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

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


Test Description

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

P. Bahadur Test: The Solid State (Old NCERT) for JEE 2024 is part of JEE preparation. The P. Bahadur Test: The Solid State (Old NCERT) questions and answers have been prepared according to the JEE exam syllabus.The P. Bahadur Test: The Solid State (Old NCERT) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for P. Bahadur Test: The Solid State (Old NCERT) below.
Solutions of P. Bahadur Test: The Solid State (Old NCERT) questions in English are available as part of our course for JEE & P. Bahadur Test: The Solid State (Old NCERT) solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt P. Bahadur Test: The Solid State (Old NCERT) | 15 questions in 20 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
P. Bahadur Test: The Solid State (Old NCERT) - Question 1

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

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 for P. Bahadur Test: The Solid State (Old NCERT) - 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 =  

1 Crore+ students have signed up on EduRev. Have you? Download the App
P. Bahadur Test: The Solid State (Old NCERT) - Question 3

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

In hexagonal close packing of sphere in three dimensions.           

Detailed Solution for P. Bahadur Test: The Solid State (Old NCERT) - 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

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 for P. Bahadur Test: The Solid State (Old NCERT) - Question 5


and aFCC = 1
 

P. Bahadur Test: The Solid State (Old NCERT) - Question 6

The following diagram shows arrangement of lattice point with a = b = c  and α = β = γ  = 900.

Choose the correct options.

Detailed Solution for P. Bahadur Test: The Solid State (Old NCERT) - 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

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

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 for P. Bahadur Test: The Solid State (Old NCERT) - Question 8

Formula  ZnAl2O4
Packing fraction = 



 

P. Bahadur Test: The Solid State (Old NCERT) - Question 9

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 for P. Bahadur Test: The Solid State (Old NCERT) - Question 9

Parallel layer are at a distance of R

P. Bahadur Test: The Solid State (Old NCERT) - Question 10

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

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

PCI5 molecule has

Detailed Solution for P. Bahadur Test: The Solid State (Old NCERT) - Question 12

Based on theory.

P. Bahadur Test: The Solid State (Old NCERT) - Question 13

Which of the following statements is correct about the conduction of electricity in pure crystal of silicon at room temperature?                       

Detailed Solution for P. Bahadur Test: The Solid State (Old NCERT) - 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

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 for P. Bahadur Test: The Solid State (Old NCERT) - Question 14

Based on Fluorite structure of CaF2

P. Bahadur Test: The Solid State (Old NCERT) - Question 15

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

Information about P. Bahadur Test: The Solid State (Old NCERT) Page
In this test you can find the Exam questions for P. Bahadur Test: The Solid State (Old NCERT) solved & explained in the simplest way possible. Besides giving Questions and answers for P. Bahadur Test: The Solid State (Old NCERT), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE