NEET-XII-Chemistry
Previous Year Paper year:2017
- #85A 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 :
(1)``C_2A_3``
(2)``C_3A_2``
(3) ``C_3A_4``
(4) ``C_4A_3``digAnsr: 3Ans : ( 3 )21
Sol. • Anions(A) are in hcp, so number of anions
(A) = 6
Cations(C) are in 75% O.V., so number of
cations (C)
=
3
6×
4
=
18
4
=
9
2
• So formula of compound will be
&implies;
9 6 9 12
2
C A C A
C
9
A
12
&implies; C
3
A
4