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JEE Exam  >  JEE Tests  >  Chemistry for JEE Main & Advanced  >  Solid State - 1 - JEE MCQ

Solid State - 1 - JEE MCQ


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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Solid State - 1

Solid State - 1 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Solid State - 1 questions and answers have been prepared according to the JEE exam syllabus.The Solid State - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Solid State - 1 below.
Solutions of Solid State - 1 questions in English are available as part of our Chemistry for JEE Main & Advanced for JEE & Solid State - 1 solutions in Hindi for Chemistry for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Solid State - 1 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Solid State - 1 - Question 1
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What type of crystal defect is indicated in the diagram below ?

[AIEEE-2004]

Detailed Solution for Solid State - 1 - Question 1

Here the crystal defect is the paired vacancies caused by the ions missing from their lattice sites. 
 

Solid State - 1 - Question 2
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An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound would be

 [AIEEE-2005]

Detailed Solution for Solid State - 1 - Question 2

Unit cell consists of A ions at the corners is 8. So we know that the 

contribution of corner is = 1/8th of corner = 1/8 x 8 = 1. 

Unit cell consists of B ions at the centre of the six faces 

Face = 1/2th of face 

= 1/2 x 6= 3 

So the empirical formula for this compound will be AB3 

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Solid State - 1 - Question 3
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Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom?

Detailed Solution for Solid State - 1 - Question 3

Radius of the atom = a ÷ a√2

=361÷2√2

=127.56

Solid State - 1 - Question 4
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In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be - 

[AIEEE 2008]
Detailed Solution for Solid State - 1 - Question 4

No. of Y atoms per unit cell in ccp lattice = 4

 No. of tetrahedral voids = 2 × 4 = 8

 No. of tetrahedral voids occupied by X = 2/3 × 8 = 16/3

 Therefore, Formula of the compound = X16/3Y4

= X16Y12

= X4 Y3

Solid State - 1 - Question 5
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Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?
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Solid State - 1 - Question 6
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If 'a' stands for the edge length of the cubic systems : simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these system will be respectively -
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Solid State - 1 - Question 7
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Solid State - 1 - Question 8
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Solid State - 1 - Question 9
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Solid State - 1 - Question 10
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What is the diameter of the largest sphere that will fit in the void at the centre of the cube edge of a BCC crystal of edge length a ?
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Solid State - 1 - Question 11
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Solid State - 1 - Question 12
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Solid State - 1 - Question 13
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Solid State - 1 - Question 14
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Solid State - 1 - Question 15
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Solid State - 1 - Question 16
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Solid State - 1 - Question 17
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Solid State - 1 - Question 18
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The packing fraction of the element that crystallizes in simple cubic arrangement is -
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Solid State - 1 - Question 19
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Solid State - 1 - Question 20
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Solid State - 1 - Question 21
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Solid State - 1 - Question 22
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A solid compound contains X, Y and Z atoms in a cubic lattice with X atoms occupying the corners, Y atoms in the body centred positions and Z atoms at the centre of the faces of the unit cell. What is the empirical formula of the compound ?
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Solid State - 1 - Question 23
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When heated above 916ºC, iron changes its bcc crystalline form to fcc without the change in the radius of atom. The ratio of density of the crystal before heating and after heating is [At . wt. Fe = 56]
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Solid State - 1 - Question 24
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The anions (A) form hexagonal closest packing and the cations (C) occupy only 2/3 of octahedral holes. The simplest formula of the ionic compound is –
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Solid State - 1 - Question 25
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Solid State - 1 - Question 26
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Solid State - 1 - Question 27
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Solid State - 1 - Question 28
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Solid State - 1 - Question 29
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A crystal 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 :
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Solid State - 1 - Question 30
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