This question and the next one appeared in the in the linked comprehension format. Let's give them a shot.
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. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarised in the scheme given below:
SCHEME:
M1, Q and R, respectively are
This question and the next one appeared in the in the linked comprehension format. Let's give them a shot.
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. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarised in the scheme given below:
SCHEME:
M1, Q and R, respectively are
The correct option is B Ni2+, HCl and KCN
This may look scary, but trust me it is not. Let's do this one step at a time. Looks like M1 with some excess compounds gives us a square planar and another complex gives us a tetrahedral geometry. So this must be an ion that has a coordiantion number of 4 right? And if a square planar complex is formed then the ligand in question must be a strong field ligand. Only then will it force pairing and give us the required dsp2 hybridisation for the square planar geometry. Clearly, R has to have a CN− anion which is the strong field ligand here. So we have eliminted options (a) and (c). This gives (b) as the answer since nickel has a coordination number of 4. Take a look at this in detail below with shematics.
Ni2+, HCl and KCN
This may look scary, but trust me it is not. Let's do this one step at a time. Looks like M1 with some excess compounds gives us a square planar and another complex gives us a tetrahedral geometry. So this must be an ion that has a coordiantion number of 4 right? And if a square planar complex is formed then the ligand in question must be a strong field ligand. Only then will it force pairing and give us the required dsp2 hybridisation for the square planar geometry. Clearly, R has to have a CN− anion which is the strong field ligand here. So we have eliminted options (a) and (c). This gives (b) as the answer since nickel has a coordination number of 4. Take a look at this in detail below with shematics.
