A solid has CCP arrangement having atoms A, B | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In a $CCP$ (cubic close packing) arrangement:$1$. Number of atoms at corners $(B)$ = $8 	imes (1/8) = 1$.$2$. Number of atoms at face centres $(A

Vedclass · Accepted Answer

$A_3 B C_4$

Vedclass · Answer

(A) In a $CCP$ (cubic close packing) arrangement:$1$. Number of atoms at corners $(B)$ = $8 	imes (1/8) = 1$.$2$. Number of atoms at face centres $(A)$ = $6 	imes (1/2) = 3$.$3$. Total number of tetrahedral voids = $2 	imes (	ext{number of atoms in } CCP) = 2 	imes 4 = 8$.$4$. Number of $C$ atoms occupying $50\%$ of tetrahedral voids = $0.50 	imes 8 = 4$.Therefore,the ratio of atoms is $A:B:C = 3:1:4$.The molecular formula is $A_3 B C_4$.

$A$ solid has $CCP$ arrangement having atoms $A, B, C$. If $A$ atoms are present at face centres,$B$ at corners and $C$ atoms occupy $50\%$ tetrahedral voids,then the molecular formula of the solid will be:

Similar Questions

In $FCC$,the distance between the nearest tetrahedral voids is:

The molecular formula of a crystalline solid is $X_3Y_2$. Atoms of $Y$ form $ccp$ lattice,and atoms of $X$ occupy $50\%$ octahedral voids and $x\%$ of tetrahedral voids. What is the percentage of unoccupied tetrahedral voids?

Which of the following lattices has the highest packing efficiency: $(i)$ simple cubic,$(ii)$ body-centred cubic,and $(iii)$ hexagonal close-packed lattice?

If the anion $(A)$ forms a hexagonal closest packing and the cation $(C)$ occupies only $2/3$ of the octahedral voids in it,then the general formula of the compound is :-

In antifluorite structure,the negative ions

Vedclass Test Series

Exam Paper Generator

Online Exam Module