \( \mathrm{SbF_{5}} \), reacting to form manganese(IV) fluoride, or \( \mathrm{MnF_{4}} \), and \(\mathrm{SbF_{6}^{-}}\). Write them as follows: \( \mathrm{MnF_{6}^{2-}} + \mathrm{SbF_{5}} \rightarrow \mathrm{MnF_{4}} + \mathrm{SbF_{6}^{-}} \)

Balance Reaction 1

On the right hand side, there are six fluorine atoms in \(\mathrm{SbF_{6}^{-}}\) and four in \(\mathrm{MnF_{4}}\), a total of ten. On the left hand side, there are five fluorine atoms in \(\mathrm{SbF_{5}}\) and six in \(\mathrm{MnF_{6}^{2-}}\), a total of eleven. Thus, the reaction is unbalanced. To balance, we add a coefficient of 2 to \(\mathrm{SbF_{5}}\). The balanced equation becomes: \( \mathrm{MnF_{6}^{2-}} + 2\mathrm{SbF_{5}} \rightarrow \mathrm{MnF_{4}} + 2\mathrm{SbF_{6}^{-}} \)

Write Reactants and Products for Reaction 2

Reaction 2 involves the disproportionation of manganese(IV) fluoride, or \( \mathrm{MnF_{4}} \), to produce manganese(III) fluoride, or \( \mathrm{MnF_{3}} \), and \(\mathrm{F_{2}(\mathrm{g})}\). Write them as follows: \( \mathrm{MnF_{4}} \rightarrow \mathrm{MnF_{3}} + \mathrm{F_{2}(\mathrm{g})}\)

Balance Reaction 2

In this reaction, there are four fluorine atoms on the left but only three plus two on the right. Add a coefficient of 2 in front of \( \mathrm{MnF_{4}} \) and \( \mathrm{MnF_{3}}\) on the left and right respectively. The balanced equation becomes: \( 2\mathrm{MnF_{4}} \rightarrow 2\mathrm{MnF_{3}} + \mathrm{F_{2}(\mathrm{g})}\)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is crucial in representing chemical reactions accurately. It ensures the conservation of mass and matter, as can be seen in the provided exercise. The first reaction involves the transformation of hexafluoromanganate(IV) ion with antimony pentafluoride.

Initially, the equation might appear straightforward, but each element must have equal numbers on both sides. In this exercise:

• Fluorine appeared imbalanced with 11 atoms on the left and only 10 on the right.
• By modifying the coefficient of antimony pentafluoride to 2, the equation becomes balanced with a perfect symmetry of 12 fluorine atoms on both sides.
• Adjusting coefficients, rather than changing formulas, preserves the identity of substances while balancing atoms.

The process may require trial and error, but patience and attention to detail are key.
Mastering this skill ensures that chemical equations not only accurately depict reactions but also respect fundamental laws of chemistry.

Disproportionation

Disproportionation is a type of redox reaction where a single substance is simultaneously oxidized and reduced, producing two different products. In our exercise, manganese(IV) fluoride undergoes a disproportionation reaction.

Here’s what happens:

• Manganese(IV) fluoride experiences a shift of oxidation states where manganese transitions from Mn(IV) in MnFe(4) to Mn(III) in MnFe(3) and releases fluorine gas, Fe(2) \( \rightarrow \) thus demonstrating the characteristic state changes of disproportionation.
• The equation initially appears imbalanced with MnFe(4) on the reactant side and insufficient fluorine on the product side.
• Balancing the equation by multiplying both the manganese-containing compounds by 2 ensures each atom type on both sides is equal.

Recognizing disproportionation reactions is vital for grasping the complex behavior of molecules and ions. It requires the ability to track electrons and oxidation states effectively.

Inorganic Chemistry

Inorganic chemistry explores a vast world of non-organic compounds, including those without carbon-hydrogen bonds. The reactions discussed in this exercise, revolving around fluorine and manganese compounds, showcase typical inorganic reactions.

Here's why they are significant:

• Inorganic reactions often involve elements from the periodic table forming compounds different from organic compounds.
• The preparation of elemental fluorine highlights the diverse methodologies in synthesizing pure elements from their compounds.
• Understanding these reactions provides insight into how compounds interact, allowing scientists to develop novel materials and chemical processes.

The principles applied here are also foundational for advances in other fields of chemistry and contribute to innovations in materials, catalysis, and more. A deep understanding of such reactions in inorganic chemistry equips students with the skills to explore the realm beyond organic molecules.