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Solid State - 2 - JEE MCQ


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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Solid State - 2

Solid State - 2 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Solid State - 2 questions and answers have been prepared according to the JEE exam syllabus.The Solid State - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Solid State - 2 below.
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Solid State - 2 - Question 1

If Z is the number of atoms in the unit cell that represents the closest packing sequence ABC ABC … , the number of tetrahedral voids in the unit cell is equal to -

Detailed Solution for Solid State - 2 - Question 1

No. of Tetrahedral Void = 2 × No. of atom

Tetrahedral Void = 2Z

Solid State - 2 - Question 2


Detailed Solution for Solid State - 2 - Question 2


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Solid State - 2 - Question 3

An element X (At. wt. = 80 g/mol) having fcc structure, calculate no. of unit cells in 8 gm of X :

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Solid State - 2 - Question 4


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Solid State - 2 - Question 5

The radius of the Na+ is 95 pm and that of Cl– ion is 181 pm. Predict the co-ordination number of Na+

Detailed Solution for Solid State - 2 - Question 5

Between 0.414 to 0.732

⇒ co-ordinaton no. = 6

Solid State - 2 - Question 6


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Solid State - 2 - Question 7


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Solid State - 2 - Question 8

If the distance between Na+ and Cl–1 ions in NaCl crystal is 265 pm, then edge length of the unit cell will be ?

Detailed Solution for Solid State - 2 - Question 8

Edge length

= 2 × distance between Na+ & Cl

= 2 × 265 = 530 pm

Solid State - 2 - Question 9


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Solid State - 2 - Question 10


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Solid State - 2 - Question 11


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Solid State - 2 - Question 12


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Solid State - 2 - Question 13


Detailed Solution for Solid State - 2 - Question 13


Solid State - 2 - Question 14

In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1/8 tetrahedral holes are occupied by Zn2+ ions and 50% of octahedral holes are occupied by Fe3+ ions. The formula of the compound is –

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Solid State - 2 - Question 15

In an FCC crystal, which of the following shaded planes contains the following type of arrangement of atoms ?


Detailed Solution for Solid State - 2 - Question 15

Shown arrangement is hexagonally closed pack plane & these types of planes are arranged perpendicular to body diagonal of fcc unit cell.

Solid State - 2 - Question 16

Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is -

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Solid State - 2 - Question 17


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Solid State - 2 - Question 18


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Solid State - 2 - Question 19


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Solid State - 2 - Question 20


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Solid State - 2 - Question 21

TlAl(SO4)2 . xH2O is bcc with 'a'= 1.22 nm. If the density of the solid is 2.32 g/cc, then the value of x is (Given : NA = 6 × 1023; at. wt. : Tl = 204, Al = 27, S = 32, O = 16, H = 1)

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Solid State - 2 - Question 22


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Solid State - 2 - Question 23


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Solid State - 2 - Question 24


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Solid State - 2 - Question 25


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Solid State - 2 - Question 26

A group IV A element with a density of 11.35 g/cm3 crystallise in a face centered cubic lattice whose unit cell edge length is 4.95Å. Calculate its atomic mass -

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Solid State - 2 - Question 27


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Solid State - 2 - Question 28


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Solid State - 2 - Question 29


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Solid State - 2 - Question 30


Detailed Solution for Solid State - 2 - Question 30


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