Solid State Test

Result

Solid State Test - 65

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following molecule has higher lattice energy?

A
NaCl.

B
$$Mg_1N_3$$.

C
Both (A) and (B) have equal lattice energy.

D
None of these

Solution

Lattice energy of a crystal is defined as the energy required to break an ionic crystal into its constituent ions in an isolated gaseous state. More ionic the compound is stronger will be its ionic bond and higher will be its lattice energy.
In $$MgN_3$$ the charge on the ionic species is greater than $$NaCl$$. So, ionic interactions in $$MgN_3$$ will be stronger.
So, lattice energy of $$MgN_3$$ is greater than $$NaCl$$
Question 2
1 / -0

A nugget of gold and quartz was found to contain x g of gold and y g of quartz and has density d if the densities of gold and quartz are $$d_{1}$$ and $$d_{2}$$ respectively then the correct relation is :

A
$$\dfrac{x}{d_{1}} + \dfrac{y}{d_{2}} = \dfrac{x + y}{d}$$

B
$$xd_{1} + yd_{2} = (x + y) d$$

C
$$\dfrac{x}{d_{2}} + \dfrac{y}{d_{1}} = \dfrac{x + y}{d}$$

D
$$\dfrac{x + y}{d} + \dfrac{x}{d_{1}} + \dfrac{x}{d_{2}} = 0$$

Solution

We know that,

$$density \ = \cfrac {Mass}{Volume} \Rightarrow Volume = \cfrac {Mass}{Density}$$

$$\Rightarrow Mass = density \times volume$$

Now, given densities of gold and quartz are respectively $$d_1$$ and $$d_2$$

Masses of gold and quartz are $$x \ g $$ and $$y \ g$$ respectively.

So,

$$Volume \ of \ gold = \cfrac {x}{d_1}$$.....(i)

$$Volume \ of \ quartz = \cfrac {y}{d_2}$$.....(ii)

$$ Volume \ of \ Nugget = \cfrac {Mass \ of \ nugget}{density \ of \ Nugget}$$

$$=\cfrac {x+y}{d}$$.....(iii)

Now, $$volume \ of \ nugget = volume \ of \ gold + volume \ of \ quartz$$

$$\Rightarrow \cfrac {x+y}{d}= \cfrac {x}{d_1} + \cfrac {y}{d_2}$$
Question 3
1 / -0

The distance between $$Na^+$$ and $$Cl^-$$ ions in $$NaCl$$ with a density $$3.165$$ g $$cm^{-3}$$ is:

A
$$300$$ pm

B
$$248.5$$ pm

C
$$248.5$$ cm

D
$$24.85$$ pm

Solution

Theoretical density of crystal, $$\rho = \cfrac {nM}{N_o a^3} \ g/cm^3$$

In $$NaCl$$ crystal, there are $$4$$ formula units per unit cell. So, $$n=4$$

$$\Rightarrow a^3 = \cfrac {nM}{N_o \rho}= \cfrac {4 \times 58.5}{6.022 \times 10^{23} \times 3.165}$$

$$\Rightarrow a= (123)^{(1/3)} \times (10^{-24})^{1/3} \ cm$$

$$\Rightarrow a = 4.97 \times 10^{-8} \ cm = 497 \ pm$$

Now, in $$NaCl$$, $$Cl^-$$ ions are present at corners and at face centers and $$Na^+$$ ions are present at edge centers and body centers.

So, distance between $$Na^+$$ and $$Cl^-= \cfrac a2= \cfrac {497}{2}=248.5 \ pm$$
Question 4
1 / -0

In the cubic close packing, this unit cell has:

A
4 tetrahedral voids each of which is shared by four adjacent unit cells.

B
4 tetrahedral voids within the unit cell.

C
8 tetrahedral voids each of which is shared by four adjacent unit cell.

D
8 tetrahedral voids within the unit cells.

Solution

Since, cube close packing is analogous to face-centered cubic (fcc) structure and number of atoms in a fcc crystal is $$4 \ per \ unit \ cell$$

So, number of tetrahedral voids $$= 2 \times Number \ of \ atoms \ in \ unit \ cell $$

$$= 2 \times 4 = 8$$
Hence, Option "D" is the correct answer.
Question 5
1 / -0

The density of metal which crystallise in $$bcc$$ lattice with unit cell edge length $$300$$ pm and molar mass $$50$$ g mol$$^{-1}$$ will be:

