Step-by-step Derivations

Step 1: Identify the tin chloride Q.

We are given three reactions involving a tin chloride compound Q. The third reaction is a redox reaction: $Q + CuC{l_2} \to Z + CuCl$ (not balanced).

1. In $CuCl_2$, the oxidation state of copper is +2.
2. In $CuCl$, the oxidation state of copper is +1.
3. Copper is reduced ($Cu^{2+} + e^- \to Cu^+$), which means it acts as an oxidizing agent.
4. Therefore, the tin chloride Q must act as a reducing agent, and the tin atom in Q must be oxidized.
5. Tin (Sn) is in Group 14 and exhibits common oxidation states of +2 and +4.
6. The possible tin chlorides are $SnCl_2$ (Sn is +2) and $SnCl_4$ (Sn is +4).
7. In $SnCl_4$, tin is already in its highest oxidation state (+4) and cannot be further oxidized.
8. In $SnCl_2$, tin is in the +2 oxidation state and can be oxidized to +4.
9. Thus, Q must be $SnCl_2$.

Step 2: Identify the product Z.

From the reaction $Q + CuC{l_2} \to Z + CuCl$, we identified Q as $SnCl_2$. The tin in $SnCl_2$ is oxidized from +2 to +4. The product Z is a neutral compound containing the oxidized tin. The stable neutral chloride of tin(IV) is $SnCl_4$. So, the reaction is $SnCl_2 + CuCl_2 \to SnCl_4 + CuCl$. (Note: the problem states the reaction is not balanced; the balanced form is $SnCl_2 + 2CuCl_2 \to SnCl_4 + 2CuCl$). Therefore, Z is $SnCl_4$.

Step 3: Identify the species X.

The first reaction is $Q + Cl^- \to X$. Since Q is $SnCl_2$, the reaction is $SnCl_2 + Cl^- \to [SnCl_3]^-$.

1. X is given as a monoanion, which means it has a charge of -1. Our product $[SnCl_3]^-$ fits this description.
2. Let's determine the geometry of X ($[SnCl_3]^-$) using VSEPR theory.
   • Central atom: Sn.
   • Valence electrons of Sn (Group 14) = 4.
   • Electrons from 3 Cl atoms = 3 × 1 = 3.
   • Add 1 electron for the negative charge.
   • Total valence electrons = 4 + 3 + 1 = 8 electrons = 4 electron pairs.
   • Number of bond pairs = 3 (from 3 Sn-Cl bonds).
   • Number of lone pairs = Total pairs - Bond pairs = 4 - 3 = 1.
3. The molecule has the general formula $AX_3E_1$, where A is the central atom, X is a bonded atom, and E is a lone pair.
4. The arrangement of 4 electron pairs is tetrahedral. Due to the presence of one lone pair, the geometry of the molecule is trigonal pyramidal.
5. This matches the description of X given in the question.
6. Therefore, X is $[SnCl_3]^-$.

Step 4: Identify the species Y.

The second reaction is $Q + Me_3N \to Y$. Since Q is $SnCl_2$, the reaction is $SnCl_2 + (CH_3)_3N \to Y$.

1. Y is a neutral compound.
2. In $SnCl_2$, the Sn atom has vacant orbitals and can accept an electron pair, making it a Lewis acid.
3. In trimethylamine, $(CH_3)_3N$, the nitrogen atom has a lone pair of electrons and can donate it, making it a Lewis base.
4. The reaction is a Lewis acid-base reaction, forming an adduct: $(CH_3)_3N: \to SnCl_2$.
5. The bond formed between the N atom and the Sn atom is a coordinate (or dative) bond.
6. The product Y is the adduct $(CH_3)_3N-SnCl_2$.

Evaluation of the Options

Now we evaluate each option based on our findings: Q = $SnCl_2$, X = $[SnCl_3]^-$, Y = $(CH_3)_3N-SnCl_2$, Z = $SnCl_4$.

A: There is a coordinate bond in Y. As determined in Step 4, Y is the adduct formed between the Lewis base $Me_3N$ and the Lewis acid $SnCl_2$. The nitrogen atom donates its lone pair to the tin atom, forming a coordinate bond. This statement is correct.

B: The central atom in Z has one lone pair of electrons. Z is $SnCl_4$. Let's find the number of lone pairs on the central Sn atom.

• Valence electrons of Sn = 4.
• Electrons from 4 Cl atoms = 4.
• Total valence electrons = 4 + 4 = 8 electrons = 4 pairs.
• Number of bond pairs = 4 (from 4 Sn-Cl bonds).
• Number of lone pairs = 4 - 4 = 0. The central Sn atom in $SnCl_4$ has zero lone pairs. This statement is incorrect.

C: The oxidation state of the central atom in Z is +2. Z is $SnCl_4$. Let the oxidation state of Sn be x. The oxidation state of Cl is -1. The overall charge of the molecule is 0. x + 4(-1) = 0 => x = +4. The oxidation state of the central Sn atom in Z is +4. This statement is incorrect.

D: The central atom in X is sp3 hybridised. X is $[SnCl_3]^-$. As determined in Step 3, the central Sn atom has 3 bond pairs and 1 lone pair. The total number of electron domains (steric number) is 3 + 1 = 4. A steric number of 4 corresponds to sp3 hybridization. This statement is correct.

Conclusion

The correct options are A and D.