Solid State Test

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Solid State Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

At zero kelvin, most of the ionic crystals possess :

A
Frenkel defect

B
Schottky defect

C
metal excess defect

D
No defect

Solution

At zero kelvin, most of the ionic crystals posses no defect. This is due to presence of highly ordered arrangement.
Question 2
1 / -0

The ABAB packing and ABCABC packing are respectively called as:

A
hexagonal close packing (hcp) and cubic close packing (ccp)

B
cubic closed packing (ccp) and hexagonal closed packing (hcp)

C
body centered cubic (bcc) and hexagonal close packing (hcp)

D
hexagonal closed packing (hcp) and body-centered cubic (bcc)

Solution

$$\text { Solution:- Packing of lattice points in a crystal is of 2-types:- } \\$$
$$\text { 1.) Cubic dose packing (CCP) } \\$$
$$\text { 2.) hexagonal close packing (HCP) } \\$$
$$\text { layers of packing sphere or lattice points } \\$$
$$\text { in ccp & hcp are } 3 \text { . }$$

$$\text { - CCP is ABCABC ... type } \\$$
$$\text { and } \\$$
$$\text { - HCP is } A B \triangle B A B \text { .... type. }$$

$$\text { Examples:- } \\$$
$$\qquad \text { - Zinc blende }(Z n S)(4: 4)-c c p \text { . } \\$$
$$\qquad \text { - Wurtzite }(Z n S)(4: 4)-h c p \text { . } \\$$
$$\text { hence correct option is (A) }$$
Question 3
1 / -0

Tetragonal crystal system has which of the following unit cell dimensions?

A
$$a = b = c$$ and $$\alpha = \beta = \gamma = 90^{\circ}$$

B
$$a = b \neq c$$ and $$\alpha = \beta = \gamma = 90^{\circ}$$

C
$$a \neq b \neq c$$ and $$\alpha = \beta = \gamma = 90^{\circ}$$

D
$$a = b \neq c$$ and $$\alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$$
Question 4
1 / -0

Solid sodium amide, $$NaNH_2$$ has a cubic closed packed arrangement of $$N{ H }_2^-$$ ions with _____ voids occupied by $$Na^+$$ ions.

A
half the number of octahedral

B
all tetrahedral

C
half the number of tetrahedral

D
double the number of octahedral

Solution

$$Z_{effective}\quad of\quad {NH_2}^{-}=6\times \cfrac {1}{2}+8\times \cfrac {1}{8}=4$$ (CCP arrangement)

To maintain the $$NaNH_2$$ formula $$Z_{effective}$$ of $$Na^{+}$$ must be $$4$$.
So, $$Na^{+}$$ will be present at alternate tetrahedral voids.

$$Z_{effective}$$ of $$Na^{+}=\cfrac {1}{2}\times 8=4$$

Answer: (C) half the number of tetrahedral
Question 5
1 / -0

Which of the following is amorphous?

A
Sodium chloride

B
Quartz glass

C
Powdered marble

D
Cane sugar

Solution

Amorphous is defined as a substance laking clear structure or shape.
In quartz glass, elements are not organized in a periodic array, so they don't have a proper shape or structure.

Hence, quartz glass is am an example of an amorphous solid.
Question 6
1 / -0

An element (atomic mass $$=100$$ g/mol) having bcc structure has unit cell edge length $$400$$ pm. The density (in g cm$$^{-3}$$) of the element is:

A
$$10.376$$

B
$$5.188$$

C
$$7.289$$

D
$$2.144$$

Solution

Given, M $$= $$ 100 g/mol, a $$= $$ 400pm

For BCC -

$$n= 8\times \dfrac{1}{8}+ 1= 2$$

As we know,

$$\text{Density}\displaystyle=\frac{n\times M}{a^3\times N_A}$$ $$\displaystyle=\frac{2\times100}{(400\times10^{-10})^3\times6.02\times10^{23}}$$ $$=5.188$$ g cm$$^{-3}$$

$$(\because 400pm= 400\times 10^{-10} cm)$$

Hence, the correct option is B.
Question 7
1 / -0

Average composition of ordinary glass is:

A
$$Na_2O\cdot CaO\cdot 6SiO_2$$

B
$$Na_2O\cdot 2CaO\cdot 3SiO_2$$

C
$$Na_2O\cdot CaO\cdot 6SiO_2\cdot Cr_2O_3$$

D
$$Na_2O\cdot CaO\cdot PbO\cdot 6SiO_2$$

Solution

$$Na_2CO_3+xSiO_2\rightarrow Na_2O\cdot xSiO_2+CO_2$$

$$Na_2SO_4+C\rightarrow Na_2O+CO+SO_2$$

$$Na_2O+ySiO_2\rightarrow Na_2O\cdot ySiO_2$$

$$CaCO_3+zSiO_2\rightarrow CaO\cdot zSiO_2+CO_2\uparrow$$

The overall reaction on simultaneous fusion may be represented as:-
$$Na_2CO_3+CaCO_3+6SiO_2\rightarrow Na_2O\cdot CaO\cdot 6SiO_2+CO_2\uparrow$$

Specifically called as soda lime glass composed of approximately $$75$$%
$$SiO_2$$,$$Na_2O$$ and $$CaO$$ (lime).

