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Chapter 19: Q42E (page 1096)

Explain how the diphosphate ion, \({\left( {{O_3}P - O - P{O_3}} \right)^{4 - }}\) can function as a water softener that prevents the precipitation of \(F{e^{2 + }}\) as an insoluble iron salt.

Short Answer

Expert verified

Due to its structure, oxygen on one side can form bonds with water and on the other one with \(F{e^{2 + }}\) as it is bonding with water, \(F{e^{2 + }}\) will remain insoluble.

Step by step solution

01

of 2: Lewis structure

The Lewis structure of the diphosphate ion is as follows

02

of 2: Explanation

  • As we could see that the negatively charged oxygen atoms on the both sides of the ion which can attract the positively charged \({\rm{F}}{{\rm{e}}^{2 + }}\)ion.
  • The other side can form ion-dipole bond with water.
  • As the bonding with water is taking place, the \(F{e^{2 + }}\)will remain insoluble in the water.

Result

Due to its structure, oxygen on one side can form bonds with water and on the other one with \(F{e^{2 + }}\)as it is bonding with water, \(F{e^{2 + }}\)will remain insoluble.

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Most popular questions from this chapter

Draw the geometric, linkage, and ionization isomers for \(\left( {CoC{l_5}CN} \right)(CN)\)

Indicate the coordination number for the central metal atom in each of the following coordination compounds:

\(\begin{aligned}{}(a)\left( {Pt{{\left( {{H_2}O} \right)}_2}B{r_2}} \right)\\(b)\left( {Pt\left( {N{H_3}} \right)(py)(Cl)(Br)} \right)\left( {py = } \right.pyridine,\left. {{C_5}{H_5}\;N} \right)\\(c)\left( {Zn{{\left( {N{H_3}} \right)}_2}C{l_2}} \right)\\(d)\left( {Zn\left( {N{H_3}} \right)(py)(Cl)(Br)} \right)\\(e)\left( {Ni{{\left( {{H_2}O} \right)}_4}C{l_2}} \right)\\(f){\left( {Fe{{(en)}_2}{{(CN)}_2}} \right)^ + }\left( {en = } \right.ethylenediamine,\left. {{C_2}{H_8}\;{N_2}} \right)\end{aligned}\)

Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)

\(\begin{aligned}(a)Fe(s) + {H_2}S{O_4}(aq) \to \\(b)FeC{l_3}(aq) + NaOH(aq) \to \\(c)Mn{(OH)_2}(s) + HBr(aq) \to \\(d)Cr(s) + {O_2}(g) \to \\(e)M{n_2}{O_3}(s) + HCl(aq) \to \\(f)Ti(s) + xs{F_2}(g) \to \end{aligned}\)

The standard reduction potential for the reaction \({(Co{({H_2}O)_6})^{3 + }}(aq) + {e^ - } \to {(Co{({H_2}O)_6})^{2 + }}(aq)\)is about 1.8 V. The reduction potential for the reaction \({(Co{(N{H_3})_6})^{3 + }}(aq) + {e^ - } \to {(Co{(N{H_3})_6})^{2 + }}(aq)\) is +0.1 V. Calculate the cell potentials to show whether the complex ions, \({(Co{({H_2}O)_6})^{2 + }}\) and/or\({(Co{(N{H_3})_6})^{2 + }}\), can be oxidized to the corresponding cobalt (III) complex by oxygen.

Which of the following is the strongest oxidizing agent: \(VO_4^3,CrO_4^{2 - },\)or \(MnO_4^ - \)?
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