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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

Would you expect the complex \(\left( {Co{{(en)}_3}} \right)C{l_3}\) to have any unpaired electrons? Any isomers?

Determine the number of unpaired electrons expected for \({\left( {{\rm{Fe}}{{\left( {{\rm{N}}{{\rm{O}}_2}} \right)}_6}} \right)^{3 - }}\)and for \({\left( {Fe{F_6}} \right)^{3 - }}\)in terms of crystal field theory.

How many unpaired electrons are present in each of the following?

\(\begin{aligned}{}(a){\left( {Co{F_6}} \right)^{3 - }}(highspin)\\(b){\left( {Mn{{(CN)}_6}} \right)^{3 - }}(lowspin)\\(c){\left( {Mn{{(CN)}_6}} \right)^{4 - }}(lowspin)\\(d){\left( {MnC{l_6}} \right)^{4 - }}(highspin)\\(e){\left( {RhC{l_6}} \right)^{3 - }}(lowspin)\end{aligned}\)

Give the coordination number for each metal ion in the following compounds:

\(\begin{aligned}{}(a){\left( {Co{{\left( {C{O_3}} \right)}_3}} \right)^{3 - }}\\(b){\left( {Cu{{\left( {N{H_3}} \right)}_4}} \right)^{2 + }}\\(c){\left( {Co{{\left( {N{H_3}} \right)}_4}B{r_2}} \right)_2}{\left( {S{O_4}} \right)_3}\\(d)\left( {Pt{{\left( {N{H_3}} \right)}_4}} \right)\left( {PtC{l_4}} \right)\\(e)\left( {Cr{{(en)}_3}} \right){\left( {N{O_3}} \right)_3}\\(f)\left( {Pd{{\left( {N{H_3}} \right)}_2}B{r_2}} \right)\\(g){K_3}\left( {Cu{{(Cl)}_5}} \right)\\(h)\left( {Zn{{\left( {N{H_3}} \right)}_2}C{l_2}} \right)\end{aligned}\)

Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:

(a) tetrahydroxozincate(II) ion (tetrahedral)

(b) hexacyanopalladate(IV) ion

(c) dichloroaurate(I) ion (note that aurum is Latin for 鈥済old鈥�)

(d) diamminedichloroplatinum(II)

(e) potassium diamminetetrachlorochromate(III)

(f) hexaamminecobalt(III) hexacyanochromate(III)

(g) dibromobis(ethylenediamine) cobalt(III) nitrate

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