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NEET CHEMISTRYMedium

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

1

C2A3C_2A_3

2

C3A2C_3A_2

3

C3A4C_3A_4

4

C4A3C_4A_3

Step-by-Step Solution

Let the number of anions (A) be NN. Number of octahedral voids = NN. Number of cations (C) = 75% of N=0.75N=34N75\% \text{ of } N = 0.75N = \frac{3}{4}N. Ratio of C:A = 34N:N=3:4\frac{3}{4}N : N = 3:4. Formula is C3A4C_3A_4.

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