Group 15 Halides Statement Analysis

Let's analyze each statement provided regarding the properties and reactions of Group 15 halides.

Statement (i): AsCl₃ complex formation

The statement says that AsCl₃ can form complexes of the type [AsCl₄]^- in the presence of a chloride source. AsCl₃ has Arsenic in the +3 oxidation state. Arsenic in AsCl₃ is bonded to three chlorine atoms and has one lone pair of electrons. It has an incomplete octet (3 bonding pairs + 1 lone pair = 4 electron pairs, but only 6 valence electrons from bonding). AsCl₃ acts as a Lewis acid due to the presence of a vacant orbital on Arsenic and its ability to accept an electron pair. A chloride ion (Cl^-) acts as a Lewis base, donating an electron pair.

When AsCl₃ reacts with a chloride source (like a salt containing Cl^- ions), it can accept a chloride ion:

\(\text{AsCl}_3 + \text{Cl}^- \rightarrow [\text{AsCl}_4]^-\)

In the complex [AsCl₄]^-, Arsenic is bonded to four chlorine atoms and has one lone pair. The electron geometry around Arsenic is trigonal bipyramidal, but the molecular geometry is tetrahedral. This complex formation is well-known for Group 15 trihalides (except Nitrogen). Thus, statement (i) is correct.

Statement (ii): PF₃ as a σ-donor ligand

The statement claims PF₃ acts as a strong σ-donor ligand towards d-metals. PF₃ is a phosphine ligand. Phosphine ligands (PR₃) are known to act as both σ-donors (through the lone pair on Phosphorus) and π-acceptors (using vacant low-lying orbitals, like $\sigma^*$ orbitals of P-F bonds or historically, d-orbitals on P, though the latter is less favoured now). The electronegative fluorine atoms in PF₃ withdraw electron density from the phosphorus atom. This withdrawal makes the lone pair on phosphorus less available for σ-donation compared to phosphines with less electronegative substituents (like PMe₃ or PPh₃). Conversely, the electron withdrawal makes the π-acceptor orbitals more accessible for receiving electron density from the metal. Therefore, PF₃ is considered a poor σ-donor but a very strong π-acceptor ligand. Describing it as a "strong σ-donor" is inaccurate. Thus, statement (ii) is incorrect.

Statement (iii): Solid SbF₅ structure

The statement says that in the solid state, SbF₅ has a trigonal bipyramidal structure. Group 15 pentafluorides generally have a trigonal bipyramidal structure in the gas phase. However, in the solid state, most of them polymerize to form more stable structures, typically involving octahedral coordination around the central atom through bridging fluorine atoms. SbF₅ in the solid state forms polymeric chains where each Sb atom is octahedrally coordinated, linked by shared fluorine atoms. This structure is far from a simple trigonal bipyramid. Thus, statement (iii) is incorrect.

Statement (iv): Reaction of SbF₅ with anhydrous HF

The statement says that the reaction of SbF₅ with anhydrous HF generates [H₂F]⁺ ions. SbF₅ is a very strong Lewis acid due to the high positive charge on Sb (+5) and the presence of available d-orbitals to accept electron density. Anhydrous HF is a weak acid, but it can act as a source of fluoride ions (F^-) or act as a base by donating a proton (in some contexts) or accepting a proton. In the presence of a strong Lewis acid like SbF₅, HF can donate a fluoride ion. The reaction proceeds as follows:

\(\text{SbF}_5 + \text{HF} \rightarrow \text{H}^+ + [\text{SbF}_6]^-\)

The proton (H⁺) generated is highly acidic and is immediately solvated by excess HF molecules, forming fluoronium ions such as [H₂F]⁺ or higher aggregates like (HF)_nH⁺. The formation of [H₂F]⁺ is a characteristic feature of very strong acid systems involving HF as the solvent and a strong fluoride acceptor like SbF₅. The mixture of HF and SbF₅ is a well-known superacid system (Fluoroantimonic acid). Thus, statement (iv) is correct.

Summary of Statement Correctness

• Statement (i): Correct
• Statement (ii): Incorrect
• Statement (iii): Incorrect
• Statement (iv): Correct

Based on the analysis, the correct statements are (i) and (iv).