The anions (A) of a C_{p}A_{q} molecule form | vedclass

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

Question

(D) In an $fcc$ unit cell,anions $(A)$ are present at $8$ corners and $6$ face centers.Effective number of $A = (\frac{1}{8} 	imes 8) + (\frac{1}{2}

Vedclass · Accepted Answer

$C_{5}A_{8}$

Vedclass · Answer

(D) In an $fcc$ unit cell,anions $(A)$ are present at $8$ corners and $6$ face centers.Effective number of $A = (\frac{1}{8} 	imes 8) + (\frac{1}{2} 	imes 6) = 1 + 3 = 4$.Cations $(C)$ are present at the body center $(1)$ and half of the edge centers ($12 	imes \frac{1}{2} = 6$ edges).Effective number of $C = 1 + (6 	imes \frac{1}{4}) = 1 + 1.5 = 2.5 = \frac{5}{2}$.The ratio of $C:A = \frac{5}{2} : 4 = 5 : 8$.Thus,the formula of the molecule is $C_{5}A_{8}$.

The anions $(A)$ of a $C_{p}A_{q}$ molecule form an $fcc$ lattice. Cations $(C)$ are positioned at the body center and half of the edge centers. The formula of the molecule is:

Similar Questions

In an ionic solid,anions $(B)$ are arranged in an $hcp$ array and cations $(A)$ occupy $\frac{2}{3}$ of the octahedral voids. What is the formula of the ionic compound?

Atoms $A$ are arranged in an $FCC$ system,and atoms $B$ occupy all the octahedral voids and half of the tetrahedral voids. The formula of the compound is:

If $Z$ is the number of atoms in the unit cell that represents the closest packing sequence $ABCABC .....,$ the number of tetrahedral voids in the unit cell is

Identify the types of voids represented by the structures $I$,$II$,and $III$ in the given image.

In an ionic crystalline solid,atoms of element $Y$ form an $hcp$ structure. The atoms of element $X$ occupy one-third of the tetrahedral voids. What is the formula of the compound?

Vedclass Test Series

Exam Paper Generator

Online Exam Module