A) \[[Co{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]\]
B) \[[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}\]
C) \[[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}\]
D) \[[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]N{{O}_{3}}\]
E) \[[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl\]
Correct Answer: C
Solution :
The number of moles of\[AgCl\]precipitated \[=\frac{4.305}{143.5}=0.03\,mol\] \[\because \]The number of moles of\[AgCl\]obtainable from mole of complex = 0.03 mol \[\therefore \]The number of moles of\[AgCl\]obtainable from mol of complex = 3 mol i.e., 3 replacable chlorines are present in complex. Thus, the formula of the complex is \[[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}\].
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