JEE Challenger
More from p-Block Elements
ChemMedAdv 2019 • Paper 1

Reactions and Structural Properties of Tin Chloride Derivatives

A tin chloride species Q\text{Q} undergoes the following unbalanced chemical transformations:

Q+Cl−→X\text{Q} + \text{Cl}^- \rightarrow \text{X} Q+Me3N→Y\text{Q} + \text{Me}_3\text{N} \rightarrow \text{Y} Q+CuCl2→Z+CuCl\text{Q} + \text{CuCl}_2 \rightarrow \text{Z} + \text{CuCl}

Given that X\text{X} is a monoanion exhibiting pyramidal geometry, and both Y\text{Y} and Z\text{Z} represent neutral molecules, choose the correct statement(s):

Options

A

The central atom in X\text{X} exhibits sp3sp^3 hybridization

Correct
B

A coordinate covalent bond is present in Y\text{Y}

Correct
C

The oxidation state of the central atom in Z\text{Z} is +2+2

D

The central atom in Z\text{Z} possesses one non-bonding lone pair of electrons

Topics & Concepts

p-Block Elementsp-Block ElementsHybridizationOxidation State

Step-by-Step Solution

To determine the correct statements, we analyze the given chemical transformations step-by-step:

1. Identification of Species Q\text{Q} and X\text{X}

A tin chloride species Q\text{Q} can be either tin(II) chloride (SnCl2\text{SnCl}_2) or tin(IV) chloride (SnCl4\text{SnCl}_4).

  • Case 1: If Q=SnCl2\text{Q} = \text{SnCl}_2 SnCl2+Cl−→[SnCl3]−\text{SnCl}_2 + \text{Cl}^- \rightarrow [\text{SnCl}_3]^- In the monoanion X=[SnCl3]−\text{X} = [\text{SnCl}_3]^-, the central tin (Sn\text{Sn}) atom has 4 valence electrons + 1 electron from the negative charge =5= 5 electrons.

    • Number of σ\sigma-bond pairs =3= 3 (with 3 chlorine atoms)
    • Number of non-bonding lone pairs =5−32=1= \frac{5 - 3}{2} = 1

    Steric Number=Bond Pairs+Lone Pairs=3+1=4\text{Steric Number} = \text{Bond Pairs} + \text{Lone Pairs} = 3 + 1 = 4

    A steric number of 44 with 11 lone pair corresponds to sp3sp^3 hybridization with a trigonal pyramidal geometry. This matches the given description of X\text{X}.

  • Case 2: If Q=SnCl4\text{Q} = \text{SnCl}_4 SnCl4+Cl−→[SnCl5]−\text{SnCl}_4 + \text{Cl}^- \rightarrow [\text{SnCl}_5]^- The anion [SnCl5]−[\text{SnCl}_5]^- has a trigonal bipyramidal geometry, which contradicts the given pyramidal geometry of X\text{X}.

Therefore, the starting compound Q\text{Q} is SnCl2\text{SnCl}_2, and X\text{X} is [SnCl3]−[\text{SnCl}_3]^-.

2. Analysis of Options

  • Option (A): Since the central Sn\text{Sn} atom in X=[SnCl3]−\text{X} = [\text{SnCl}_3]^- has a steric number of 44 (33 bond pairs + 11 lone pair), its hybridization is sp3sp^3.
    Option (A) is correct.

  • Option (B): SnCl2\text{SnCl}_2 acts as a Lewis acid because Sn\text{Sn} has an incomplete octet/vacant dd-orbitals, while trimethylamine (Me3N\text{Me}_3\text{N}) acts as a Lewis base. SnCl2+Me3N→Me3N→SnCl2\text{SnCl}_2 + \text{Me}_3\text{N} \rightarrow \text{Me}_3\text{N} \rightarrow \text{SnCl}_2 In the adduct Y=Me3N⋅SnCl2\text{Y} = \text{Me}_3\text{N}\cdot\text{SnCl}_2, the nitrogen atom donates a lone pair to the tin atom, forming a coordinate covalent bond (N→Sn\text{N} \rightarrow \text{Sn}).
    Option (B) is correct.

  • Option (C): When SnCl2\text{SnCl}_2 reacts with CuCl2\text{CuCl}_2, a redox reaction occurs where Sn(II)\text{Sn(II)} is oxidized to Sn(IV)\text{Sn(IV)} and Cu(II)\text{Cu(II)} is reduced to Cu(I)\text{Cu(I)}: SnCl2+2CuCl2→SnCl4+2CuCl\text{SnCl}_2 + 2\text{CuCl}_2 \rightarrow \text{SnCl}_4 + 2\text{CuCl} Thus, Z\text{Z} is SnCl4\text{SnCl}_4. The oxidation state of the central atom Sn\text{Sn} in Z\text{Z} is +4+4 (not +2+2).
    Option (C) is incorrect.

  • Option (D): In Z=SnCl4\text{Z} = \text{SnCl}_4, the central tin atom utilizes all 44 of its valence electrons to form four single covalent bonds with chlorine atoms. Consequently, Sn\text{Sn} has zero non-bonding lone pairs of electrons in Z\text{Z}.
    Option (D) is incorrect.

Conclusion

The correct options are A and B.

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