$[P t (N H_{3})_{5} C l] C l_{3}$ is the true formula of the complex $P t C l_{4} \cdot 5 N H_{3}$ . One mole of it on reaction with excess of $A g N O_{3}$ yields three moles of white precipitate of $A g C l$ .
If true enter 1, else enter 0.

Open in App

Solution

Given that the complex $P t C l_{4} .5 N H_{3}$ yiels three moles of white precipitate of AgCl on reaction with excess of $A g N O_{3}$ .
The Cl atoms present outside the complex only react with $A g N O_{3}$ . Since three moles are produced, this means 3 $C l$ atoms has to be present outside. So, the formula of the complex is $[P t (N H_{3})_{5} C l] C l_{3}$ .
Hence, the correct answer is 1.

Suggest Corrections

Similar questions

C o C l_{3} \cdot 4 N H_{3}

A g N O_{3}

[C o (N H_{3})_{4} C l_{2}] C l

1

C r C l_{3} \cdot 6 H_{2} O

A g N O_{3}

A g C l

P t C l_{4} .6 H_{2} O

A g N O_{3}

2

A g C l

P t C l_{4} .6 H_{2} O

[P t C l_{2} (H_{2} O)_{4}] C l_{2} . 2 H_{2} O

N a_{2} C O_{3}

C O_{2}

H_{2} O

Join BYJU'S Learning Program