A
$$10$$ g cm$$^{-3}$$

B
$$14.2$$ g cm$$^{-3}$$

C
$$6.15$$ g cm$$^{-3}$$

D
$$9.31$$ g cm$$^{-3}$$

Solution

Theoretical density of crystal, $$\rho = \cfrac {nM}{N_o a^3} \ g/cm^3$$

Given for bcc $$n=2; \ a=300 \times 10^{-10} \ cm, \ M=50 \ g$$

$$\therefore \rho = \cfrac {2 \times 50}{6.022 \times 10^{23} \times (300)^3 \times (10^{-10})^3}$$

$$= 6.15 \ g/cm^3$$
Hence, Option "C" is the correct answer.
Question 6
1 / -0

An element with atomic mass 100 has a $$bcc$$ structure and edge length 400 pm. The density of element is:

A
10.37 g $$cm^{-3}$$

B
5.19 g $$cm^{-3}$$

C
7.29 g $$cm^{-3}$$

D
2.14 g $$cm^{-3}$$

Solution

Theoretical density of crystal, $$\rho = \cfrac {nM}{N_o a^3} \ g/cm^3$$

Given for bcc $$n=2; \ a=400 \times 10^{-10} \ cm, \ M=100 \ g$$

$$\therefore \rho = \cfrac {2 \times 100}{6.022 \times 10^{23} \times (400)^3 \times (10^{-10})^3}$$

$$= 5.19 \ g/cm^3$$
Hence, Option "B" is the correct answer.
Question 7
1 / -0

Which of the following points defects are shown by $$AgBr_{(s)}$$ crystals?
(I) Schottky defect
(II) Frenkel defect
(III) Metal defect
(IV) Metal deficiency defect

A
(I) and (II)

B
(III) and (IV)

C
(I) and (III)

D
(II) and (IV)

Solution

$$AgBr$$ shows both Schottky and Frenkel defects. In $$AgBr$$, both $$Ag^+$$ and $$Br^-$$ ions are absent from the lattice causing a Schottky defect. However, $$Ag^+$$ ions are mobile so they have a tendency to move aside the lattice and trapped in the interstitial site, hence cause Frenkel defect.

Hence, Option "A" is the correct answer.

Question 8

1 / -0

Match the types of packing given in column I with the items given in column II.

	Column I		Column II
(i)	Square close packing in two dimension	(p)	Triangular voids
(ii)	Hexagonal close packing in two dimensions	(q)	Pattern of spheres is repeated in every fourth layer
(iii)	Hexagonal close packing in three dimension	(r)	Coordination number=4
(iv)	Cubic close packing in three dimension	(s)	Pattern of spheres is repeated in alternate layers

(i)- (p), (ii)- (r), (iii)- (q), (iv)- (s)

(i)- (q), (ii)- (s), (iii)- (p), (iv)- (r)

(i)- (r), (ii)- (p), (iii)- (s), (iv)- (q)

(i)- (r), (ii)- (p), (iii)- (q), (iv)- (s)

Solution

Square close packing in 2D: In this kind of arrangement the AAA type of pattern of stacking is followed. The coordination number of each sphere in this arrangement will be $$4$$.

Hexagonal close packing in 2D has triangular voids as shown in the figure.

The pattern of arrangement in hcp and ccp in 3D is shown in the figure.

Question 9

1 / -0

Match the defects given in column I with statements given in column II and mark the appropriate choice.

	Column I		Column II
(A)	Simple vacancy defect	(i)	Shown by non-ionic solids and increases the density of the solid.
(B)	Simple interstitial defect	(ii)	Shown by ionic solution and decreases the density of the solid
(C)	Frenkel defect	(iii)	Shown by non-ionic solids and decreases the density of the solid
(D)	Schottky defect	(iv)	Shown by ionic solids and density of the solid remains the same

(A)$$\rightarrow$$(iv), (B)$$\rightarrow$$(iii), (C)$$\rightarrow$$(ii), (D)$$\rightarrow$$(i)

(A)$$\rightarrow$$(iii), (B)$$\rightarrow$$(iv), (C)$$\rightarrow$$(i), (D)$$\rightarrow$$(ii)

(A)$$\rightarrow$$(iii), (B)$$\rightarrow$$(i), (C)$$\rightarrow$$(iv), (D)$$\rightarrow$$(ii)

(A)$$\rightarrow$$(i), (B)$$\rightarrow$$(iii), (C)$$\rightarrow$$(iv), (D)$$\rightarrow$$(ii)

