Download CBSE Sample Paper 2024-25 for class 12th to 8th

Solid State Test - 49

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Solid State Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

Select the wrong statement (s) :-

A
A NaCl type AB crystal lattice can be interpreted to be made up of two individual fcc type unit lattice of $$A^+$$ and $$B^-$$ fused together in such a manner that the corner of one unit lattice becomes the edge centre of the other

B
In a fcc unit cell, the body center is an octahedral void

C
In an scc lattice, there can be no octahedral void

D
In an scc lattice, the body center is the octahedral void

Solution

A NaCl type AB crystal lattice can be interpreted to be made up of two individual fcc type unit lattice of A+ and B− fused together in such a manner that the corner of one unit lattice becomes the edge center of the other
In an fcc unit cell, the body center is an octahedral void
In an SCC lattice, the body center is the octahedral void
These are the correct statement. So Option A, B, D are correct
and C is correct
Question 2
1 / -0

In a unit cell, atoms A, B, C and D are present at half of total corners, all face-centres, body-centre and one third of all edge-centres respectively. Then formula of unit cell is ?

A
$$ AB_{3}CD_{3} $$

B
ABCD

C
$$ AB_{6}C_{2}D_{4} $$

D
$$ AB_{6}C_{2}D_{2} $$

Solution

Atoms at the corner in a unit cell is 1
No of atoms of A in one unit cell = $$\frac{1}{2}$$

Atoms at the corner in a unit cell is 3
No of atoms of B in one unit cell = 3

Atoms at the body center in a unit cell is 1
No of atoms of C in one unit cell = 1

Atoms at the edge center in a unit cell is 3
No of atoms of D in one unit cell = 1

Formula is $$A_{\frac{1}{2}}B_3CD or AB_6C_2D_2$$
Question 3
1 / -0

If the radius of $${Cl}^{-}$$ ion is $$181pm$$, and the radius of $${Na}^{+}$$ ion is $$101pm$$ then the edge length of unit cell is:

A
$$282pm$$

B
$$285.71pm$$

C
$$512pm$$

D
$$564pm$$

Solution

Edge length of unit cell,
$$=2\times (\text{radius of }Cl^-+\text{radius of } Na^+)\\=2\times (181+ 101)\\=2 \times 282\\=564pm$$
Question 4
1 / -0

Total number of unit cells present in a cube shaped ideal crystal of $$1g \ NaCl(s)$$ is (Avogdro's number = $$6\times10^{23}$$).

A
$$2.56\times10^6$$

B
$$2.56\times10^{21}$$

C
$$2.56\times10^{23}$$

D
$$4.2\times10^{22}$$

Solution

The number of $${Na}^+$$ ions present in one unit cell $$= 12 \times \frac{1}{4} + 1 = 4$$ atoms per unit cell.

The number of $${Cl}^-$$ ions present in one unit cell $$= 6 \times \frac{1}{2} + 8 \times \frac{1}{8} = 4$$ atoms per unit cell.

So, each unit cell has 4 atoms of each $${Na}^+$$ and $${Cl}^-$$. Hence there are total of 4 molecules of NaCl in each unit cell.

As the mass of -Avogadro number of molecules of NaCl is 58.5g.

Then, number of molecules in one gram $$= \frac{6.022 \times {10}^{23}} {58.5} = 1.025 \times {10}^{22}$$ molecules

As we found above that each unit cell contains 4 molecules of NaCl hence we can now easily find the number of unit cells containing $$1.025 \times {10}^{22}$$ molecules $$= \frac{1.03 \times {10}^{22}} {4}$$ $$=$$ $$2.56 \times {10}^{21}$$ unit cells

So, $$2.56 \times {10}^{21}$$ unit cells contains 1gm of NaCl molecules.
Question 5
1 / -0

(i) Frenkel defect create a vacancy defect at its original site and an interstitial defect at its new position.
(ii) Frenkel defect is stoichiometric point defect.
(iii) Frenkel defect is also called dislocation defect.
(iv) Silver halide shows Frenkel defect
Select the correct statements :

A
$$(ii),(iii),(iv)$$ only

B
$$(i),(ii),(iii)$$ only

C
$$(i),(ii),(iii),(iv)$$

D
$$(i),(ii),(iv)$$ only

Solution

Imperfections in the solid state compounds are known as defects. In the Frenkel defect, the cation moves from its site and occupies an interstitial site. In this type of defect, density remains constant whereas dielectric constant increases.

