Iron(\({\rm{III}}\)) sulfate (\({\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}_{\rm{3}}}\)) is composed of \({\rm{F}}{{\rm{e}}^{{\rm{3 + }}}}\) and \({\rm{S}}{{\rm{O}}_{\rm{4}}}^{{\rm{2 - }}}\) ions. Explain why a sample of iron(\({\rm{III}}\)) sulfate is uncharged.

A chemical bond is a long-term attraction between atoms, ions, or molecules that allows chemical compounds to form.

The number of \({\rm{F}}{{\rm{e}}^{{\rm{3 + }}}}\) ions is said to be \({\rm{2}}\).

The charge on \({\rm{F}}{{\rm{e}}^{{\rm{3 + }}}}\) ions is said to be \({\rm{ + 3}}\).

The total charge then corresponding to \({\rm{F}}{{\rm{e}}^{{\rm{3 + }}}}\) ions is obtained as \({\rm{ + 6}}\).

The number of \({\rm{S}}{{\rm{O}}_{\rm{4}}}^{{\rm{2 - }}}\) ions is said to be \({\rm{3}}\).

The charge on \({\rm{S}}{{\rm{O}}_{\rm{4}}}^{{\rm{2 - }}}\) ions is said to be \({\rm{ - 2}}\).

The total charge then corresponding to \({\rm{S}}{{\rm{O}}_{\rm{4}}}^{{\rm{2 - }}}\) ions is obtained as \({\rm{ - 6}}\).

So, the overall charge will be \({\rm{ - 6 + 6 = 0}}\).

It is then said that the sample of Iron(\({\rm{III}}\)) Sulfate is neutral.

Therefore, sample of Iron(\({\rm{III}}\)) Sulfate is neutral since its total charge is zero.