The p-Block Elements JEE 2021: Cl2O6 Oxidation State

What was the numerical question on average oxidation state of chlorine from ozonolysis of ClO2 in JEE Advanced 2021 Paper 2?

The question asked for the average oxidation state of chlorine in the higher oxide produced by the reaction of chlorine dioxide with ozone. It was a numerical-answer format question from p-Block Elements in JEE Advanced 2021 Paper 2, of medium difficulty with roughly 60 seconds expected solving time.

The reaction of chlorine dioxide with ozone produces a higher oxide of chlorine. The task is to find the average oxidation state of chlorine in the oxide formed.

What is the official step-by-step solution to the JEE Advanced 2021 oxidation state numerical?

Ozonolysis refers to reaction with O3 that oxidises ClO2. The balanced equation is $2 C l O_{2} + 2 O_{3} \to C l_{2} O_{6} + 2 O_{2}$ . The product identified is Cl2O6, where the average oxidation state of each chlorine is +6.

Let average oxidation state of each Cl be x. Oxygen is assigned an oxidation state of -2 in neutral oxides. For the neutral molecule the equation is $2 x + 6 (- 2) = 0$ . This solves to $2 x = 12$ , so $x = + 6$ .

Structural formula of Cl2O6 showing two chlorine atoms linked by an oxygen bridge with additional terminal oxygens, labels indicating possible +6 and +6 average on each chlorine

What common method mistake leads to +7 instead of +6 in this oxidation state question?

The common method mistake is assuming the product is Cl2O7 (the highest oxide) instead of verifying the specific ozonolysis product Cl2O6. This produces the wrong equation $2 x + 7 (- 2) = 0$ which solves to $x = + 7$ .

The error arises from skipping the balanced reaction and jumping to the highest possible oxidation state of chlorine. Identify the exact product first so the average stays at +6.

What core concepts on chlorine oxides and oxidation states does the 2021 question reinforce?

Oxygen is assigned -2 in neutral oxides. Chlorine exhibits +4 in ClO2 and average +6 in Cl2O6.

Average oxidation state must be used when the molecule contains multiple chlorine atoms that may not be equivalent. The balanced equation is mandatory to identify the correct oxide before assigning states.

What related p-block questions on chlorine oxides and oxidation states use the same logic?

Question 1: What is the oxidation state of chlorine in Cl2O7? Compare it with the average value in Cl2O6 from the 2021 question.

Oxygen is -2 and the molecule is neutral, so $2 x + 7 (- 2) = 0$ gives $x = + 7$ . This is one unit higher than the +6 average in Cl2O6 and confirms Cl2O7 as chlorine's maximum oxidation state.

Question 2: When ClO2 is passed through NaOH it forms NaClO2 and NaClO3. What are the oxidation states of chlorine in these products?

In ClO2^- let oxidation state of Cl be y: $y + 2 (- 2) = - 1$ solves to $y = + 3$ . In ClO3^- the equation $y + 3 (- 2) = - 1$ solves to $y = + 5$ . Chlorine disproportionates from +4 in ClO2 to +3 and +5.

Question 3: In the disproportionation of Cl2O6 with water, chlorine changes to +5 and +7 oxidation states. Verify these values.

For ClO3^- the equation $x + 3 (- 2) = - 1$ gives $x = + 5$ . For ClO4^- the equation $x + 4 (- 2) = - 1$ gives $x = + 7$ . The values match only after the balanced equation identifies the products.

How should you prepare for oxidation state numericals in the p-block for JEE Advanced?

Memorise specific reactions of ClO2 with ozone and other reagents rather than all p-block trends. You can expect faster recall in future papers because these exact reactions have repeated since 2007.

Always write the balanced equation before assigning oxidation numbers in numerical questions. This habit eliminates the +7 trap.

Use the free past-paper archive to locate every p-block numerical from 2007 onward. The expected time of 60 seconds means the reaction and equation must be recalled instantly.

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Frequently asked questions

What is the average oxidation state of chlorine in Cl2O6 in JEE Advanced 2021?

The average oxidation state of each chlorine in Cl2O6 is +6. For the neutral molecule, the equation 2x + 6(-2) = 0 solves to x = +6. This comes from the balanced reaction of ClO2 with ozone producing Cl2O6 as the higher oxide.

What is the balanced equation for ozonolysis of ClO2 in JEE 2021?

The balanced equation is 2ClO2 + 2O3 → Cl2O6 + 2O2. This reaction produces Cl2O6 where chlorine has an average oxidation state of +6. Always identify the exact product from the balanced equation before calculating oxidation states.

Why do students get +7 instead of +6 in the p-block JEE 2021 question?

Many students assume the product is Cl2O7 without checking the specific reaction of ClO2 with ozone. This leads to the equation 2x + 7(-2) = 0 giving x = +7. The actual product is Cl2O6, so the average oxidation state remains +6.

What is the oxidation state of chlorine in Cl2O7?

Chlorine has +7 oxidation state in Cl2O7. Solving 2x + 7(-2) = 0 for the neutral molecule gives x = +7. This is the maximum oxidation state of chlorine, one unit higher than the average +6 found in Cl2O6 from the 2021 question.

The p-Block Elements JEE 2021: Oxidation State of Cl in Cl2O6