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Chapter 7: Q2E (page 389)

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.

Short Answer

Expert verified

As, the overall charge is obtained as zero, then, we can say that a sample of Iron(\({\rm{III}}\)) Sulfate is neutral.

Step by step solution

01

Define Chemical Bonding

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

02

Explanation

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.

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Most popular questions from this chapter

Question: Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulphur double bond in \({\rm{C}}{{\rm{S}}_{\rm{2}}}\).

Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: (a)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{5}}}\)(b)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}\)(c)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}\)(d)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{p}}^{\text{4}}}\)(e)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{p}}^{\text{1}}}\).

Determine the formal charge of each element in the following:

(a) \({{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}\)

(b) \({\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}\)

(c) \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)

(d) \({\rm{O}}_{\rm{2}}^{{\rm{2 - }}}\)

(e) \({{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\)

Write Lewis structures for: (a) \({\rm{P}}{{\rm{O}}_{\rm{4}}}^{{\rm{3 - }}}\) (b) \({\rm{IC}}{{\rm{I}}_{\rm{4}}}^{\rm{ - }}\) (c) \({\rm{S}}{{\rm{O}}_{\rm{3}}}^{{\rm{2 - }}}\) (d) \({\rm{HONO}}\) .

The reaction of a metal, \({\rm{M}}\), with a halogen, \({{\rm{X}}_{\rm{2}}}\), proceeds by an exothermic reaction as indicated by this equation: \({\rm{M(s) + }}{{\rm{X}}_{\rm{2}}}{\rm{(g)}} \to {\rm{M}}{{\rm{X}}_{\rm{2}}}{\rm{(s)}}\). For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.

(a) a large radius vs. a small radius for \({{\rm{M}}^{{\rm{ + 2}}}}\)

(b) a high ionization energy vs. a low ionization energy for \({\rm{M}}\)

(c) an increasing bond energy for the halogen

(d) a decreasing electron affinity for the halogen

(e) an increasing size of the anion formed by the halogen
See all solutions

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