A solution containing 0.319 g of complex `CrCl_(3).6H_(2)O` was passed through cation exchanger and the solution given out was neutralised by 28.5 ml of 0.125 m NaOH. The correct formula of the complex will be: [ molecular weight of complex =266.5]

Options

A. `[Cr(H_(2)O)_(6)]Cl_(3)`
B. `[Cr(H_(2)O)_(5)Cl]H_(2)O.Cl_(2)`
C. `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O`
D. `[Cr(H_(2)O)_(3)Cl_(3)].3 H_(2)O`

Answer

A

Explanation

Suppose the number of ionizable `Cl^(-)` ions (present outside the coordination sphere = n). When the solution of the complex passes through the cation exchanger, these `Cl^(-)` ions will combine with `H^(+)` ions of the cation exchanger to form HCl
`nCl^(-)+nH^(+)rarr nHCl`
Thus, 1 mole of the complex will form n moles of HCl.
1 mole of the complex `-=` n moles of HCl `-=` n moles of NaOH
Moles of the comples `= (0.319)/(266.5)` =0.0012 mole
Moles of NaOH used `= (28.5xx0.125)/(1000)`
=0.0036 mole
0.0012 mole of the complex = 0.0036 mole NaOH `-=` 0.0036 mole HCl
`therefore` 1 mole of the complex `= (0.0036)/(0.0012)=3` mole of
HCl `-=` 3 moles of `Cl^(-)` ions
Thus, all the `Cl^(-)` ions are outside the coordination sphere. Hence, the complex is `[Cr(H_(2)O)_(6)]Cl_(3)`.