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 &lsquo; x &rsquo; is (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

PtBr4.xNH3 0.01mole+AgNO3Solution Excess arrow AgBr downarrow 0.03moles 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 [.PtBrN 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

$PtB r_{4} . x N H_{3} 0.01 m o l e + A g N O_{3} S o l u t i o n E x cess \to A g B r ↓ 0.03 m o l es$
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

PtBr4​.xNH3​0.01mole+AgNO3​SolutionExcess→AgBr↓0.03moles 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 [PtBrNH3​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

$PtB r_{4} . x N H_{3} 0.01 m o l e + A g N O_{3} S o l u t i o n E x cess \to A g B r ↓ 0.03 m o l es$
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 $[PtB r N H_{3}_{5}] B r_{3}$