Option A is the answer.
Question 8
1 / -0

A compound $$AB$$ has a rock type structure with $$A:B =1:1$$. The formula mass of $$AB$$ is $$6.023$$ $$y^{1/3}$$ y amu and the closest $$A-B$$ distance is $$y^{1/3}$$ nm.Calculate the density of lattice.

A
$$5.0 kg/m^3$$

B
$$10 kg/m^3$$

C
$$50kg/m^3$$

D
$$20 kg/m^3$$

Solution

$$AB$$ has rock-salt $$(A:B: :1:1)$$ structure, i.e., fcc structure $$(n=4)$$ and formula mass of $$AB$$ is 6.023 y g having closest distance $$A-B\:y^{1/3}$$ nm. Therefore, edge length of unit cell
$$(a)=2(A^++B^-)=2\times y^{1/3}\times10^{-9}m$$

$$\therefore$$ $$Density of AB=\displaystyle\frac{n\times\:formula\:mass}
{Av.no.\times V}=\frac{n\times\:formula\:mass}{6.023\times10^{23}\times a^3}$$

$$=\displaystyle\frac{4\times6.023\times y\times10^{-3}}{6.023\times10^{23}\times(2y1/3\times10^{-9})^3}$$ (formula mass in kg)

$$=5.0kg/m^3$$
Question 9
1 / -0

Crystal system in which maximum number of Bravais lattices are possible is:

A
cubic

B
triclinic

C
orthorhombic

D
rhombohedral

Solution

Orthorhombic crystal system has all four lattices i.e. primitive, body-centered, face-centered and end-centered.
Question 10
1 / -0

The maximum proportion of available volume that can be filled by hard spheres in diamond is:

A
$$0.52$$

B
$$0.34$$

C
$$0.32$$

D
$$0.68$$

Solution

The maximum packing or the maximum proportion of volume filled by hard spheres in various arrangements are as follows:

(a) Simple cubic $$\displaystyle=\frac{\pi}{6}=0.52$$

(b) Body-centered cubic $$\displaystyle=\frac{\pi\sqrt{3}}{8}=0.68$$

(c) Face-centered cubic $$\displaystyle=\frac{\pi\sqrt{2}}{6}=0.74$$

(d) Hexagonal closed packing $$\displaystyle=\frac{\pi\sqrt{2}}{6}=0.74$$

(e) Diamond $$\displaystyle=\frac{\pi\sqrt{3}}{6}=0.34$$

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Solid State Test - 32

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Solid State Test - 32

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Question 1

Frenkel defect

Schottky defect

metal excess defect

No defect

Solution

Question 2

hexagonal close packing (hcp) and cubic close packing (ccp)

cubic closed packing (ccp) and hexagonal closed packing (hcp)

body centered cubic (bcc) and hexagonal close packing (hcp)

hexagonal closed packing (hcp) and body-centered cubic (bcc)

Solution

Question 3

$$a = b = c$$ and $$\alpha = \beta = \gamma = 90^{\circ}$$

$$a = b \neq c$$ and $$\alpha = \beta = \gamma = 90^{\circ}$$

$$a \neq b \neq c$$ and $$\alpha = \beta = \gamma = 90^{\circ}$$

$$a = b \neq c$$ and $$\alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$$

Question 4

half the number of octahedral

all tetrahedral

half the number of tetrahedral

double the number of octahedral

Solution

Question 5

Sodium chloride

Quartz glass

Powdered marble

Cane sugar

Solution

Question 6

$$10.376$$

$$5.188$$

$$7.289$$

$$2.144$$

Solution

Question 7

$$Na_2O\cdot CaO\cdot 6SiO_2$$

$$Na_2O\cdot 2CaO\cdot 3SiO_2$$

$$Na_2O\cdot CaO\cdot 6SiO_2\cdot Cr_2O_3$$

$$Na_2O\cdot CaO\cdot PbO\cdot 6SiO_2$$

Solution

Question 8

$$5.0 kg/m^3$$

$$10 kg/m^3$$

$$50kg/m^3$$

$$20 kg/m^3$$

Solution

Question 9

cubic

triclinic

orthorhombic

rhombohedral

Solution

Question 10

$$0.52$$

$$0.34$$

$$0.32$$

$$0.68$$

Solution

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