On adding excess of AgNO3 solution into 0.01 mole complex compound PtBr4 . xNH3,0.03 moles yellow precipitate was obtained, then value of ‘ x ’ is

Question

On adding excess of AgNO3 solution into 0.01 mole complex compound PtBr4 . xNH3,0.03 moles yellow precipitate was obtained, then value of ‘ x ’ is (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/1f262736b8a2e596f6a204109ca06f7a-.png" /> As 3 AgBr molecules are formed it means 3Br- are outside the coordination sphere and only one is inside so the value of x=5 i.e., complex is [.PtBr(N H3) .5]Br3

Q. On adding excess of $A g N O_{3}$ solution into 0.01 mole complex compound $PtB r_{4}$ . $x N H_{3}, 0.03$ moles yellow precipitate was obtained, then value of ‘ $x$ ’ is

2

3

4

5

As 3 AgBr molecules are formed it means $3 B r^{-}$ are outside the coordination sphere and only one is inside so the value of $x = 5$

i.e., complex is $[PtB r (N H_{3})_{5}] B r_{3}$

Q. On adding excess of AgNO3​ solution into 0.01 mole complex compound PtBr4​ . xNH3​,0.03 moles yellow precipitate was obtained, then value of ‘ x ’ is

2

3

4

5

As 3 AgBr molecules are formed it means 3Br− are outside the coordination sphere and only one is inside so the value of x=5 i.e., complex is [PtBr(NH3​)5​]Br3​

Q. On adding excess of $A g N O_{3}$ solution into 0.01 mole complex compound $PtB r_{4}$ . $x N H_{3}, 0.03$ moles yellow precipitate was obtained, then value of ‘ $x$ ’ is

As 3 AgBr molecules are formed it means $3 B r^{-}$ are outside the coordination sphere and only one is inside so the value of $x = 5$

i.e., complex is $[PtB r (N H_{3})_{5}] B r_{3}$