(B) In a $ccp$ structure,there are $8$ corners and $6$ face centres. Number of $A$ atoms initially at corners $= 8 \times \frac{1}{8} = 1$. Since $2$

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

(B) In a $ccp$ structure,there are $8$ corners and $6$ face centres. Number of $A$ atoms initially at corners $= 8 \times \frac{1}{8} = 1$. Since $2$ $A$ atoms are missing from the corners,the number of $A$ atoms remaining $= 1 - (2 \times \frac{1}{8}) = 1 - 0.25 = 0.75 = \frac{3}{4}$. Number of $B$ atoms at face centres $= 6 \times \frac{1}{2} = 3$. The ratio of $A : B = \frac{3}{4} : 3 = 3 : 12 = 1 : 4$. Therefore,the composition of the alloy is $AB_4$.

Class 12 Chemistry Solid State Crystal structure and Coordination number

Medium

The unit cell of a binary compound of $A$ and $B$ metals has a $ccp$ structure with $A$ atoms occupying the corners and $B$ atoms occupying the centres of each face of the cubic unit cell. If during the crystallisation of this alloy,two $A$ atoms are missing from the unit cell,the overall composition per unit cell is:

A
$AB_6$
B
$AB_4$
C
$AB_8$
D
$A_6B_{24}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solid State

Subtopic

Crystal structure and Coordination number

$A$ solid $AB$ crystallises in $NaCl$ structure. The anion $B$ forms $fcc$ lattice,while the cation $A^{+}$ occupies all the octahedral holes. If all the particles along a diagonal plane (as shown in the figure) are removed,then the formula of the solid becomes

View Solution

$A^{+}B^{-}$ shows rock salt type arrangement. Edge length of unit cell is $6 \ \mathring{A}$. If radius of $A^{+}$ is $1 \ \mathring{A}$,then the incorrect option is:

Difficult

View Solution

In a crystal of substance $A_xB_y$,$A$ atoms are present at each corner of the face-centered cubic $(FCC)$ unit cell,and $B$ atoms are present at each face center. What is the correct formula of the substance $A_xB_y$?

Medium

View Solution

How many atoms are there in a cube-based unit cell,having one atom on each corner and $2$ atoms on each body diagonal of the cube?

EasyKCET 2022

View Solution

$A$ solid $AB$ has a $NaCl$ structure. $B$ atoms are at the corners and all face-centered positions of the unit cell,while $A$ atoms occupy all the $O.H.V.$ (Octahedral Voids). If all the atoms from the face-centered positions along one axis are removed,what will be the resulting stoichiometry of the solid?

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ solid $AB$ crystallises in $NaCl$ structure. The anion $B$ forms $fcc$ lattice,while the cation $A^{+}$ occupies all the octahedral holes. If all the particles along a diagonal plane (as shown in the figure) are removed,then the formula of the solid becomes

$A^{+}B^{-}$ shows rock salt type arrangement. Edge length of unit cell is $6 \ \mathring{A}$. If radius of $A^{+}$ is $1 \ \mathring{A}$,then the incorrect option is:

In a crystal of substance $A_xB_y$,$A$ atoms are present at each corner of the face-centered cubic $(FCC)$ unit cell,and $B$ atoms are present at each face center. What is the correct formula of the substance $A_xB_y$?

How many atoms are there in a cube-based unit cell,having one atom on each corner and $2$ atoms on each body diagonal of the cube?

Vedclass Test Series

Exam Paper Generator

Online Exam Module