The number of terminal oxygen atoms present in the product B obtained from the following reaction is ______. $\mathrm{FeCr}_{2}\mathrm{O}_{4}+\mathrm{Na}_{2}\mathrm{CO}_{3}+\mathrm{O}_{2}\rightarrow \mathrm{A}+\mathrm{Fe}_{2}\mathrm{O}_{3}+\mathrm{CO}_{2}$ $\mathrm{A}+\mathrm{H}^{+}\rightarrow \mathrm{B}+\mathrm{H}_{2}\mathrm{O}+\mathrm{Na}^{+}$

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · d and f Block Elements

The number of terminal oxygen atoms present in the product B obtained from the following reaction is ______.

FeCr_{2} O_{4} + Na_{2} CO_{3} + O_{2} \to A + Fe_{2} O_{3} + CO_{2}

A + H^{+} \to B + H_{2} O + Na^{+}

Answer: 6

First reaction:

4 FeCr_{2} O_{4} + 8 Na_{2} CO_{3} + 7 O_{2} \to 8 Na_{2} CrO_{4} + 2 Fe_{2} O_{3} + 8 CO_{2}

Product A is sodium chromate (

Na_{2} CrO_{4}

).

Second reaction:

2 Na_{2} CrO_{4} + 2 H^{+} \to Na_{2} Cr_{2} O_{7} + 2 Na^{+} + H_{2} O

Product B is sodium dichromate (

Na_{2} Cr_{2} O_{7}

).

The structure of dichromate ion has 6 terminal oxygen atoms (3 on each Cr atom) and 1 bridging oxygen. Therefore, the answer is 6.

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer