Question

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 reactionn with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarised in the scheme given below:
SCHEME:

Reagent S is

• A. $K_4\left[Fe\right(CN{)}_6]$
• B. $N{a}_{2}HP{O}_4$
• C. $K_{2}Cr{O}_4$
• D. $KOH$

Answer 1

The correct option is D

$KOH$

It is important we do the previous question before this, only then can we attempt thiswith ease. We know what Q and R are. One produces a strong field and the other a weak field ligand.
Now this is interesting, the M2 ion reacts with both Q and R to give us the SAME geometry. Which means that it the ion M2 must have a full electron shell i.e. a $d^{1}0$ configuration right? Only then will the effect of pairing not matter.
The $d^{1}0$ ion we are most familiar with would be $Z{n}^{2+}$ right?
So which of these options do you think forms a white precipitate with it?
$KOH$ right? remember from salt analysis? $Zn(OH{)}_2$ forms a white precipitate right? Wait !
Does this precipitate dissolve when there is excess of $KOH$? Yes! Check out the scheme and the reactions below.