Understanding Dissolving in Liquid BrF3

Liquid bromine trifluoride (\(\text{BrF}_3\)) is a non-aqueous solvent that undergoes autoionization. This means it slightly dissociates into ions, similar to how water autoionizes into \(\text{H}^+\) and \(\text{OH}^-\).

The autoionization equilibrium for \(\text{BrF}_3\) is:

\(\text{2BrF}_3 \rightleftharpoons \text{BrF}_2^+ + \text{BrF}_4^-\)

In this solvent system, a substance that accepts fluoride ions (\(\text{F}^-\)) acts as an acid, increasing the concentration of the cationic species, \(\text{BrF}_2^+\). A substance that donates fluoride ions acts as a base, increasing the concentration of the anionic species, \(\text{BrF}_4^-\).

Dissolving SnF4 in BrF3

When tin(IV) fluoride (\(\text{SnF}_4\)) is dissolved in liquid \(\text{BrF}_3\), \(\text{SnF}_4\) acts as a Lewis acid. It accepts a fluoride ion (\(\text{F}^-\)) to form a complex anion, \([\text{SnF}_5]^-\).

\(\text{SnF}_4 + \text{F}^- \rightarrow [\text{SnF}_5]^-\)

Where does the fluoride ion come from? It comes from the autoionization of \(\text{BrF}_3\). The consumption of \(\text{F}^-\) (or effectively taking \(\text{F}^-\) from \(\text{BrF}_3\) which then re-establishes equilibrium) shifts the autoionization equilibrium to the right, producing more \(\text{BrF}_2^+\) and \(\text{BrF}_4^-\). The overall reaction can be written as:

\(\text{SnF}_4 + \text{BrF}_3 \rightarrow \text{BrF}_2^+ + [\text{SnF}_5]^-\)

So, when \(\text{SnF}_4\) is dissolved, the cation formed is \(\text{BrF}_2^+\).

Dissolving AuF3 in BrF3

Similarly, when gold(III) fluoride (\(\text{AuF}_3\)) is dissolved in liquid \(\text{BrF}_3\), \(\text{AuF}_3\) also acts as a Lewis acid. It accepts a fluoride ion (\(\text{F}^-\)) to form a complex anion, \([\text{AuF}_4]^-\).

\(\text{AuF}_3 + \text{F}^- \rightarrow [\text{AuF}_4]^-\)

Again, the fluoride ion comes from the autoionization of \(\text{BrF}_3\). This acidic behavior shifts the autoionization equilibrium to the right, increasing the concentration of \(\text{BrF}_2^+\). The overall reaction can be written as:

\(\text{AuF}_3 + \text{BrF}_3 \rightarrow \text{BrF}_2^+ + [\text{AuF}_4]^-\)

Thus, when \(\text{AuF}_3\) is dissolved, the cation formed is also \(\text{BrF}_2^+\).

Summary of Cations Formed

In both cases, dissolving \(\text{SnF}_4\) and \(\text{AuF}_3\) in liquid \(\text{BrF}_3\) causes the solvent's autoionization equilibrium to shift, leading to an increase in the concentration of the solvent's characteristic cation.

• When \(\text{SnF}_4\) is dissolved, the cation formed is \(\text{BrF}_2^+\).
• When \(\text{AuF}_3\) is dissolved, the cation formed is \(\text{BrF}_2^+\).

Therefore, the cations formed upon dissolving \(\text{SnF}_4\) and \(\text{AuF}_3\) in liquid \(\text{BrF}_3\) separately, respectively are \(\text{BrF}_2^+\) and \(\text{BrF}_2^+\).

Looking at the options, the one that correctly identifies the cation formed in both scenarios is the one stating \(\text{BrF}_2^+\) only, indicating that for both solutes, the cation is \(\text{BrF}_2^+\).