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Solutions

Question
CBSEENCH12010766

On treatment of 100 mL of 0.1 M solution of CoCl3 . 6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is

  • [Co(H2O)4 Cl2]Cl.2H2O

  • [Co(H2O)3Cl3].3H2O

  • [Co(H2O)6]Cl3

  • [Co(H2O)5Cl]Cl2.H2O

Solution

D.

[Co(H2O)5Cl]Cl2.H2O

Moles space of space complex space equals space fraction numerator Molarity space straight x space volume space left parenthesis ml right parenthesis over denominator 1000 end fraction
space equals space fraction numerator 100 space straight x space 0.1 over denominator 1000 end fraction space equals space 0.01 space mole
Moles space of space ions space precipitated space with space excess space of space
AgNO subscript 3 space equals space fraction numerator 1.2 space straight x space 10 to the power of 22 over denominator 6.02 space straight x space 10 to the power of 23 end fraction space equals space 0.02 space moles
Number of Cl present in ionization sphere = 
fraction numerator Mole space of space ion space precipitated space with space exess space AgNO subscript 3 over denominator Mole space of space complex end fraction space equals space fraction numerator 0.02 over denominator 0.01 end fraction space equals space 2

It means 2Cl ions present in ionization sphere
∴ complex is [Co(H2O)5Cl]Cl2.H2O