Practice Start Test

Solid State

NEET Chemistry · 84 questions · Page 4 of 9 · Click an option or "Show Solution" to reveal answer

Q31
Percentage of free space in a body centred cubic unit cell is
A 34%
B 28%
C 30%
D 32%
Correct Answer
Option D
Solution

The ratio of volumes occupied by atoms in unit cell to the total volume of the unit cell is called as packing fraction or density of packing.

For body centred cubic structure, packing fraction = 0.68 i.e., 68% of the unit cell is occupied by atoms and 32% is empty.

Q32
If NaCl is doped with 10 −-4 mol% of SrCl 2 , the concentration of cation vacancies will be ( N A = 6.02 × \times 10 23 mol −-1 )
A 6.02 × \times 10 16 mol −-1
B 6.02 × \times 10 17 mol −-1
C 6.02 × \times 10 14 mol −-1
D 6.02 × \times 10 15 mol −-1
Correct Answer
Option B
Solution

As each Sr 2+ ion introduces one cation vacancy, therefore concentration of cation vacancies = mole % of SrCl 2 added.

∴\therefore Concentration of cation vacancies = 10 –4 mole % =

10−4100×6.023×1023{{{{10}^{ - 4}}} \over {100}} \times 6.023 \times {10^{23}}

=

6.023×10236.023 \times {10^{23}}

×\times 10 -6 =

6.023×10176.023 \times {10^{17}}
Q33
The fraction of total volume occupied by the atoms present in a simple cube is
A π32{\pi \over {3\sqrt 2 }}
B π42{\pi \over {4\sqrt 2 }}
C π4{\pi \over 4}
D π6{\pi \over 6}
Correct Answer
Option D
Solution

The maximum properties of the available volume which may be filled by hard sphere in simple cubic arrangement is

π6{\pi \over 6}

or 0.52.

Q34
CsBr crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 × \times 10 23 mol −-1 , the density of CsBr is
A 4.25 g/cm 3
B 42.5 g/cm 3
C 0.425 g/cm 3
D 8.25 g/cm 3
Correct Answer
Option A
Solution

Density of CsBr =

Z×MV×NA{{Z \times M} \over {V \times {N_A}}}

=

1×213(436.6×10−10)3×6.023×1023{{1 \times 213} \over {{{\left( {436.6 \times {{10}^{ - 10}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}

= 4.25 g/cm 3

Q35
The appearance of colour in solid alkali metal halides is generally due to
A interstitial positions
B F .-centres
C Schottky defect
D Frenkel defect
Correct Answer
Option B
Solution

F-centres are the sites where anions are missing and instead electrons are present. They are responsible for colours.

Q36
In a face-centered cubic lattice, a unit cell is shared equally by how many unit cells?
A 2
B 4
C 6
D 8
Correct Answer
Option C
Solution

In given unit cell it is shared equally by six faces of different unit cells.

Q37
A compound formed by elements X and Y crystallizes in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face-centers. The formula of the compound is :-
A XY 3
B X 3 Y
C XY
D XY 2
Correct Answer
Option A
Solution

To determine the formula of the compound formed by elements X and Y based on their positions in a cubic structure, we need to count the number of atoms per unit cell for each type of atom and then simplify the ratio.

For element X: Corners of a cube: Each corner atom is shared by 8 adjacent cubes, so each corner contributes 1/8 of an atom to the cube.

Since there are 8 corners in a cube, the total contribution from corners is 8×18=18 \times \dfrac{1}{8} = 1 atom of X per unit cell.

For element Y: Face-centers of a cube: Each face-centered atom is shared by 2 adjacent cubes, so each face-center contributes 1/2 of an atom to the cube.

Since there are 6 faces on a cube, the total contribution from face-centers is 6×12=36 \times \dfrac{1}{2} = 3 atoms of Y per unit cell.

Therefore, within one unit cell, we have 1 atom of X and 3 atoms of Y, giving us the simplest ratio of X:Y=1:3X:Y = 1:3.

Hence, the formula of the compound is XY3XY_3.

So, the correct answer is: Option A: XY 3

Q38
The pyknometric density of sodium chloride crystal is 2.165 × \times 10 3 kg m −-3 while its X-ray density is 2.178 × \times 10 3 kg m −-3 . The fraction of unoccupied sites in sodium chloride crystal is
A 5.96
B 5.96 × \times 10 −-2
C 5.96 × \times 10 −-1
D 5.96 × \times 10 −-3
Correct Answer
Option D
Solution

Molar volume from pyknometric density =

M2.165×103{M \over {2.165 \times {{10}^3}}}

m 3 Molar volume from X-ray density =

M2.178×103{M \over {2.178 \times {{10}^3}}}

m 3 Volume unoccupied =

M103(12.165−12.178){M \over {{{10}^3}}}\left( {{1 \over {2.165}} - {1 \over {2.178}}} \right)

m 3 Fraction unoccupied =

(0.013M×10−32.165×2.178)(M×10−32.165){{\left( {{{0.013M \times {{10}^{ - 3}}} \over {2.165 \times 2.178}}} \right)} \over {\left( {{{M \times {{10}^{ - 3}}} \over {2.165}}} \right)}}

= 5.96 ×\times 10 -3

Q39
When Zn converts from melted state to its solid state, it has hcp structure, then find the number of nearest atoms.
A 6
B 8
C 12
D 4
Correct Answer
Option C
Solution

hcp is a closed packed arrangement in which the unit cell is hexagonal and coordination number is 12.

Q40
Cation and anion combines in a crystal to form following type of compound
A ionic
B metallic
C covalent
D dipole-dipole.
Correct Answer
Option A
Solution

We know that electrostatic force is that binds the oppositely charged ions which are formed by transfer of electron from one atom to another is called ionic bond.

We also know that cation and anion are oppositely charged particles therefore they form ionic bond in crystal.

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