Animated Solution for Chemistry - s and p-Block Elements: A tin chloride $\text{Q}$ undergoes the following reactions (not balanced) $\text{Q}+{\text{Cl}}^{-}\to \text{X}$ $\text{Q}+{\text{Me}}_3\text{N}\to \text{Y}$ $\text{Q}+{\text{CuCl}}_2\to \text{Z}+\text{CuCl}$ $\text{X}$ is a monoanion having pyramidal geometry. Both $\text{Y}$ and $\text{Z}$ are neutral compounds. Choose the correct option(s).

Imagine you are a chemical detective, and you are handed a mysterious tin chloride, labeled simply as Q. Your first clue is its reaction with a chloride ion to form a monoanion, X, which has a pyramidal geometry.

Let's think about the common chlorides of tin. Tin primarily exists in two oxidation states, $+2$ and $+4$, giving us ${\text{SnCl}}_2$ and ${\text{SnCl}}_4$. If our mystery compound Q were ${\text{SnCl}}_4$, adding a chloride ion would result in $[{\text{SnCl}}_5{]}^{-}$ or $[{\text{SnCl}}_6{]}^{2-}$, neither of which fits the description of a pyramidal monoanion. However, if Q is ${\text{SnCl}}_2$, the addition of ${\text{Cl}}^{-}$ yields the $[{\text{SnCl}}_3{]}^{-}$ ion.

Let's put $[{\text{SnCl}}_3{]}^{-}$ under the VSEPR microscope. Tin, being in Group 14, has $4$ valence electrons. In this anion, it forms $3$ single bonds with the chlorine atoms. The negative charge brings in an additional electron, leaving tin with $1$ lone pair.

With $3$ bond pairs and $1$ lone pair, the steric number is $4$. This dictates an ${\text{sp}}^3$ hybridization. Because of the lone pair pushing down on the bonds, the molecular geometry is indeed trigonal pyramidal. This perfectly matches our first clue! Therefore, Q is definitely ${\text{SnCl}}_2$, and X is $[{\text{SnCl}}_3{]}^{-}$. This confirms that Option (A) is correct.

Now, let's look at the second reaction where Q (${\text{SnCl}}_2$) meets trimethylamine (${\text{Me}}_3\text{N}$). ${\text{SnCl}}_2$ is an interesting molecule; it has an incomplete octet and empty orbitals, making it electron-deficient. In the language of chemistry, it is a Lewis acid.

On the other hand, trimethylamine has a nitrogen atom with a non-bonding lone pair of electrons, making it a willing donor, or a Lewis base. When they meet, the nitrogen atom gracefully donates its lone pair to the tin atom, forming a coordinate covalent bond. The resulting adduct, ${\text{Me}}_3\text{N}\to {\text{SnCl}}_2$, is our compound Y. Since this structure clearly features a coordinate bond, Option (D) is correct.

The final act involves the reaction of Q (${\text{SnCl}}_2$) with cupric chloride (${\text{CuCl}}_2$). Here, we must recall a crucial property of ${\text{SnCl}}_2$: it is a powerful reducing agent. Tin prefers the $+4$ oxidation state over the $+2$ state (unlike lead, where the inert pair effect is dominant).

Driven by this thermodynamic preference, ${\text{SnCl}}_2$ eagerly gives up electrons to reduce the copper in ${\text{CuCl}}_2$ from $+2$ to $+1$, forming $\text{CuCl}$. In the process, tin is oxidized to its $+4$ state, resulting in the formation of stannic chloride, ${\text{SnCl}}_4$. This is our compound Z.

With Z identified as ${\text{SnCl}}_4$, we can easily evaluate the remaining options. In ${\text{SnCl}}_4$, the oxidation state of tin is clearly $+4$, not $+2$. Thus, Option (B) is incorrect. Furthermore, tin uses all $4$ of its valence electrons to form the four covalent bonds with chlorine, leaving zero lone pairs on the central atom. This makes Option (C) incorrect as well.

In conclusion, the only correct statements are that the central atom in X is ${\text{sp}}^3$ hybridized, and there is a coordinate bond in Y.