12th Grade > Chemistry
SOLID STATE MCQs
Total Questions : 30
| Page 3 of 3 pages
Question 21. Assume a hypothetical cubic crystal lattice, named JEE-centered cubic (jcc) with the following characteristics:
I. An atom is present at all the corners of the cube
II. An atom is present at the center of two pairs of opposite faces
III. An atom is present at the center of all the edges of the cube
IV. One atom is present at its body-center
An element having the jcc lattice structure has a cell edge of 120 pm. The density of the element is 6.8 g/cm3. How many atoms are present in 408 g of the element?
I. An atom is present at all the corners of the cube
II. An atom is present at the center of two pairs of opposite faces
III. An atom is present at the center of all the edges of the cube
IV. One atom is present at its body-center
An element having the jcc lattice structure has a cell edge of 120 pm. The density of the element is 6.8 g/cm3. How many atoms are present in 408 g of the element?
Answer: Option C. -> 2.43×1026
:
C
Volume of unit cell =(120pm)3=(120×10−12)3m3=(123×10−33)m3
Volume of 408 g of the element =massdensity=4086.8=60cm3=6×10−5m3
So, number of unit cells present in 408 g of the elements =6×10−5123×10−33=3.472×1025 unit cells
Since each jcc unit cell consist of 7 atoms,
therefore the total number of atoms presents in 408 g of the given element
=7×3.472×1025=2.43×1026atoms
:
C
Volume of unit cell =(120pm)3=(120×10−12)3m3=(123×10−33)m3
Volume of 408 g of the element =massdensity=4086.8=60cm3=6×10−5m3
So, number of unit cells present in 408 g of the elements =6×10−5123×10−33=3.472×1025 unit cells
Since each jcc unit cell consist of 7 atoms,
therefore the total number of atoms presents in 408 g of the given element
=7×3.472×1025=2.43×1026atoms
Answer: Option C. -> c
:
C
Choose the correct option:
Crystal systemAxial distanceAxial angles(a)Tetragonala=b≠cα=β=γ=90∘(b)Monoclinica≠b≠cα=γ=90∘β≠90∘(c)Rhombohedrala = b = cα=β=γ≠90∘(d)Triclinica≠b≠cα≠β≠γ≠90∘
:
C
Choose the correct option:
Crystal systemAxial distanceAxial angles(a)Tetragonala=b≠cα=β=γ=90∘(b)Monoclinica≠b≠cα=γ=90∘β≠90∘(c)Rhombohedrala = b = cα=β=γ≠90∘(d)Triclinica≠b≠cα≠β≠γ≠90∘
Answer: Option D. -> All of above
:
D
Polar and non-polar molecular solids are soft and non-conductors of electricity.
Ionic solids are non-conductors of electricity in solid state, as the constituent ions are not free to move (but in molten state or when dissolved in water the ions are free to move and hence they conduct electricity)
:
D
Polar and non-polar molecular solids are soft and non-conductors of electricity.
Ionic solids are non-conductors of electricity in solid state, as the constituent ions are not free to move (but in molten state or when dissolved in water the ions are free to move and hence they conduct electricity)
Question 24. Question 11-12 is based on the following given information:
Assume a hypothetical cubic crystal lattice, named JEE-centered cubic (jcc) with the following characteristics:
I. An atom is present at all the corners of the cube
II. An atom is present at the center of two pairs of opposite faces
III. An atom is present at the center of all the edges of the cube
IV. One atom is present at its body-center
How many atoms are effectively present per unit cell in this hypothetical crystal lattice?
Assume a hypothetical cubic crystal lattice, named JEE-centered cubic (jcc) with the following characteristics:
I. An atom is present at all the corners of the cube
II. An atom is present at the center of two pairs of opposite faces
III. An atom is present at the center of all the edges of the cube
IV. One atom is present at its body-center
How many atoms are effectively present per unit cell in this hypothetical crystal lattice?
Answer: Option B. -> 7
:
B
I. 8 corner atom ×(18) atom per unit cell = 1 atom
II. 4 face-centered atoms ×(12)atom per unit cell = 2 atom
III. 12 edge-centered atoms ×(14) atom per unit cell = 3 atom
IV. 1 body centered atom ×1 atom per unit cell = 1 atom
So, Total number of atoms per unit cell =1+2+3+1=7atoms
:
B
I. 8 corner atom ×(18) atom per unit cell = 1 atom
II. 4 face-centered atoms ×(12)atom per unit cell = 2 atom
III. 12 edge-centered atoms ×(14) atom per unit cell = 3 atom
IV. 1 body centered atom ×1 atom per unit cell = 1 atom
So, Total number of atoms per unit cell =1+2+3+1=7atoms
Answer: Option D. -> FeCl2
:
D
Fe+2 have unpaired electrons
:
D
Fe+2 have unpaired electrons
Answer: Option C. -> CsI
:
C
Cs+andI− have the largest sizes.
:
C
Cs+andI− have the largest sizes.
Answer: Option C. -> KBr
:
C
KBr exhibits Schottky defect and not Frenkel defect
AgBr shows both Schottky as well as Frenkel defect
Frenkel defect is shown by compounds with large difference in size of cations and anions, low coordination number and low ionic nature.
:
C
KBr exhibits Schottky defect and not Frenkel defect
AgBr shows both Schottky as well as Frenkel defect
Frenkel defect is shown by compounds with large difference in size of cations and anions, low coordination number and low ionic nature.
Answer: Option B. -> If both assertion and reason are true but reason is not the correct explanation of the assertion.
:
B
On heating, the metal atoms deposit on the surface and finally they diffuse into the crystal and after ionization the alkali metal ion occupies cationic vacancy whereas electron occupies anionic vacancy.
:
B
On heating, the metal atoms deposit on the surface and finally they diffuse into the crystal and after ionization the alkali metal ion occupies cationic vacancy whereas electron occupies anionic vacancy.
Answer: Option D. -> If assertion Is false but reason is true.
:
D
Electrical conductivity of semiconductors increases with increasing temperature, as with increase in temperature large number of electrons from the valence band can jump to the conduction band.
:
D
Electrical conductivity of semiconductors increases with increasing temperature, as with increase in temperature large number of electrons from the valence band can jump to the conduction band.
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