Lingard Substitution Colours and Equations

Question 1

Cu 2+ & NH3

NH3 precipitate, excess NH3 & equation

Answer 1

some NH3= light blue
excess NH3= dark blue
equation=[Cu(H2O)6]2+ + 4NH3 = [Cu(NH3)4(H2O)2]2+ + 4H2O

Question 2

Cr 3+ & NH3

initial colour, NH3 precipitate & excess NH3

Answer 2

initial= purple
some NH3= light green
excess NH3= dark green

Question 3

Fe 2+ & NH3

initial colour, NH3 precipitate, & equation

Answer 3

initial= pale yellow
some NH3= green
equation= [Fe(H2O)6]2+ + 2NH3 = [Fe(OH)2(H2O)4] + 2NH4

Question 4

Fe 3+ & NH3

initial colour, NH3 precipitate, & equation

Answer 4

initial= dark yellow
some NH3= orange
equation= [Fe(H2O)6]3+ + 3NH3 = [Fe(OH)3(H2O)3] + 3NH4

Question 5

Mn 2+ & NH3

initial colour, NH3 precipitate, & equation

Answer 5

initial= colourless
some NH3= brown
equation= [Mn(H2O)6]2+ + 2NH3 = [Mn(OH)2(H2O)4] + 2NH4

Question 6

Cu 2+ & HCl

initial colour, HCl precipitate & excess Hcl

Answer 6

initial= blue
some HCl= green
excess= yellow
equation=[Cu(H2O)6]2+ + 4Cl- = [CuCl4]2- + 6H2O

Question 7

Cu 2+ & OH-

initial colour, OH- precipitate & equation

Answer 7

initial= blue
some= blue
equation= [Cu(H2O)6]2+ + 2OH- = [Cu(OH)2(H2O)4] + 2H2O

Question 8

Fe 2+ & OH-

initial colour, OH- precipitate & equation

Answer 8

initial=pale yellow
some=green
equation=Fe2+ + 2OH- = [Fe(OH)2]

Question 9

Fe 3+ & OH-

initial colour, OH- precipitate & equation

Answer 9

initial=dark yellow
some=orange
equation=Fe3+ + 3OH- = [Fe(OH)3]

Question 10

Mn 2+ & OH-

initial colour, OH- precipitate & equation

Answer 10

initial=colourless
some=brown
equation=[Mn(H2O)6]2+ + 2OH- = [Mn(OH)2(H2O)4] + 2H2O

Question 11

Cr 3+ & OH-

initial colour, OH- precipitate, excess OH- precipitate & equations

Answer 11

initial=purple
some=pale green
excess=dark green
initial equation=[Cr(H2O)6]2+ + 2OH- = [Cr(OH)3(H2O)3] + 3H2O
excess equation=[Cr(OH)3(H2O)3] + 3OH- = [Cr(OH)6] + 3H2O