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Test: Types of Solids and Properties - JEE MCQ


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10 Questions MCQ Test Chemistry for JEE Main & Advanced - Test: Types of Solids and Properties

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

A crystalline solid

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In crystalline solid there is perfect arrangement of the constituent particles only at 0K. They have sharp M.P.

Test: Types of Solids and Properties - Question 2

Which of the following is not a crystalline solid?

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Glass is an amorphous solid.

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Test: Types of Solids and Properties - Question 3

Which one of the following forms a molecular solid when solidified?

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form molecular solid.
Test: Types of Solids and Properties - Question 4
An example for network solid is
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is a covalent network solid that forms a network of tetrahedral units. and are ionic solids.
Test: Types of Solids and Properties - Question 5
What is the energy gap between valence band and conduction band in crystal of insulators?
Detailed Solution for Test: Types of Solids and Properties - Question 5
When insulators (non metal atoms) interact to form a solid, their atomic orbitals mix to form two bunch of orbitals, separated by a large band gap. Electrons cannot therefore be promoted to an empty level, where they could move freely.
Test: Types of Solids and Properties - Question 6

In a solid 'AB' having the NaCl structure, 'A' atoms occupy the corners in cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

Detailed Solution for Test: Types of Solids and Properties - Question 6

There were atoms on the face-centres removing face-centred atoms along one of the
axes means removal of atoms.
Now, number of atoms per unit cell

Number of atoms per unit cell
Hence the resultant stoichiometry is

Test: Types of Solids and Properties - Question 7

Assertion (A): White tin is an example of tetragonal system.
Reasoning (R): For a tetragonal system a = b = c and α = β = γ ≠ 90. The correct answer is

Detailed Solution for Test: Types of Solids and Properties - Question 7

White tin, an allotropic form of tin is an example of tetragonal system and for tetragonal system, one side is not equal to other two while all the angles are equal, ie, a = b ≠ c and α = β = γ = 90.
Hence, Assertion (A) is true but Reason (R) is not.

Test: Types of Solids and Properties - Question 8
How many number of atoms are there in a cube based unit cell having one atom on each corner and two atoms on each body diagonal of cube?
Detailed Solution for Test: Types of Solids and Properties - Question 8
There are four body diagonals. Atoms on the body diagonals are not shared by any other unit cell.
Contribution by atoms on corners
Contribution by atoms on body diagonal = 2 x 4 = 8
& & & & Total number of atoms = 1 + 8 = 9.
Test: Types of Solids and Properties - Question 9

The unit cell dimensions of a cubic lattice (edges a, b, c and the angles between them, α, β and γ) are

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Test: Types of Solids and Properties - Question 10

A crystal made up of particles X, Y, and Z. X forms FCC& packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. If all particles along one body diagonal are removed, then the formula of the crystal is

Detailed Solution for Test: Types of Solids and Properties - Question 10

For FCC, number of X atoms = 4/unit cell
Number of Tetrahedral Voids = Z = 8
Number of Octahedral Voids = Y = 4
Number of atoms removed along one body diagonal = 2X (corner) and 2Z (TVs) and 1 Y (OV at body centre)
∴ Number of X atoms left
Number of Y atom left = 4 - (1 × 1) = 3
Number of Z atom left = 8 - (2 × 1) = 6
The simplest formula

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