On treatment of 100 mL of 0.1 M solution of `COCl_(3).6H_(2)O` with excess of `AgNO_(3), 1.2 xx 10^(22)` ions are precipitated. The complex is

Explanation

Number of moles of the complex reacted = `((M xx (Vml))/(1000 mL))`
`((0.1M) xx (100 mL))/(1000 mL)` = 0.01 mol
Number of mole of `Cl^(-)` ions precipitated = `(1.2 xx 10^(22))/(6.022 xx 10^(23))` = 0.02 mol ions
Number of moles of `Cl^(-)` ions present in the ionisation sphere = `("Moles of ions precipated")/("Moles of complex")`
`(0.02 mol)/(0.01 mol)` = 2
Hence, the formula of the complex = `[Co(H_(2)O_(3))_(5)Cl] Cl_(2).H_(2)O`.