Animated Solution for Chemistry - Coordination Compounds: Comprehension Passage

Welcome to a classic puzzle from Coordination Chemistry! This problem is a beautiful blend of crystal field theory and qualitative salt analysis. It tests your ability to deduce the identity of metal ions based on their geometric preferences and chemical reactions. Let's dive into the scheme and decode it step by step.

We are presented with a reaction scheme involving two mysterious metal ions, $M_1$ and $M_2$.

The first clue is that $M_1$ reacts with two different reagents, $Q$ and $R$, to form complexes with entirely different geometries: one is tetrahedral and the other is square planar.

On the other hand, $M_2$ is steadfast. Regardless of whether it reacts with $Q$ or $R$, it always forms a tetrahedral complex. Furthermore, $M_2$ has a specific reaction with a reagent $S$, forming a white precipitate that dissolves when excess $S$ is added.

Let's focus on $M_1$. A metal ion that can switch between tetrahedral and square planar geometries depending on the ligand is a classic signature of a $d^8$ system. The most common and important $d^8$ ion in our syllabus is $N{i}^{2+}$.

When $N{i}^{2+}$ encounters a weak field ligand like the chloride ion ($C{l}^{-}$), the crystal field splitting energy is small. The electrons do not pair up against Hund's rule, and the ion utilizes outer $4s$ and $4p$ orbitals to undergo $s{p}^3$ hybridization. This results in a tetrahedral complex, $[NiC{l}_4{]}^{2-}$.

However, when $N{i}^{2+}$ encounters a strong field ligand like the cyanide ion ($C{N}^{-}$), the crystal field splitting energy is large. This forces the unpaired $3d$ electrons to pair up, freeing up one inner $3d$ orbital. The ion now undergoes $ds{p}^2$ hybridization, resulting in a square planar complex, $[Ni(CN{)}_4{]}^{2-}$.

Thus, we can confidently deduce that $M_1$ is $N{i}^{2+}$, reagent $Q$ provides $C{l}^{-}$ (like $HCl$), and reagent $R$ provides $C{N}^{-}$ (like $KCN$).

Now, let's look at $M_2$. This ion forms tetrahedral complexes regardless of the ligand's strength. This behavior is typical of a metal ion with a completely filled d-subshell, a $d^{10}$ system.

The most prominent $d^{10}$ ion is $Z{n}^{2+}$. Because all its $3d$ orbitals are fully occupied, it cannot form inner orbital complexes. It has no choice but to use its outer $4s$ and $4p$ orbitals, always undergoing $s{p}^3$ hybridization. Therefore, $Z{n}^{2+}$ consistently forms tetrahedral complexes, such as $[ZnC{l}_4{]}^{2-}$ and $[Zn(CN{)}_4{]}^{2-}$.

This confirms that $M_2$ is $Z{n}^{2+}$.

The final piece of the puzzle is reagent $S$. We know $M_2$ ($Z{n}^{2+}$) reacts with $S$ to form a white precipitate that dissolves in excess $S$.

This is a textbook qualitative analysis test for $Z{n}^{2+}$. When a base like $KOH$ or $NaOH$ is added to a $Z{n}^{2+}$ solution, a white precipitate of Zinc Hydroxide is formed:

Zinc hydroxide is amphoteric. This means it can react with both acids and bases. When excess base ($O{H}^{-}$) is added, the precipitate dissolves to form a soluble complex ion, tetrahydroxozincate(II):

Therefore, reagent $S$ must be a strong base, such as $KOH$.

By carefully analyzing the geometric preferences and chemical properties, we have successfully unmasked all the unknowns. $M_1$ is $N{i}^{2+}$, $Q$ is $HCl$, and $R$ is $KCN$. $M_2$ is $Z{n}^{2+}$, and $S$ is $KOH$. This perfectly matches the options provided in the question.