Consider the following reactions A + NaCl + H2SO4 -> CrO2Cl2 + Side Products Little amount CrO2Cl2 (Vapour) + NaOH -> B + NaCl + H2O B + H+ -> C + H2O The number of terminal 'O' present in the compound 'C' is _______.

The correct answer is Option 6. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q73 – JEE-MAIN 2025

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Home Chemistry d- & f-Block ElementsQuestion #367

Consider the following reactions

A + N a C l + H_{2} S O_{4} \to C r O_{2} C l_{2} + Side Products

Little amount

C r O_{2} C l_{2} (Vapour) + N a O H \to B + N a C l + H_{2} O

B + H^{+} \to C + H_{2} O

The number of terminal 'O' present in the compound 'C' is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Identify the reactions

The first reaction is the chromyl chloride test, which helps identify the initial chromium compound.

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Step 1: Identify compounds A, B, and C✦ Active

The first reaction, $A + N a C l + H_{2} S O_{4} \to C r O_{2} C l_{2} + Side Products$ , is the chromyl chloride test. This test indicates that A is a chromate or dichromate salt (e.g., $K_{2} C r_{2} O_{7}$ ). Chromyl chloride ( $C r O_{2} C l_{2}$ ) is a reddish-brown vapour.

Chromyl chloride reacts with a strong base like $N a O H$ to form a chromate. The balanced reaction is:

C r O_{2} C l_{2} + 4 N a O H \to N a_{2} C r O_{4} + 2 N a C l + 2 H_{2} O

Therefore, compound B is $N a_{2} C r O_{4}$ (sodium chromate).

When chromate ions ( $C r O_{4}^{2 -}$ ) are acidified ( $H^{+}$ ), they convert to dichromate ions ( $C r_{2} O_{7}^{2 -}$ ). The reaction is:

2 N a_{2} C r O_{4} + 2 H^{+} \to N a_{2} C r_{2} O_{7} + H_{2} O + 2 N a^{+}

Therefore, compound C is $N a_{2} C r_{2} O_{7}$ (sodium dichromate).

Step 2: Determine the structure of compound C and count terminal oxygen atoms○ Expand

Compound C is sodium dichromate, $N a_{2} C r_{2} O_{7}$ . We need to find the number of terminal oxygen atoms in the dichromate ion, $C r_{2} O_{7}^{2 -}$ .

The dichromate ion consists of two $C r O_{4}$ tetrahedra sharing one oxygen atom. The structure can be represented as $O_{3} C r - O - C r O_{3}$ . Each chromium atom is in a +6 oxidation state and is tetrahedrally coordinated.

In this structure, one oxygen atom acts as a bridge between the two chromium atoms ( $C r - O - C r$ ). The remaining six oxygen atoms are bonded to only one chromium atom. These are the terminal oxygen atoms.

💡 Teacher's Secret Hint

Remember the bridging oxygen in dichromate.

Step 3: Final Answer○ Expand

The number of terminal oxygen atoms in the dichromate ion ( $C r_{2} O_{7}^{2 -}$ ) is 6.

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Chemistry Question 73 – JEE-MAIN 2025

Clue 1 of 3: Identify the reactions

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