Q.

Select the species in which the precipitate formed on adding NaOH further dissolves in excess of NaOH.

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a

ZnCl2

b

AlCl3

c

FeCl3

d

CrCl3

answer is A.

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Detailed Solution

(a) ZnCl2+2NaOH→⟶ZnOH2↓white ppt+2NaClZn(OH)2+2NaOH⟶Na2ZnO2(soluble sodium zincate)+2H2O (b) AlCl3+3NaOH→⟶ZnOH2↓white ppt+3NaClAl(OH)3+NaOH⟶NaAIO2(sodium metaaluminate)+2H2ONote :NaAIO2 is also written as NaAl(OH)4 (c) FeCl3+3NaOH⟶Fe(OH)3↓ Brown ppt +3NaClFe(OH)3 is insoluble in excess of NaOH.  (d) CrCl3+3NaOH⟶Cr(OH)3↓ Green ppt +3NaCl Cr(OH)3 is insoluble in excess NaOH. Note :Cr(OH)3 can be made soluble in NaOH  if H2O2Cr(OH)3+2NaOH+H2O2⟶Na2CrO4 Soluble  (yellow) +H2OThis is redox reaction.
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