Question

A compound formed by cation c and anion a the anions form hcp lattice and the cations occupy 75% of octahedral voids the formula of the compound is

Answer 1

coordination number of hexagonal close packed structure (HCP) = 6

and also co-ordination number of octahedral voids = 6

a/c to question,

A form hcp lattice

so, number of A in compound = 6

C occupy 75% of octahedral voids ,

so, number of C in compound = 75% of 6

= 3/4 × 6 = 9/2

now, formula will be $C_{9/2}A_6$ or, $C_3A_4$

hence, required formula of compound is C4A4.

Answer 2

Answer:

C3A4

Explanation:

since A forms hcp that means it is located at corners as well as at faces of lattice. hence adding the contributions of both its 4.......[corners=1/8*8=1......(i)

                                                                                              face=1/2*6=3..........(ii) ]

as we know octahedral voids has the same value  as that of z which is 4.

and here it occupies 75%.so 75/100*4=3.

so,

A4C3