Question

In which of the following all the given characteristics are present?

(I) Vacant orbitals involved in hybridization.

(II) Octet of underlined atom is complete.

(III) Geometry at underlined atom is tetrahedral.

• A. B₂H₆
• B. Si₂H₆
• C. Al₂Cl₆
• D. I₂Cl₆

Answer 1

Answer: (C)

Al₂Cl₆

Answer 2

The problem requires us to identify the molecule among the given options that satisfies all three specified characteristics for the underlined (central) atom:

(I) Vacant orbitals involved in hybridization.

(II) Octet of underlined atom is complete.

(III) Geometry at underlined atom is tetrahedral.

Let's analyze each option:

A) B₂H₆ (Diborane)

• Structure: B₂H₆ has two boron atoms bridged by two hydrogen atoms, forming two 3-center-2-electron bonds (banana bonds). Each boron atom is also bonded to two terminal hydrogen atoms.
• Underlined Atom: Boron (B).
• Hybridization: Each B atom is bonded to 4 atoms (2 terminal H, 2 bridging H) and has no lone pairs. Thus, it is sp³ hybridized.
• (I) Vacant orbitals involved in hybridization: Boron is in Group 13 with 3 valence electrons ($2s^2 2p^1$). To achieve sp³ hybridization, it uses its $2s$ and three $2p$ orbitals. One of the $2p$ orbitals is initially empty in the ground state. This empty $p$-orbital is crucial for the formation of the 3-center-2-electron bonds. Therefore, a vacant orbital is involved. (Met)
• (II) Octet of underlined atom is complete: Each B atom is surrounded by 6 electrons (2 from each terminal B-H bond, and 1 from each 3-center-2-electron bridging bond). The octet is not complete. (Not Met)
• (III) Geometry at underlined atom is tetrahedral: Each B atom is sp³ hybridized and bonded to 4 atoms. The electron geometry and molecular geometry around each boron are approximately tetrahedral. (Met)
• Conclusion for B₂H₆: Fails condition (II).

B) Si₂H₆ (Disilane)

• Structure: Si₂H₆ is analogous to ethane, with a Si-Si single bond and each Si atom bonded to three H atoms.
• Underlined Atom: Silicon (Si).
• Hybridization: Each Si atom is bonded to 4 atoms (1 Si, 3 H) and has no lone pairs. Thus, it is sp³ hybridized.
• (I) Vacant orbitals involved in hybridization: Silicon is in Group 14 with 4 valence electrons ($3s^2 3p^2$). In its excited state ($3s^1 3p^3$), it has four half-filled orbitals. These four orbitals combine to form four sp³ hybrid orbitals. No initially empty orbital needs to be involved in the hybridization process for Si to form 4 covalent bonds. (Not Met)
• (II) Octet of underlined atom is complete: Each Si atom forms 4 single bonds, meaning it is surrounded by $4 \times 2 = 8$ electrons. The octet is complete. (Met)
• (III) Geometry at underlined atom is tetrahedral: Each Si atom is sp³ hybridized and bonded to 4 atoms. The geometry around each Si atom is tetrahedral. (Met)
• Conclusion for Si₂H₆: Fails condition (I).

C) Al₂Cl₆ (Aluminium Chloride Dimer)

• Structure: Al₂Cl₆ is a dimer of AlCl₃, with two bridging chlorine atoms.
• Underlined Atom: Aluminium (Al).
• Hybridization: Each Al atom is bonded to 4 chlorine atoms (2 terminal, 2 bridging) and has no lone pairs. Thus, it is sp³ hybridized.
• (I) Vacant orbitals involved in hybridization: Aluminium is in Group 13 with 3 valence electrons ($3s^2 3p^1$). To form 4 bonds and achieve sp³ hybridization, the Al atom accepts a lone pair from a bridging chlorine atom into an initially vacant $3p$ orbital. Therefore, a vacant orbital is involved. (Met)
• (II) Octet of underlined atom is complete: Each Al atom forms 4 bonds, meaning it is surrounded by $4 \times 2 = 8$ electrons. The octet is complete. (Met)
• (III) Geometry at underlined atom is tetrahedral: Each Al atom is sp³ hybridized and bonded to 4 atoms. The geometry around each Al atom is tetrahedral. (Met)
• Conclusion for Al₂Cl₆: All three conditions are met.

D) I₂Cl₆ (Iodine Chloride Dimer)

• Structure: I₂Cl₆ is a dimer of ICl₃, with two bridging chlorine atoms.
• Underlined Atom: Iodine (I).
• Hybridization: Each I atom in I₂Cl₆ is bonded to 4 chlorine atoms (2 terminal, 2 bridging). Iodine is in Group 17 with 7 valence electrons. In I₂Cl₆, each I atom has 4 bond pairs and 2 lone pairs (total 6 electron domains). Thus, it is sp³d² hybridized.
• (I) Vacant orbitals involved in hybridization: Iodine is in Period 5 and has vacant 5d orbitals. Since the hybridization is sp³d², vacant d-orbitals are explicitly involved. (Met)
• (II) Octet of underlined atom is complete: Each I atom is surrounded by 4 bonding pairs ($4 \times 2 = 8$ electrons) and 2 lone pairs ($2 \times 2 = 4$ electrons). The total number of electrons around Iodine is $8 + 4 = 12$ electrons. This is an expanded octet, so the octet is not complete. (Not Met)
• (III) Geometry at underlined atom is tetrahedral: The hybridization is sp³d² (steric number 6), which corresponds to an octahedral electron geometry. With 4 bond pairs and 2 lone pairs, the molecular geometry around Iodine is square planar. It is not tetrahedral. (Not Met)
• Conclusion for I₂Cl₆: Fails conditions (II) and (III).

Based on the analysis, only Al₂Cl₆ satisfies all three given characteristics.

The final answer is $\boxed{\text{C}}$.

Core Solution:

1. B₂H₆ (Diborane): Boron is sp³ hybridized. Vacant p-orbital involved in 3c-2e bonding. Octet is not complete (6 electrons). Geometry is tetrahedral. (Fails II)
2. Si₂H₆ (Disilane): Silicon is sp³ hybridized. No vacant orbitals are involved in the sense of accepting electron pairs or expanding octet. Octet is complete. Geometry is tetrahedral. (Fails I)
3. Al₂Cl₆ (Aluminium Chloride Dimer): Aluminium is sp³ hybridized. An empty p-orbital accepts a lone pair from bridging Cl. Octet is complete (8 electrons). Geometry is tetrahedral. (All conditions Met)
4. I₂Cl₆ (Iodine Chloride Dimer): Iodine is sp³d² hybridized. Vacant d-orbitals are involved. Octet is not complete (12 electrons, expanded octet). Geometry is square planar, not tetrahedral. (Fails II and III)

Therefore, Al₂Cl₆ is the correct answer.

The final answer is $\boxed{C}$