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D and F Block Elements question

2022 · 29 Jun · Shift 1 · Q15

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D and F Block Elements
2022 · 29 Jun · Shift 1 · Q15

D and F Block Elements question

2022 · 29 Jun · Shift 1 · Q15

JEE MainChemistryD and F Block ElementsNumerical+4 / −1

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

\underline{\hspace{2cm}}

.

FeCr_2O_4

+

Na_2CO_3

+

O_2

\to

A +

Fe_2O_3

+

CO_2

A +

H^+

\to

B +

H_2O

+

Na^+

Numerical answer

View written solutionFree

Correct answer: 6

Identify compound $A$ formed in the roasting reaction

Chromite ore is $FeCr_2O_4$ . On fusion with sodium carbonate in presence of oxygen, chromium is converted into sodium chromate.

The reaction is: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$

Hence, $A = Na_2CrO_4$
Acidification of sodium chromate

On acidification, chromate ions convert into dichromate ions: $2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O$

Therefore, product $B$ is dichromate species, i.e. chromic acid form / dichromate ion corresponding to acidified medium.

In terms of structure, the dichromate unit is: $[O_3Cr-O-CrO_3]^{2-}$
Count terminal oxygen atoms in $B$

In dichromate, there are total 7 oxygen atoms:
- 1 oxygen is bridging between the two Cr atoms.
- Remaining $7-1=6$ oxygen atoms are terminal oxygen atoms.
So, number of terminal oxygen atoms in $B$ is: $6$
Final answer

$\boxed{6}$

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