Solution

Simple vacancy defect: It is a kind of point defect in which an atom is missing from one of the lattice sites. In this, the density of the crystal decreases. It is generally shown by non-ionic solids.
Simple interstitial defect: It is a kind of point defect in which an atom moves from its regular lattice site to one of the interstitial sites. The density of the crystal remains same in the crystals showing this defect. It is generally shown by non-ionic solids.
Frenkel Defect: It is a kind o defect in crystalline solids in which atoms are displaced from their lattice position to interstitial site creating a vacancy at the lattice point. It usually occurs in ionic solid with a large difference in the size of ions. The density of crystal remains unchanged.
Schottky Defect: This defect occurs when oppositely charged ions leave their lattice site creating vacancies in such a way that electrical neutrality of crystal is maintained. It is generally seen in highly ionic compounds where the difference in size of cation and anion is small. So. An equal number of cation and anion vacancies are present in the crystals with Schottky Defect. The density of the crystal decreases in this defects.

Question 10
1 / -0

Experimentally it was found that a metal oxide has formula $$M_{0.98}O.$$ Metal $$M$$, is present as $$M^{2+}$$ and $$M^{3+}$$ in its oxide. Fraction of the metal which exists as $$M^{3+}$$ would be:

A
5.08%

B
7.01%

C
4.08%

D
6.05%

Solution

Since the oxidation state of oxygen is $$-2$$. So, for $${ M }_{ 0.98 }0$$ to be neutral, the total oxidation state of $${ M }_{ 0.98 }$$ has to be $$+2$$.

Let the fraction of $${ M }^{ 3+ }$$ be $$x$$.

Then fraction of $${ M }^{ 2+ }$$ will be $$(0.98-x)$$.

Now for the compound to be neutal,

$$3x+2(0.98-x)=2\\ 3x+1.96-2x=2\\ x=2-1.96\\ x=.04$$

So, fraction of $${ M }^{ 3+ }$$ will be 4%.

So, the correct answer is option $$C$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Solid State Test - 65

Result

Solid State Test - 65

Score

-

Rank

-

SHARING IS CARING

Question 1

NaCl.

$$Mg_1N_3$$.

Both (A) and (B) have equal lattice energy.

None of these

Solution

Question 2

$$\dfrac{x}{d_{1}} + \dfrac{y}{d_{2}} = \dfrac{x + y}{d}$$

$$xd_{1} + yd_{2} = (x + y) d$$

$$\dfrac{x}{d_{2}} + \dfrac{y}{d_{1}} = \dfrac{x + y}{d}$$

$$\dfrac{x + y}{d} + \dfrac{x}{d_{1}} + \dfrac{x}{d_{2}} = 0$$

Solution

Question 3

$$300$$ pm

$$248.5$$ pm

$$248.5$$ cm

$$24.85$$ pm

Solution

Question 4

4 tetrahedral voids each of which is shared by four adjacent unit cells.

4 tetrahedral voids within the unit cell.

8 tetrahedral voids each of which is shared by four adjacent unit cell.

8 tetrahedral voids within the unit cells.

Solution

Question 5

$$10$$ g cm$$^{-3}$$

$$14.2$$ g cm$$^{-3}$$

$$6.15$$ g cm$$^{-3}$$

$$9.31$$ g cm$$^{-3}$$

Solution

Question 6

10.37 g $$cm^{-3}$$

5.19 g $$cm^{-3}$$

7.29 g $$cm^{-3}$$

2.14 g $$cm^{-3}$$

Solution

Question 7

(I) and (II)

(III) and (IV)

(I) and (III)

(II) and (IV)

Solution

Question 8

(i)- (p), (ii)- (r), (iii)- (q), (iv)- (s)

(i)- (q), (ii)- (s), (iii)- (p), (iv)- (r)

(i)- (r), (ii)- (p), (iii)- (s), (iv)- (q)

(i)- (r), (ii)- (p), (iii)- (q), (iv)- (s)

Solution

Question 9

(A)$$\rightarrow$$(iv), (B)$$\rightarrow$$(iii), (C)$$\rightarrow$$(ii), (D)$$\rightarrow$$(i)

(A)$$\rightarrow$$(iii), (B)$$\rightarrow$$(iv), (C)$$\rightarrow$$(i), (D)$$\rightarrow$$(ii)

(A)$$\rightarrow$$(iii), (B)$$\rightarrow$$(i), (C)$$\rightarrow$$(iv), (D)$$\rightarrow$$(ii)

(A)$$\rightarrow$$(i), (B)$$\rightarrow$$(iii), (C)$$\rightarrow$$(iv), (D)$$\rightarrow$$(ii)

Solution

Question 10

5.08%

7.01%

4.08%

6.05%

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.