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Test: Solid State - 1


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20 Questions MCQ Test Physical Chemistry | Test: Solid State - 1

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Test: Solid State - 1 - Question 1

Which of the units given can be used to build the structure of covalent crystal, diamond:

Detailed Solution for Test: Solid State - 1 - Question 1

Diamond and Graphite, Two allotropes of carbon are covalent network solids which differ in the bonding geometry of the carbon atoms.

In diamond, the bonding occurs in the tetrahedral geometry, while in graphite the carbons bond with each other in the trigonal planar arrangement. This difference accounts for the drastically different appearance and properties of these two forms of carbon.

Test: Solid State - 1 - Question 2

Which of the following expressions is true in case of a sodium chloride unit cell?

Detailed Solution for Test: Solid State - 1 - Question 2

In a face-centered cubic system, the interionic distance is given by rc + ra = a/2

rc = Radius of the cation
ra = Radius of the Anion.

Test: Solid State - 1 - Question 3

Solid CO2 is an example of the crystal type:

Detailed Solution for Test: Solid State - 1 - Question 3

Dry ice, or solid carbon dioxide, is a perfect example of a molecular solid. The van der Waals forces holding the CO2 molecules together are weak enough that dry ice sublimes, i.e. it passes directly from the solid to the gas phase, at -78oC.

Test: Solid State - 1 - Question 4

Name of the crystal class for which all the four types of unit cells are possible:

Detailed Solution for Test: Solid State - 1 - Question 4

Every crystal system does not have all the four types of unit cells (simple, face centered, end centered, and body centered). But orthorhombic crystal class has all four types of unit cells- primitive, face centered, end centered and body centered.

Test: Solid State - 1 - Question 5

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is

Detailed Solution for Test: Solid State - 1 - Question 5

Test: Solid State - 1 - Question 6

Which of the following is a molecular crystal?

Detailed Solution for Test: Solid State - 1 - Question 6

Molecular crystals are substances that have relatively weak intermolecular binding, such as Dry Ice (solidified carbon dioxide), solid forms of the rare gases (e.g., argon, krypton, and xenon), and crystals of numerous organic compounds.

Test: Solid State - 1 - Question 7

An element exists as hexagonal close packed structure as well as cubic closed packed structure. In which case the element would have higher density:

Detailed Solution for Test: Solid State - 1 - Question 7

The density of the element will be same in both the cases as the two structures will have the same coordination number and hence the same packing fraction.

Test: Solid State - 1 - Question 8

The molar volume of KCl and NaCl are 37.46 ml and 27.94 ml. The ratio of unit cell edges of the two crystals is:

Detailed Solution for Test: Solid State - 1 - Question 8

Molar volume =N4/3πr³ where N is Avogadro number ,

Now Molar volume of KCl is 37.46cm³ So N4/3πR³ = 37.46 cm³

On solving this gives R=2.458 x 10-8 cm

And for NaCl by using the same formula it gives R=2.22 x 10-8 cm

Now the ratio of edge length must be equal to ratio of radii as both belong to same crystal system so KCl/NaCl = 2.458/2.22 = 1.108

Test: Solid State - 1 - Question 9

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3. The molar mass of the metal is (NA Avogadro's constant= 6.02 x 1023 mol-1)

Detailed Solution for Test: Solid State - 1 - Question 9

Given, cell is fcc, So Z =  4
Edge length, a = 404 pm = 4.04 x 10-8 cm
Density of metal, d = 2.72 g cm-3
NA = 6.02 x 1023 mol-1
Molar mass ofg the metal, M =?
We know that

Test: Solid State - 1 - Question 10

In a crystalline solid, anions B are arranged in CCP lattice and cations A occupy 50% of the octahedral voids and 50% of the tetrahedral voids. What is the formula of the solid:

Detailed Solution for Test: Solid State - 1 - Question 10

No of atoms in a CCP unit cell = similar to FCC = 4

Therefore no of B atoms present = 4

No of A atom = tetrahedral void and octohedral void(4+2) = 6

The formula of the compound  = A3​B2​

Test: Solid State - 1 - Question 11

If ‘a’ stand for the edge length of the cubic system: scc, bcc and fcc, then the ratio of the radii of the sphere in the system will be:

Detailed Solution for Test: Solid State - 1 - Question 11

Test: Solid State - 1 - Question 12

The packing efficiency of 2D square unit cell shown is:

Detailed Solution for Test: Solid State - 1 - Question 12

Test: Solid State - 1 - Question 13

In which of the crystal lattice, alternate tetrahedral voids are occupied:

Detailed Solution for Test: Solid State - 1 - Question 13

In ZnS (zinc blends), Zn2+ ions are present in alternate tetrahedral voids. 

Test: Solid State - 1 - Question 14

2D representation of the fcc unit cell is:

Test: Solid State - 1 - Question 15

Na crystallizes in fcc lattice having edge length 500 pm. Mass of the Na particle is 4.151 × 10– 23g. The density of the lattice is:

Test: Solid State - 1 - Question 16

Formula of the compound is:

Detailed Solution for Test: Solid State - 1 - Question 16

The centre atom will be considered as whole.

The face centre atom will be considered as half..

The corner atoms will be considered 1/8.

Test: Solid State - 1 - Question 17

Schottky defect generally appears in:

Detailed Solution for Test: Solid State - 1 - Question 17

Schottky defect is a type of point defect in a crystal lattice named after Walter H. Schottky. In non-ionic crystals it means a lattice valency defect. In ionic crystals, this point defect forms when oppositely charged ions leave their lattice sites, creating vacancies.

Test: Solid State - 1 - Question 18

Na2SeO4 and Na2SO4 show:

Detailed Solution for Test: Solid State - 1 - Question 18

 Na2SO4 and Na2SeO4 are different substances possessing similar formula, they are called isomorphous systems.

Test: Solid State - 1 - Question 19

The melting point of RbBr is 682°C, while that of NaF is 988°C. The principal reason that melting point of NaF is much higher than that of RbBr is that:

Detailed Solution for Test: Solid State - 1 - Question 19

In ionic compounds as internuclear distance incrreases, melting point decreases so as we move down the group, melting point decreases. So, as r+ ra  is greater for RbBr than for NaF so NaF have higher melting point.

Test: Solid State - 1 - Question 20

An alloy of Copper , Silver and Gold is found to have Copper constituting the C.C.P lattice. If Silver atoms occupy the edge centres and Gold is present at body centre , the alloy will have the formula:

Detailed Solution for Test: Solid State - 1 - Question 20

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