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JEE Main 202229 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Morning Shift
Subject: Chemistry
Chapter: d and f Block Elements

JEE Main 2022 · 29 June · Morning Shift · Chemistry · d and f Block Elements

The number of terminal oxygen atoms present in the product B...

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

${FeCr}_{2}{O}_{4} + {Na}_{2}{CO}_{3} + {O}_{2} \to A + {Fe}_{2}{O}_{3} + {CO}_{2}$

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

The number of terminal oxygen atoms present in the product B obtained from the following reaction is FeCr _ 2 O _ 4 +

Correct answer

6

Step-by-step solution

$4{FeCr}_{2}{O}_{4} + 8{Na}_{2}{CO}_{3} + 7{O}_{2} \to 8{Na}_{2}{CrO}_{4} + 2{Fe}_{2}{O}_{3} + 8{CO}_{2}$

$2{Na}_{2}{CrO}_{4} + 2{H}^{+} \to \mathop{\mathop{{Na}_{2}{Cr}_{2}{O}_{7}}_{⏟}}_{B} + 2{Na}^{+} + {H}_{2}O$

The number of terminal oxygen atoms present in the product B obtained from the following reaction is FeCr _ 2 O _ 4 +

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