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

Question

A compound is formed by cation C and anion A. The anions form hexagonal close packed (hep) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is : (A) C3A4 (B) C4A3 (C) C2A3 (D) C3A2

Tardigrade Admin · Accepted Answer

bullet Anions(A) are in hcp, so number of anions (A) = 6 Cations(C) are in 75% O.V., so number of cations (C) = 6 × (3/4) = (18/4) = (9/2) bullet So formula of compound will be C(9/2) A6 ⇒ C9A12 C9 A12 ⇒ C3A4

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

$C_{3} A_{4}$

$C_{4} A_{3}$

$C_{2} A_{3}$

$C_{3} A_{2}$

$∙$ Anions(A) are in hcp, so number of anions (A) = 6
Cations(C) are in 75% O.V., so number of cations (C)
$= 6 \times \frac{3}{4}$
$= \frac{18}{4}$
$= \frac{9}{2}$
$∙$ So formula of compound will be
$C_{\frac{9}{2}} A_{6} \Rightarrow C_{9} A_{12}$
$C_{9} A_{12} \Rightarrow C_{3} A_{4}$

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

C3​A4​

C4​A3​

C2​A3​

C3​A2​

∙ Anions(A) are in hcp, so number of anions (A) = 6 Cations(C) are in 75% O.V., so number of cations (C) =6×43​ =418​ =29​ ∙ So formula of compound will be C29​​A6​⇒C9​A12​ C9​A12​⇒C3​A4​

$C_{3} A_{4}$

$C_{4} A_{3}$

$C_{2} A_{3}$

$C_{3} A_{2}$

$∙$ Anions(A) are in hcp, so number of anions (A) = 6
Cations(C) are in 75% O.V., so number of cations (C)
$= 6 \times \frac{3}{4}$
$= \frac{18}{4}$
$= \frac{9}{2}$
$∙$ So formula of compound will be
$C_{\frac{9}{2}} A_{6} \Rightarrow C_{9} A_{12}$
$C_{9} A_{12} \Rightarrow C_{3} A_{4}$