A) \[[Pt{{(N{{H}_{3}})}_{4}}C{{l}_{3}}]\,Cl\]
B) \[[Pt{{(N{{H}_{3}})}_{4}}Cl]\,C{{l}_{2}}\]
C) \[[Pt{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]\,C{{l}_{3}}\]
D) \[[Pt{{(N{{H}_{3}})}_{4}}C{{l}_{4}}]\]
Correct Answer: B
Solution :
\[\Delta {{T}_{f}}=im{{k}_{f}}\]; \[0.0054=i\times 1.8\times 0.001\] \[i=3\] so it is \[[Pt{{(N{{H}_{3}})}_{4}}Cl]C{{l}_{2}}\].
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