Question

Choose the correct statements.

• A. Conc. H\textsubscript{2}SO\textsubscript{4} on reaction with potassium chlorate gives chloric acid
• B. Marshall's acid on partial hydrolysis gives two acids 'X' and 'Y' both having 'S' in sp\textsuperscript{3} hybridisation state
• C. Aqueous solution of SO\textsubscript{2} decolorises I\textsubscript{2} solution
• D. I\textsubscript{2} solution is decolorised by Na\textsubscript{2}S\textsubscript{2}O\textsubscript{3}, giving a sulphur containing compound 'X', where the sum of oxidation number of all the S-atoms is +10

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

The question asks us to identify the correct statements among the given options. Let's analyze each statement:

A. Conc. H₂SO₄ on reaction with potassium chlorate gives chloric acid

Potassium chlorate (KClO₃) reacts with concentrated sulfuric acid (H₂SO₄) to produce chlorine dioxide (ClO₂) and potassium persulfate (K₂S₂O₈) or potassium sulfate (K₂SO₄) and potassium perchlorate (KClO₄). The reaction is known to be quite violent and can be explosive. A common representation of the reaction is: $3\text{KClO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{KHSO}_4 + \text{KClO}_4 + 2\text{ClO}_2 + \text{H}_2\text{O}$

Chloric acid (HClO₃) is not the primary product; rather, chlorine dioxide (ClO₂) gas is formed. Therefore, statement A is incorrect.

B. Marshall's acid on partial hydrolysis gives two acids 'X' and 'Y' both having 'S' in sp³ hybridisation state

Marshall's acid is peroxodisulfuric acid (H₂S₂O₈). Its structure is $\text{HO}_3\text{S-O-O-SO}_3\text{H}$.

In H₂S₂O₈, each sulfur atom is bonded to four oxygen atoms (two double-bonded oxygens, one oxygen in an -OH group, and one oxygen in the peroxide linkage). Since sulfur forms four sigma bonds and has no lone pairs, its steric number is 4, indicating sp³ hybridization.

Partial hydrolysis of H₂S₂O₈ gives: $\text{H}_2\text{S}_2\text{O}_8 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_5 + \text{H}_2\text{SO}_4$

Here, 'X' is Caro's acid (peroxomonosulfuric acid, H₂SO₅) and 'Y' is sulfuric acid (H₂SO₄).

Let's check the hybridization of sulfur in H₂SO₅ and H₂SO₄:

• H₂SO₅ (Caro's acid): Structure is $\text{HO-O-SO}_3\text{H}$. The sulfur atom is bonded to four oxygen atoms (two double-bonded oxygens, one oxygen in an -OH group, and one oxygen in the peroxide linkage). Sulfur forms four sigma bonds and has no lone pairs, so it is sp³ hybridized.
• H₂SO₄ (Sulfuric acid): Structure is $(\text{HO})_2\text{SO}_2$. The sulfur atom is bonded to four oxygen atoms (two double-bonded oxygens and two oxygens in -OH groups). Sulfur forms four sigma bonds and has no lone pairs, so it is sp³ hybridized.

Since both H₂SO₅ and H₂SO₄ have sulfur in sp³ hybridization, statement B is correct.

C. Aqueous solution of SO₂ decolorises I₂ solution

Sulfur dioxide (SO₂) acts as a reducing agent in aqueous solution. It reacts with iodine (I₂) solution, reducing I₂ to iodide ions (I⁻), which are colorless.

The reaction is: $\text{SO}_2(\text{aq}) + \text{I}_2(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_4(\text{aq}) + 2\text{HI}(\text{aq})$

In this reaction, iodine (oxidation state 0) is reduced to iodide (-1), causing the brown/yellow I₂ solution to decolorize. Sulfur (oxidation state +4 in SO₂) is oxidized to sulfur (+6 in H₂SO₄). Therefore, statement C is correct.

D. I₂ solution is decolorised by Na₂S₂O₃, giving a sulphur containing compound 'X', where the sum of oxidation number of all the S-atoms is +10

This is a standard reaction used in iodometric titrations. Sodium thiosulfate (Na₂S₂O₃) reacts with iodine (I₂): $2\text{Na}_2\text{S}_2\text{O}_3(\text{aq}) + \text{I}_2(\text{aq}) \rightarrow \text{Na}_2\text{S}_4\text{O}_6(\text{aq}) + 2\text{NaI}(\text{aq})$

Iodine (I₂) is reduced to iodide (I⁻), leading to the decolorization of the iodine solution. The sulfur-containing compound 'X' is sodium tetrathionate (Na₂S₄O₆).

Let's determine the sum of oxidation numbers of all sulfur atoms in Na₂S₄O₆. The tetrathionate ion (S₄O₆²⁻) has the structure: $\text{O}_3\text{S-S-S-SO}_3^{2-}$

There are two terminal sulfur atoms and two central sulfur atoms.

Using the formal charge/bond method for oxidation states:

• The two central sulfur atoms are bonded only to other sulfur atoms. Since S-S bonds are non-polar, the oxidation state of these two central sulfur atoms is 0.
• The two terminal sulfur atoms are each bonded to three oxygen atoms and one sulfur atom.
  • Each S=O double bond contributes +2 to the sulfur's oxidation state. There are two S=O bonds for each terminal sulfur, so +4.
  • Each S-O single bond contributes +1 to the sulfur's oxidation state. There is one S-O single bond for each terminal sulfur, so +1.
  • The S-S bond contributes 0 to the sulfur's oxidation state.
  • Thus, the oxidation state of each terminal sulfur atom is +4 + 1 = +5.

The oxidation states of the four sulfur atoms are +5, 0, 0, and +5.

The sum of the oxidation numbers of all S-atoms = (+5) + (0) + (0) + (+5) = +10.

Therefore, statement D is correct.

Conclusion: Statements B, C, and D are correct.