Question

Bismuth forms an ion with the formula ${\text{Bi}}_5^{3+}$. Arsenic and fluorine form a complex ion$\left[{\text{AsF}}_6\right]{}^{-}$ , with fluorine atoms arranged around a central arsenic atom. Assign oxidation numbers to each of the atoms in the bright yellow crystalline solid with the formula ${\text{Bi}}_{\text{5}}{\left({\text{AsF}}_{\text{6}}\right)}_{\text{3}}{\text{.2SO}}_{\text{2}}$.

Short answer

The oxidation numbers to each of the atoms in the bright yellow crystalline solid are:

$\stackrel{+3}{{\mathrm{Bi}}_5}{\left(\stackrel{+5}{{\mathrm{A}}_{\mathrm{s}}}\stackrel{-6}{{\mathrm{F}}_6}\right)}^{-}{\left(\stackrel{+5}{{\mathrm{A}}_{\mathrm{s}}}\stackrel{-6}{{\mathrm{F}}_6}\right)}^{-}{\left(\stackrel{+5}{{\mathrm{A}}_{\mathrm{s}}}\stackrel{-6}{{\mathrm{F}}_6}\right)}^{-}.2\stackrel{+4}{S}\stackrel{-4}{O_2}$

Step-by-step solution

The oxidation state of arsenic in the complex

$$\begin{gathered} \left[{\text{AsF}}_6\right]{}^{-} \\ x+6(-1)=-1 \\ x=-1+6 \\ x=+5 \end{gathered}$$

The oxidation numbers for each of the atoms in the bright yellow crystalline solid.

As given, the charge on bismuth is +3.

${\mathrm{Bi}}_5^3+$

Now, the oxidation state of all the atoms in the given complex is:

$\stackrel{+3}{{\mathrm{Bi}}_5}{\left(\stackrel{+5}{{\mathrm{A}}_{\mathrm{s}}}\stackrel{-6}{{\mathrm{F}}_6}\right)}^{-}.2\stackrel{+4}{S}\stackrel{-4}{O_2}$

$\stackrel{+3}{{\mathrm{Bi}}_5}{\left(\stackrel{+5}{{\mathrm{A}}_{\mathrm{s}}}\stackrel{-6}{{\mathrm{F}}_6}\right)}^{-}{\left(\stackrel{+5}{{\mathrm{A}}_{\mathrm{s}}}\stackrel{-6}{{\mathrm{F}}_6}\right)}^{-}{\left(\stackrel{+5}{{\mathrm{A}}_{\mathrm{s}}}\stackrel{-6}{{\mathrm{F}}_6}\right)}^{-}.2\stackrel{+4}{S}\stackrel{-4}{O_2}$

The number of lost electrons and the number of gained electrons are the same.

Here, the -3 charge of the three arsenic complexes is balanced by the +3 charge of the bismuth.