P. Bahadur Test: The Solid State (Old NCERT) - JEE MCQ

In fcc, a total of 4 octahedral voids are present.
So, it contains 4 A and 4 B atoms but  all the particles along the plane as shown in the figure given in the question are removed.This plane contains 3/2 octahedral void atoms (1 at body center and 2 at edge center) and 3/2 lattice atoms (2 face centers and 4 corners).
So, the formula will be A3/2 B3/2 = AB.

(D)
When all particle along are body diagonal one removed, these 2 X atoms from corner are removed, one Y particle removed & 2 Z particle removed.
 
Hence new arrangement, X particle = 
Hence formula =  

From the sizes of octahedral and tetrahedral voids, it is clear that the atoms occupying these voids will not touch each other as we have along body diagonal of FCC.

(B)
HCP = AB AB AB . pattem repeat For calculating voids between two layers A and B. Total tetrahedral voids = 12 (represented by dots) out of which 8 are completely inside but rest are shared by other unit cells. Total octahedral voids = 6 (represented by cross). All are completely inside. 

and a_FCC = 1
 

The correct answer is option A

 In a simple cubic structure, the spheres are not packed as closely as they could be, and they only “fill” about 52% of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. A solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom
 directly below it in the layer below. The number of other particles that each particle in crystalline solid contact is known as its coordination number. For a polonium atom in a simple cubic array, the coordination number is, therefore, six.

Cube which contains 8 unit cell.
By observing face ABCP ,we can say that the face of the cube look like in given figure by this P Q R S can be considered as a face of a unit cell which is a simple cubic arrangement.
So n = 8 for this cube.
Since only one atom is present in simple cubic.
In simple cubic 2r = a.

Formula  ZnAl₂O₄
Packing fraction = 

Parallel layer are at a distance of R

In 1 mole of M_xO,
mole of M²⁺ ion = y
∴ mole of M³⁺ ion = x - y.
Applying charge balance, 2y + 3(x-y)-2 = 0
⇒ y = 3x - 2

Total charge of nickel.
(0.96 x 2) + (0.04 x 3) = 2.04.

Formula of solid = NiO_1.02 = Ni_0.98O.

Based on theory.

At room temperature, thermal energy causes some electrons of the Si-Si bonds to leave.The electron deficient bond becomes a positive hole. Both the positive holes and electrons released account for the conduction of electricity.

Based on Fluorite structure of CaF₂

The correct answer is option  A

NaCl type of crystal lattice is a fcc arrangement in which Cl ions and present at the centre of the lattice and at each corner whereas Sodium and is present at each of it's centre.
The radius ratio b/n Na⁺ and Cl- ion is 0.414 to 0.731