Question: Select the CORRECT statement(s) for the hydrolysis of SF₄...

Question

Select the CORRECT statement(s) for the hydrolysis of SF₄

Answer 1

Solution

The hydrolysis of SF₄ with excess water proceeds as follows:

SF₄(g) + 2H₂O(l) → SO₂(g) + 4HF(aq)

Let's analyze each statement:

Statement 1: In the transition state sulphur atom gets sp³d hybridised state.

The hydrolysis of SF₄ involves nucleophilic attack by water on the sulfur atom. This typically leads to an increase in the coordination number of sulfur in the transition state or intermediate. SF₄ has a seesaw geometry with sulfur being sp³d hybridized (4 bonds + 1 lone pair, steric number 5). When water attacks, the sulfur atom forms a bond with the oxygen atom of water while simultaneously an S-F bond starts to break. This transition state would involve sulfur interacting with 4 F atoms and 1 O atom, plus the lone pair. This gives a steric number of 6 (5 ligands + 1 lone pair), which corresponds to sp³d² hybridization. Therefore, the statement that sulfur is sp³d hybridized in the transition state is incorrect.

Statement 2: 'S' does not shows it's maximum oxidation state in product.

In SF₄, the oxidation state of sulfur is +4. In the product SO₂, the oxidation state of sulfur is also +4. The maximum oxidation state that sulfur can exhibit is +6 (e.g., in H₂SO₄ or SF₆). Since the oxidation state of sulfur in the product SO₂ (+4) is less than its maximum oxidation state (+6), this statement is correct.

Statement 3: In one of the product pπ-dπ bond(s) is/are present.

The products are SO₂ and HF. In HF, the bond is a single covalent bond between H and F. There are no π bonds. In SO₂, sulfur is bonded to two oxygen atoms. The structure of SO₂ is bent. The bonding in SO₂ involves sigma bonds and pi bonds. Sulfur is in the third period and has vacant 3d orbitals. Oxygen is in the second period and has filled 2p orbitals. The pi bonding in SO₂ involves the overlap of p orbitals on S and O, and also the overlap of filled p orbitals on O with vacant d orbitals on S (pπ-dπ bonding). Therefore, SO₂ contains pπ-dπ bonds. This statement is correct.

Statement 4: Hydrolysis product(s) obtained by 1 mole SF₄, requires 12 litre, 0.5M NaOH solution for displacement of all acidic hydrogen.

From the hydrolysis of 1 mole of SF₄, we obtain 1 mole of SO₂ and 4 moles of HF. SO₂ dissolves in water to form sulfurous acid, H₂SO₃. H₂SO₃ is a diprotic acid.

H₂SO₃ + 2NaOH → Na₂SO₃ + 2H₂O

1 mole of H₂SO₃ requires 2 moles of NaOH for complete neutralization. HF is a weak acid.

HF + NaOH → NaF + H₂O

4 moles of HF require 4 moles of NaOH for complete neutralization. Total moles of NaOH required for complete neutralization of the products from 1 mole of SF₄ = Moles of NaOH for H₂SO₃ + Moles of NaOH for HF = 2 moles + 4 moles = 6 moles. The volume of 0.5 M NaOH solution required is given by:

Volume (L) = Moles of solute / Molarity (mol/L)

Volume (L) = 6 moles / 0.5 mol/L = 12 L. So, 12 litres of 0.5 M NaOH solution are required. This statement is correct.

The correct statements are 2, 3, and 4.