Hard INTEGER +4 / -1 PYQ · JEE Mains 2020

Consider the following reactions :

$NaCl{\rm{ }} + {\rm{ }}{K_2}C{r_2}{O_7} + \mathop {{H_2}S{O_4}}\limits_{(conc.)}$$\to$ (A) + side products

(A) + NaOH $\to$ (B) + Side products

$$\left( B \right){\rm{ }} + \mathop {{H_2}S{O_4}}\limits_{(dilute)} + {\rm{ }}{H_2}{O_2}$$ $\to$ (C) + Side products

The sum of the total number of atoms in one molecule each of (A) and (B) and (C) is

Answer (integer) 18

Solution

$NaCl{\rm{ }} + {\rm{ }}{K_2}C{r_2}{O_7} + \mathop {{H_2}S{O_4}}\limits_{(conc.)}$$\to$ <br>2CrO<sub>2</sub>Cl<sub>2</sub>(A) + 4NaHSO<sub>4</sub> + 2KHSO<sub>4</sub> + 3H<sub>2</sub>O <br><br>CrO<sub>2</sub>Cl<sub>2</sub>(A) + 4NaOH $\to$ Na<sub>2</sub>CrO<sub>4</sub>(B) + 2NaCl + 2H<sub>2</sub>O <br><br>Na<sub>2</sub>CrO<sub>4</sub>(B) + 2H<sub>2</sub>SO<sub>4</sub> + 2H<sub>2</sub>O<sub>2</sub> $\to$ <br>CrO<sub>5</sub>(C) + 2NaHSO<sub>4</sub> + 3H<sub>2</sub>O <br><br>A = CrO<sub>2</sub>Cl<sub>2</sub> <br><br>B = Na<sub>2</sub>CrO<sub>4</sub> <br><br>C = CrO<sub>5</sub> <br><br>Total number of atom in A + B + C = 18

About this question

Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals

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