Molecular formula of chromite is:

F e C r_{2} O_{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F e_{2} C r_{3} O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e C r O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C u_{2} H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $F e C r_{2} O_{4}$
The only commercially important ore of $C r$ is chromite, $F e C r_{2} O_{4}$ .
Chromite has a spinel structure. In this structure the $O$ atoms are arranged in a cubic close-packed lattice with $F e^{I I}$ in one eighth of the available tetrahedral voids and $C r^{I I I}$ in one quarter of the octahedral voids.

Suggest Corrections