A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is:

C_{2} A_{3}

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

C_{3} A_{2}

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

C_{3} A_{4}

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

C_{4} A_{3}

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

Open in App

Solution

The correct option is C $C_{3} A_{4}$
$Anions (A) are in hcp, so the number of anions (A) = 6 Cations (C) are in 75 % of octahedral voids, so the number of cations (C) = 6 \times \frac{3}{4} = \frac{18}{4} = \frac{9}{2} So the formula of the compound will be C_{9 / 2} A_{6} \Rightarrow C_{9} A_{12} \Rightarrow C_{3} A_{4}$

Suggest Corrections

Similar questions

C

A

75 %

(A)

(C)

\frac{2}{3}

(A)

(C)

2 / 3^{r d}

Join BYJU'S Learning Program