1. Interpret the coordination behavior of $M_1$

We need a metal ion $M_1$ that forms:

• a tetrahedral complex with reagent $Q$ in excess,
• a square planar complex with reagent $R$ in excess.

A very characteristic case is $\mathrm{Ni^{2+}}$:

• with excess $\mathrm{Cl^-}$ (from $\mathrm{HCl}$), it forms tetrachloronickelate(II), $[\mathrm{NiCl_4}]^{2-}$ which is tetrahedral.
• with excess $\mathrm{CN^-}$ (from $\mathrm{KCN}$), it forms tetracyanonickelate(II), $[\mathrm{Ni(CN)_4}]^{2-}$ which is square planar.

So for $M_1$, the pattern strongly suggests: $M_1 = \mathrm{Ni^{2+}}, \quad Q = \mathrm{HCl}, \quad R = \mathrm{KCN}.$

This already matches Option B.

2. Check the behavior of $M_2$

The question says another metal ion $M_2$ always forms tetrahedral complexes with reagents $Q$ and $R$.

If $Q = \mathrm{HCl}$ and $R = \mathrm{KCN}$, then a suitable ion is $\mathrm{Zn^{2+}}$:

• with excess chloride: $[\mathrm{ZnCl_4}]^{2-}$ tetrahedral
• with excess cyanide: $[\mathrm{Zn(CN)_4}]^{2-}$ tetrahedral

This is consistent.

3. Use reagent $S$

The problem states that aqueous solution of $M_2$ gives a white precipitate with reagent $S$, which dissolves in excess of $S$.

This is the classic behavior of $\mathrm{Zn^{2+}}$ with aqueous ammonia: $\mathrm{Zn^{2+} + 2OH^- \to Zn(OH)_2 \downarrow}$ (white precipitate)

In excess ammonia, it dissolves due to complex formation: $\mathrm{Zn(OH)_2 + 4NH_3 \to [Zn(NH_3)_4]^{2+}}$ So $M_2 = \mathrm{Zn^{2+}}$ and $S$ can be $\mathrm{NH_4OH/NH_3}$, which fits the scheme.

Thus the identification remains consistent with: $M_1 = \mathrm{Ni^{2+}}, \quad Q = \mathrm{HCl}, \quad R = \mathrm{KCN}.$

4. Evaluate options

Option A: $\mathrm{Zn^{2+}}, KCN, HCl$

For $\mathrm{Zn^{2+}}$:

• with $\mathrm{KCN}$: tetrahedral $[\mathrm{Zn(CN)_4}]^{2-}$
• with $\mathrm{HCl}$: tetrahedral $[\mathrm{ZnCl_4}]^{2-}$

It does not give one tetrahedral and one square planar complex. So A is incorrect.

Option B: $\mathrm{Ni^{2+}}, HCl, KCN$

For $\mathrm{Ni^{2+}}$:

• with excess $\mathrm{HCl}$: $[\mathrm{NiCl_4}]^{2-}$, tetrahedral
• with excess $\mathrm{KCN}$: $[\mathrm{Ni(CN)_4}]^{2-}$, square planar

This matches perfectly. So B is correct.

Option C: $\mathrm{Cd^{2+}}, KCN, HCl$

$\mathrm{Cd^{2+}}$ generally forms tetrahedral complexes with both $\mathrm{CN^-}$ and $\mathrm{Cl^-}$, not square planar with one of them. So C is incorrect.

Option D: $\mathrm{Co^{2+}}, HCl, KCN$

$\mathrm{Co^{2+}}$ does not fit this characteristic tetrahedral/square planar pair as cleanly as $\mathrm{Ni^{2+}}$. So D is incorrect.

5. Final answer

Therefore, $\boxed{M_1 = \mathrm{Ni^{2+}},\; Q = \mathrm{HCl},\; R = \mathrm{KCN}}$

So the correct option is: $\boxed{\text{B}}$

6. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They agree.