When 0.01 mole of a cobalt complex is treated with excess of silver nitrate solution 4.035 g of silver chloride is precipitated. The formula of the complex is :

Options

A. `[Co(NH_(3))_(3)Cl_(3)]`
B. `[Co(NH_(3))_(5)Cl]Cl_(2)`
C. `[Co(NH_(3))_(6)]Cl_(3)`
D. `[Co(NH_(3))_(4)Cl_(2)]NO_(3)`

Answer

C

Explanation

4.305 g AgCl = `(4.305)/(143.5)` mole = 0.03 mole
As 0.01 mole of the complex gives 0.03 mole of AgCl, this shows that there are 3 ionizable Cl. Hence, formula is (c ).