R.C. Mukherjee Test: The Solid State (Old NCERT) - JEE MCQ

Total body diagonal = 4
two atoms in each body diagonal so total number of atoms = 4×2=8 and 8 atom on each corner, So its contribution will be

so total number of atoms = 8+1=9

In Fcc
Face contribution 1/2
total 6 faces so its contribution  atom and contribution by its corner will be

So total number of atom in F.C.C will be 4.

There are 7 atoms at the corner. so contribution by 7 atoms form corner will be

B will be at face so total contribution by it will be 3.

Rubber, P lastics and glass, all are amorphous solids so

In B.C.C
4r = a √3
2r = 1.73.
2×1.73 = a √3
Q = 200×10^–12 m = 200 pm

For N spheres, the number of tetrahedral holes is  2N

NaCl has octahredral structure

6 face centred atom contbriution

and 3 atom and 3 atoms are enclosed so total number of enclosed atom in the unit cell = 3 +3 = 6 atom

Z is ccp so its total number 4.
X atom occupy tetrahedral void so its total number will be 8
X₈Z₄ = X₂Z

Close packing is maximum in either face centrered ork in hexagonal

B.C.C. has the packing efficieny is 68%
so the vacant space in BCCk is 32%

option B 

a = 5.14 A°
structure is like NaCl

r^– = 1.815 A°

The total number of octahedral void(s) per atom present in a cubic close packed structure is 4. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4​th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is 1.
12 octahedral voids located at each edge and shared between four unit cells= 
Total number of octahedral voids =4
We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to 4/4=1.

In a face centred cubic lattice the number of nearest neighour for a given lattice point 12

Distance between two oppositely charge ions for BCC

= 335.15 pm

putting the value we get 180 pm

Zns - Zinc blende structure
B^- ions consisting the lattice so 4
A here will be 2
A₂ B₄ = AB₂.

Simple cubic lattice → B.C.C
So the Z will increase
so density will increase

Centre of edge is octehedral vold so

Zns is ionic erystal

LiF iso an ionic crstal

A solid having no definite shape is called Amorphous

CsCl has b.c.c. structure
CS⁺ will be at the centre of the cente of the and 8 Cl^— will occupy at the corner whch contribution will be 
lhus, 1 CS⁺ & 1 Cl^—

Close packing is maximum in F.C.C.

The correct answer is Option C.

For sodium chloride, there are 6 anions around each cation (coordination number of Na=6) because of the 1:1 stoichiometry, there must also be 6 cations around each Cl^-  ion.
Similarly, Mg²⁺O^2- also has the octahedral structure and follows NaCl type geometry. 
Therefore, the coordination number of Mg is also 6.
 

12 Na⁺ will find at centre of the edge and one will at centre of body so 13 Na atoms.