Discussion Forum



1)

As an aqueous solution of metal ion M1 reacts separately with reagents  Q and R  in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2  always forms tetrahedral complexes with these reagents. An aqueous solution of M2 on  reaction with  reagent S gives white precipitate  which dissolves in  excess of S. The  reactions are summarized in the scheme given below

   2732021467_m10.PNG

$ M_{1}$, Q, R  respectively are


A) $Zn^{2+},KCN $and $HCl$

B) $Ni^{2+},KCN $and $HCl$

C) $Cd^{2+},KCN $and $HCl$

D) $Co^{2+},KCN $and $HCl$

Answer:

Option B

Explanation:

 Plan This problem can be solved by using the concept of chemical reaction of transition metal ions (.)  colour and structure of transition metal compounds.

   Here, among given four option Ni2+ and Zn2+ has  ability to form tetrahedral as well as square planar complex depending upon types of reagent used.

 Ni2+ on reaction with KCN forms square planar complex [Ni(CN)4]2- due to strong field strength of CN.

Ni2+ + KCN $\rightarrow$ [Ni(CN)4]2- (Square Planar)

while on reaction with HCl.Ni2+  forms a stable tetrahedral complex [Ni(Cl)4]2-. 

Zn2+, on the other hand on reaction with KCN as well as HCl, produces tetrahedral complex because of its d10 electronic configuration.

 2732021599_p13.JPG

Complete reaction sequence can be shown as

2732021452_p14.JPG