There are four stoichiometric defects namely, Vacancy defect, Interstitial defect, Schottky defect and frenkel defects.

Frenkel defects is also known as dislocation effect because a cation is dislocated. Silver (I) halides show frenkel defects.

So, all the statements are true.

Hence, opt. C is correct.
Question 6
1 / -0

Lattice energy of $$NaCl$$ is $$'X'$$. If the ionic size of $$A^{ +2 }$$ is equal to that of $$Na^{ + }$$ and $$B^{ -2 }$$ is equal to $$Cl^{ - }$$, then lattice energy associated with the crystal $$AB$$ is:

A
$$X$$

B
$$2X$$

C
$$4X$$

D
$$8X$$

Solution

The lattice energy of crystal $$AB$$ is $$4X$$.

Lattice energy is the energy associated to the crystal lattice of a compound.

Mathematically,

$$E_{lattice}=k\dfrac{Q_1Q_2}{d}$$

where, $$k =$$ proportionality constant

$$Q_1$$ and $$Q_2$$ are the charges on the ions in coulombs.

$$d =$$ distance between the ion charges.

Lattice energy of $$NaCl = X$$

$$E_{lattice}(NaCl)=\dfrac{Q_1Q_2}{d}=X$$

As the ionic size remains same when $$AB$$ crystal is formed, d remains same

Lattice energy of $$AB$$,

$$E_{lattice}(AB)=\dfrac{(2Q_1)(2Q_2)}{d}$$

$$E_{lattice}(AB)=4\left(\dfrac{Q_1Q_2}{d}\right )=4X$$

The lattice energy of crystal $$AB$$ is $$4X$$.
Question 7
1 / -0

Which of the following expression is correct for packing fraction of $$NaCI$$ if the ions along with face are diagonally removed ?

A
$$\dfrac{\dfrac{13}{3} \pi r^{3}_{-}+\dfrac{16}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$$

B
$$\dfrac{\dfrac{13}{3} \pi r^{3}_{-}+\dfrac{4}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$$

C
$$\dfrac{\dfrac{16}{3} \pi r^{3}_{-}+\dfrac{13}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$$

D
$$\dfrac{\dfrac{3}{4} \pi r^{3}_{-}+\dfrac{13}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$$

Solution
Question 8
1 / -0

An element having a face-centered cubic structure has a density of $$6.23g\ cm^{-3}$$. The atomic mass of elements is $$60$$. The edge length of the unit cell is:

A
$$400\ pm$$

B
$$300\ pm$$

C
$$425\ pm$$

D
$$370\ pm$$

Solution

$$a^3=\large\dfrac{z\times M}{N_A\times d}$$

$$\;\;\;\;=\large\dfrac{4\times 60}{6.023\times 10^{23}\times 6.23}$$

$$\Rightarrow a^3=64\times 10^{-24}cm^3$$

$$\Rightarrow a=(4\times 10^{-8}\times 10^{10})pm=400pm$$
Question 9
1 / -0

Which of the following solid substance(s) will have the same refractive index when measured in different directions?

A
Rubber

B
$$NaCl$$

C
Plastic

D
Graphite

Solution

Isotropic solids have same refractive index in all the directions. As amorphous solids are isotropic in nature. Hence, amorphous solids like$$NaCl$$ will show same refractive index when measured in different directions.

Hence, the correct option is $$B$$
Question 10
1 / -0

Lattice energy of an ionic compound depends upon?

A
Charge on the ion and size of the ion

B
Packing of Ions only

C
Size of the ion

D
Charge on the Ion only

Solution