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Consider the sequence of reactions: FeCr2O4 + Na2CO3 + O2 -> A + Fe2O3 + CO2, and then A + H+ -> B + H2O + Na+. How many terminal oxygen atoms are present in product B?

1. 4
2. 6
3. 8
4. 2

Correct answer: 6

Solution

Roasting chromite gives sodium chromate (A = Na2CrO4); on acidification it converts to dichromate (B = Cr2O7²-), which has one bridging O and six terminal O atoms (three on each Cr).

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