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  • When 1 mol {CrCl}_{3}. {6H}_{2} {O} is treated with an excess of {AgNO}_{3}, 3 mol of {AgCl} are obtained. The formula of the complex is:

When 1 mol \text {CrCl}_{3}.\text {6H}_{2}\text {O} is treated with an excess of \text {AgNO}_{3}, 3 mol of \text {AgCl} are obtained. The formula of the complex is:

(i) \left [ \text {CrCl}_{3}\left ( \text {H}_{2}\text {O}\right )_{3} \right ].\text {3H}_{2}\text {O}

(ii) \left [ \text {CrCl}_{2}\left ( \text {H}_{2}\text {O}\right )_{4} \right ]\text {Cl}.\text {2H}_{2}\text {O}

(iii) \left [ \text {CrCl}\left ( \text {H}_{2}\text {O}\right )_{5} \right ]\text {Cl}_{2}.\text {H}_{2}\text {O}

(iv) \left [ \text {Cr}\left ( \text {H}_{2}\text {O}\right )_{6} \right ]\text {Cl}_{3}

Answers (1)

Option (iv) is the correct answer.

Explanation: 3 mol of \text {AgCl}  indicates that \text {3Cl}^{-} ions are given in the solution. Therefore, the formula should be \left [ \text {Cr}\left ( \text {H}_{2}\text {O}\right )_{6} \right ]\text {Cl}_{3}.

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