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JEE Exam  >  JEE Test  >  Chapter-wise Tests Main & Advanced  >  P. Bahadur Test: The Solid State (Old NCERT) - JEE MCQ

The Solid State (Old NCERT) - Free MCQ Practice Test with solutions, JEE


MCQ Practice Test & Solutions: P. Bahadur Test: The Solid State (Old NCERT) (15 Questions)

You can prepare effectively for JEE Chapter-wise Tests for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "P. Bahadur Test: The Solid State (Old NCERT)". These 15 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 15

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P. Bahadur Test: The Solid State (Old NCERT) - Question 1
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Acrystal is made of particles A and B. A forms FCC packing and B occupies all the octahedral voids. If all the particles along the plane as shown in figure are removed, then, the formula of the crystal would be:

Detailed Solution: Question 1

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.

P. Bahadur Test: The Solid State (Old NCERT) - Question 2
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A crystal is made of particle X, Y & Z. X forms FCC packing, Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X, if all the particles along one body diagonal are removed then the formula of the crystal would be –

Detailed Solution: Question 2

(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 =  

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P. Bahadur Test: The Solid State (Old NCERT) - Question 3
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In a hypothetical solid C atoms are found to form cubical close packed lattice. A atoms occupy all tetrahedral vioids & B atoms occupy all octahedral voids. A and B atoms are of appropriate size, so that there is no distortion in CCP lattice of C atoms. Now if a plane as shown in the following figure is cut.Then the cross section of this plane will look like.

Detailed Solution: Question 3

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.

P. Bahadur Test: The Solid State (Old NCERT) - Question 4
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In hexagonal close packing of sphere in three dimensions.           

Detailed Solution: Question 4

(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. 

P. Bahadur Test: The Solid State (Old NCERT) - Question 5
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Diamond has face-centred cubic lattice. There are two atoms at (0,0,0) and  coordinates.The ratio of the carbon-carbon bond distance to the edge of the unit cell is

Detailed Solution: Question 5


and aFCC = 1
 

P. Bahadur Test: The Solid State (Old NCERT) - Question 6
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The following diagram shows arrangement of lattice point with a = b = c  and α = β = γ  = 900.

Choose the correct options.

Detailed Solution: Question 6

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.

P. Bahadur Test: The Solid State (Old NCERT) - Question 7
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Consider a cube containing n unit cells of a cubic system. A plane ABCD obtained by joining the mid points of the edges on one of its identical faces had atoms arranged as shown. Let p be the packing fraction. Choose the correct option:

Detailed Solution: Question 7

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.

P. Bahadur Test: The Solid State (Old NCERT) - Question 8
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Spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has 1/ 8 of the tetrahedral holes occupied by one type of metal ion and 1/2 of the octahedral holes occupied by another type of metal ion. Such a spinel is formed by Zn2+, A13x and 02- with Zn2+ in the tetrahedral holes. If CCP arrangement of oxide ions remains undistorted in the presence of all the cations, formula of spinel and fraction of the packing fraction of crystal are respectively :

Detailed Solution: Question 8

Formula  ZnAl2O4
Packing fraction = 



 

P. Bahadur Test: The Solid State (Old NCERT) - Question 9
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What is the maximum number of layers of atoms in close packed planes that will lie within two imaginary parallel planes having a distance between them of  R (where R is the radius of atom) in the copper crystal (FCC) ? Consider the atoms to be within the parallel planes if their centres are on or within the two parallel planes.                                               

Detailed Solution: Question 9

Parallel layer are at a distance of R

P. Bahadur Test: The Solid State (Old NCERT) - Question 10
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A transition Metal M can exist in two oxidation states +2 and +3.It forms an oxide whose experimental formula is given by MxO where x < 1. Then the ratio of metal ions in +3 state to those in +2 state in oxide is given by :

Detailed Solution: Question 10

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


 

P. Bahadur Test: The Solid State (Old NCERT) - Question 11
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Analysis show that nickel oxide consist of nickel ion with 96% ions having d8 configuration and 4% having d7 configuration. Which amongst the following best represents the formula of the oxide.     

Detailed Solution: Question 11


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

Formula of solid = NiO1.02 = Ni0.98O.

P. Bahadur Test: The Solid State (Old NCERT) - Question 12
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PCI5 molecule has

Detailed Solution: Question 12

Based on theory.

P. Bahadur Test: The Solid State (Old NCERT) - Question 13
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Which of the following statements is correct about the conduction of electricity in pure crystal of silicon at room temperature?                       

Detailed Solution: Question 13

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.

P. Bahadur Test: The Solid State (Old NCERT) - Question 14
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Statement-1 : An important feature of Fluorite structures is that cations being large in size occupy FCC lattice points whereas anions occupy all the tetrahedral voids giving the formula unit AB2 (A : cation, B : anion).
Statement-2 : There are 6 cations and 12 anions per FCC unit cell of the Fluorite structure.

Detailed Solution: Question 14

Based on Fluorite structure of CaF2

P. Bahadur Test: The Solid State (Old NCERT) - Question 15
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Statement-1 : In NaCI crystal each Na ion is touching 6 CI– ions but these CI- ions do not touch each other.
Statement--2 : The radius ratio roe rNa+/rCI- is greater than 0.414, required for exact fitting

Detailed Solution: Question 15

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

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