JEE Main 2025 Apr 3 Shift 1, Chemistry Q25: d-Block Elements

JEE Main2025Apr 3, Shift 1Chemistry

Consider the following reactions

$A + NaCl + H_{2} SO_{4} \to CrO_{2} Cl_{2} + Side Products$ (little amount)

$CrO_{2} Cl_{2} (Vapour) + NaOH \to B + NaCl + H_{2} O$

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

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

Your answer

Solution

This is the chromyl chloride test sequence.

A is a dichromate (or chromate) salt; reaction with NaCl in conc. $H_{2} SO_{4}$ generates the deep red vapour $CrO_{2} Cl_{2}$ .

Passing $CrO_{2} Cl_{2}$ into NaOH gives sodium chromate: $B = Na_{2} CrO_{4}$ (yellow).

Acidification of chromate gives dichromate (orange): $C = Na_{2} Cr_{2} O_{7}$ , i.e. the $Cr_{2} O_{7}^{2 -}$ ion.

The structure of $Cr_{2} O_{7}^{2 -}$ has two $CrO_{3}$ tetrahedra sharing one bridging oxygen. Of the 7 oxygens, 1 is bridging and the remaining $7 - 1 = 6$ are terminal (three on each Cr).

$∴$ Number of terminal O atoms in C $= 6$ .

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JEE Main 2025 Apr 3 Shift 1, Chemistry Q25: d-Block Elements

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