Chapter 19: Q31 E (page 814)

Consider aqueous solutions of the following coordination compounds: $C o {(N H_{3})}_{6} I_{3}, P t {(N H_{3})}_{4} I_{4}, N a_{2} P t I_{6}, a n d C r (N H_{3}) 4 I_{3}$ If $A g N O_{3}$ aqueous is added to separate beakers containing solutions of each coordination compound, how many moles of AgI will precipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex.

Short Answer

Expert verified

$C o {(N H_{3})}_{6} I_{3} - t h r e e m o l e s o f A g I w i l l f o r m . P t {(N H_{3})}_{4} I_{4} - I n t h i s t w o m o l e s o f A g I w i l l f o r m . N a_{2} P t I_{6} - Z e r o m o l e o f A g I w i l l f o r m C r (N H_{3}) 4 I_{3} - O n e m o l e o f A g I w i l l f o r m .$

Step by step solution

Coordination sphere

One mole of reacts with I and one AgI will precipitate. Condition for this iodine should be present outside the co-ordination sphere as secondary valency. So that it is in ionizable form and can form precipitates. Then only it reacts with silver nitrate.

Octahedral primary sphere

As we have an octahedral primary sphere, only six ligands will be present inside the coordination sphere and the remaining will be outside the sphere in ionizable form. As in $C o {(N H_{3})}_{6} I_{3}$ six amine ligands are there, so three iodines are present outside the sphere. The same rule applies to other compounds also.

$[C o {(N H_{3})}_{6}] I_{3} - t h r e e m o l e s o f A g I w i l l f o r m . [P t {(N H_{3})}_{4} I_{2}] I_{2} - I n t h i s t w o m o l e s o f A g I w i l l f o r m . N a_{2} [P t I_{6}] - Z e r o m o l e o f A g I w i l l f o r m, [C r (N H_{3}) 4 I_{2}] I - O n e m o l e o f A g I w i l l f o r m .$