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Question: Match the Column **Column-I** (A) $\text{ICl}_3$ (B) $\text{AlCl}_3$ (C) $\text{AlF}_3$ (D) $\text...

Match the Column

Column-I

(A) ICl3\text{ICl}_3 (B) AlCl3\text{AlCl}_3 (C) AlF3\text{AlF}_3 (D) NO2\text{NO}_2

Column-II

(p) Hybridisation of central atom is similar in both dimer and monomer form. (q) Both monomer and dimer forms are planar (r) In dimer form all atoms are sp3\text{sp}^3 hybridised. (s) Does not exist in dimer form

A

Hybridisation of central atom is similar in both dimer and monomer form.

B

Both monomer and dimer forms are planar

C

In dimer form all atoms are sp3\text{sp}^3 hybridised.

D

Does not exist in dimer form

Answer

A - (p), (q) B - (r) C - (s) D - (p), (q)

Explanation

Solution

The question asks to match the compounds in Column-I with their properties in Column-II. We will analyze each compound individually.

Analysis of Column-I Compounds:

(A) ICl₃:

  • Monomer (ICl₃):
    • Central atom: Iodine (I). Valence electrons = 7.
    • Number of bond pairs = 3 (with 3 Cl atoms).
    • Number of lone pairs = (7 - 3)/2 = 2.
    • Steric Number (SN) = 3 (bp) + 2 (lp) = 5.
    • Hybridization: sp³d.
    • Electron geometry: Trigonal bipyramidal.
    • Molecular geometry: T-shaped.
    • Planarity: A T-shaped molecule is planar.
  • Dimer (I₂Cl₆):
    • ICl₃ dimerizes to form I₂Cl₆, which has a structure similar to Al₂Cl₆, but it is planar (D₂h symmetry).
    • In I₂Cl₆, each Iodine atom is bonded to 4 Chlorine atoms (2 terminal, 2 bridging).
    • For each Iodine, considering 4 bond pairs and 1 lone pair (to satisfy the octet expansion and diamagnetism), the Steric Number = 4 (bp) + 1 (lp) = 5.
    • Hybridization: sp³d.
    • Planarity: The I₂Cl₆ dimer is planar.
  • Matching Properties:
    • (p) Hybridisation of central atom is similar in both dimer and monomer form: Yes, I is sp³d in both ICl₃ and I₂Cl₆.
    • (q) Both monomer and dimer forms are planar: Yes, ICl₃ (T-shaped) and I₂Cl₆ (D₂h) are both planar.
    • (r) In dimer form all atoms are sp³ hybridised: No, Iodine is sp³d.
    • (s) Does not exist in dimer form: No, it exists as I₂Cl₆.
  • Conclusion for (A): (p), (q)

(B) AlCl₃:

  • Monomer (AlCl₃):
    • Central atom: Aluminum (Al). Valence electrons = 3.
    • Number of bond pairs = 3 (with 3 Cl atoms).
    • Number of lone pairs = 0.
    • Steric Number (SN) = 3.
    • Hybridization: sp².
    • Molecular geometry: Trigonal planar.
    • Planarity: Planar.
  • Dimer (Al₂Cl₆):
    • AlCl₃ dimerizes to form Al₂Cl₆, which has a bridged structure.
    • In Al₂Cl₆, each Aluminum atom is bonded to 4 Chlorine atoms (2 terminal, 2 bridging).
    • For each Al, Steric Number = 4 (bp) + 0 (lp) = 4.
    • Hybridization: sp³.
    • Planarity: The Al₂Cl₆ dimer is non-planar (tetrahedral geometry around each Al).
    • Hybridization of bridging Cl atoms: Each bridging Cl forms 2 bonds and has 2 lone pairs. SN = 2 (bp) + 2 (lp) = 4. Hybridization: sp³.
    • Hybridization of terminal Cl atoms: Each terminal Cl forms 1 bond and has 3 lone pairs. SN = 1 (bp) + 3 (lp) = 4. Hybridization: sp³.
  • Matching Properties:
    • (p) Hybridisation of central atom is similar in both dimer and monomer form: No, Al is sp² in monomer and sp³ in dimer.
    • (q) Both monomer and dimer forms are planar: No, the dimer is non-planar.
    • (r) In dimer form all atoms are sp³ hybridised: Yes, Al, bridging Cl, and terminal Cl are all sp³ hybridized.
    • (s) Does not exist in dimer form: No, it exists as Al₂Cl₆.
  • Conclusion for (B): (r)

(C) AlF₃:

  • Nature: AlF₃ is a highly ionic compound. In the solid state, it forms a giant polymeric network structure where each Al³⁺ ion is octahedrally coordinated by 6 F⁻ ions. It does not exist as discrete monomeric AlF₃ molecules or dimeric Al₂F₆ molecules in the gas phase under normal conditions.
  • Matching Properties:
    • (p) Hybridisation of central atom is similar in both dimer and monomer form: Not applicable, as it doesn't form discrete monomer/dimer.
    • (q) Both monomer and dimer forms are planar: Not applicable.
    • (r) In dimer form all atoms are sp³ hybridised: Not applicable.
    • (s) Does not exist in dimer form: Yes, it exists as a polymeric network, not a discrete dimer.
  • Conclusion for (C): (s)

(D) NO₂:

  • Monomer (NO₂):
    • Central atom: Nitrogen (N). Valence electrons = 5.
    • It's an odd-electron molecule. It has one double bond and one single bond to oxygen atoms, and one unpaired electron.
    • Hybridization: sp² (N forms 2 sigma bonds and has 1 unpaired electron, and a p-orbital for pi bonding).
    • Molecular geometry: Bent.
    • Planarity: A bent molecule is planar.
  • Dimer (N₂O₄):
    • NO₂ dimerizes to form N₂O₄.
    • In N₂O₄, each Nitrogen atom is bonded to 2 Oxygen atoms and 1 other Nitrogen atom.
    • For each N, it forms 3 sigma bonds.
    • Steric Number (SN) = 3.
    • Hybridization: sp².
    • Planarity: N₂O₄ is a planar molecule.
  • Matching Properties:
    • (p) Hybridisation of central atom is similar in both dimer and monomer form: Yes, N is sp² in both NO₂ and N₂O₄.
    • (q) Both monomer and dimer forms are planar: Yes, NO₂ (bent) and N₂O₄ (planar) are both planar.
    • (r) In dimer form all atoms are sp³ hybridised: No, Nitrogen is sp².
    • (s) Does not exist in dimer form: No, it exists as N₂O₄.
  • Conclusion for (D): (p), (q)

Final Match: (A) - (p), (q) (B) - (r) (C) - (s) (D) - (p), (q)

The final answer is A−(p,q);B−(r);C−(s);D−(p,q)\boxed{A-(p,q); B-(r); C-(s); D-(p,q)}

Explanation of the solution:

  • ICl₃: Monomer (T-shaped, planar, I is sp³d). Dimer (I₂Cl₆, planar, I is sp³d). Thus, hybridization is similar, and both are planar.
  • AlCl₃: Monomer (Trigonal planar, Al is sp²). Dimer (Al₂Cl₆, non-planar, Al is sp³). In the dimer, all atoms (Al, bridging Cl, terminal Cl) are sp³ hybridized.
  • AlF₃: Forms a polymeric network in solid state, does not exist as discrete monomer or dimer molecules.
  • NO₂: Monomer (Bent, planar, N is sp²). Dimer (N₂O₄, planar, N is sp²). Thus, hybridization is similar, and both are planar.