Solid State Test

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Solid State Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

The unit cell with the given structure and a = b = c, represents ____________ crystal system.

A
cubic

B
orthorhombic

C
tetragonal

D
trigonal

Solution

If $$a=b=c$$ and $$\alpha = \beta = \gamma = 90^o$$, the given structure will be a cubic crystal system.

Image I : $$a = b = c $$
all angles $$= 90^o$$ (Isometric ) cubic

Image II : $$a \neq b \neq c$$
all angles $$= 90^o$$
Orthorhombic
Question 2
1 / -0

The pink colour of lithium chloride crystal is due to:

A
frenkel defect

B
metal excess defect

C
metal deficiency defect

D
impurity defect

Solution

$$LiCl$$ has non-stochiometric metal excess defect due to anion vacancies. The negative ions $$(Cl^-)$$ are missing from their lattice sites leaving the holes in which electrons are entrapped so that electrical neutrality is maintained.
When $$LiCl$$ is heated, $$Li$$ atoms gets deposited on the surface of the crystal. The $$Cl^-$$ ions diffuse into the surface and combine with $$Li$$ atoms to give $$LiCl$$. This is so because of loss of electrons by $$Li$$ atoms to form $$Li^+$$. The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of $$Li$$. The anionic sites occupied by unpaired electrons are F-centers which imparts a pink color to $$LiCl$$ crystals. The color is observed as a result of excitation of these electrons when they absorb energy from visible light falling on crystals.
Question 3
1 / -0

Which of the following is an example for interstitial solid solution?

A
$$SiC$$

B
$$TiC$$

C
$$Li_{2}C_{2}$$

D
$$Al_{4}C_{3}$$

Solution

$$ \begin{array}{l} \text { Interstitial solid solution - Interstitial solid solution } \\ \text { forms when the solute atom is small enough to } \\ \text { fit at interstitial sites between the solvent atoms. } \\ \text { example- TiC } \end{array} $$
Question 4
1 / -0

In fcc, a tetrahedral void is formed by atoms at

A
3 corners and 1 face centre

B
3 face centres and 1 corner

C
2 face centres and 2 corners

D
2 face centres, 1 corner and 1 body centre

Solution

1 from corner and 3 adjacent face centres form tetrahedral void.. hence option B is correct.
Question 5
1 / -0

$$N{a_2}O$$ has:

A
NaCl structure

B
ZnS structure

C
fluorite structure

D
antifluorite structure

Solution

$$Na_2O$$ structure is antifluorite type structure in which $$O^{2-}$$ ions forms ccp (cubic close packing) and $$Na^{+}$$ occupy all tetrahedral voids.
Question 6
1 / -0

$$KF$$ has $$NaCl$$ structure. What is the distance (in $$A^o$$) between $$K^+$$ and $$F^-$$ in $$KF$$, if density of $$KF$$ is $$2.48$$ $$gm/cm^3$$? $$[K=39, F=19]$$

A
$$2.86$$

B
$$2.68$$

C
$$3.16$$

D
$$3.68$$

Solution

Effective number of KF molecules is 4.
$$\rho=\cfrac{Z\times M_A}{N_A\times a^3}$$
$$\implies 2.48=\cfrac{4\times 58}{6.023\times 10^{23}\times a^3}$$
$$\implies a^3=155.3\times 10^{-24}$$
$$\implies a=5.36\times 10^{-8}$$ $$cm=5.36\mathring A$$
In $$NaCl$$ type stucture distance between cation and anion $$=\cfrac{a}{2}$$
$$\implies$$ Distance between cation and anion $$=\cfrac{a}{2}=2.68\mathring A$$
Question 7
1 / -0

Radii of $$A^+$$ and that of $$X^-$$ and $$Y^-$$ have been given as
$$A^+=1.00$$ pm
$$X^-=1.00$$ pm
$$Y^-=2.00$$ pm
Determine the volume of unit cells of AX and AY crystals.

A
$$123.2\times10^{-36} m^3, 2160\times10^{-36} m^3$$

B
$$12.32\times10^{-36} m^3, 216\times10^{-36} m^3$$

C
$$22.62\times10^{-36} m^3, 76.36\times10^{-36} m^3$$

D
$$93.23\times10^{-36} m^3, 142\times10^{-36} m^3$$

Solution

$$i)$$ For $$A^+X^-$$
$$r_{A+}=1.00$$ $$pm$$
$$r_{A-}=1.00$$ $$pm$$
If $$\cfrac{r_{A^+}}{r_{X^-}}=1$$, then it is in CsCl type structure
In CsCl type structure, $$A$$ is present at body centre and $$B$$ is present at corner.
$$2(r_{A^+}+r_{X^-})=a\sqrt{3}$$
$$a=\cfrac{4}{\sqrt{3}}$$ $$pm$$
Volume $$=a^3=\cfrac{64}{3\sqrt{3}}\times 10^{-36}$$ $$m^3$$ = $$ 12.32\times 10^{-36}$$ $$m^3$$
$$ii)$$ For $$A^+Y^-$$
$$r_{A^+}=1.00$$ $$pm$$
$$r_{Y^-}=2.00$$ $$pm$$
$$\cfrac{r_{A^+}}{r_{Y^-}}=\cfrac{1.00}{2.00}=0.5$$. Then the crystal will have octahedral structure. ($$NaCl$$ type)
$$2(r_{A^+}+r_{Y^-})=a$$
$$a=6$$ $$pm$$
Volume $$=(6)^3=216\times 10^{-36}$$ $$m^3$$
Question 8
1 / -0

Packing fraction in a body-centered cubic cell of crystals is:

A
$$\frac{{\sqrt 3 }}{8}\pi $$

B
$$\frac{\pi }{6}$$

C
$$\frac{{\sqrt 2 }}{6}\pi $$

D
$$\frac{1}{{3\sqrt 2 }}\pi $$

Solution

We know that,
$$\begin{array}{l}APF = \dfrac{{{N_{atoms}}{V_{atom}}}}{{{V_{unit\_cell}}}} = \frac{{6.\dfrac{4}{3}\pi {r^3}}}{{\dfrac{{3\sqrt 3 }}{2}{a^2}c}}\\ = \dfrac{{6.\frac{4}{3}\pi {r^3}}}{{\dfrac{{3\sqrt 3 }}{2}{{(2r)}^2}\sqrt {\dfrac{2}{3}} .4r}} = \dfrac{{6.\frac{4}{3}\pi {r^3}}}{{\dfrac{{3\sqrt 3 }}{2}.\sqrt {\dfrac{2}{3}} .16{r^3}}}\\ = \dfrac{\pi }{{\sqrt {18} }} = \dfrac{\pi }{{3\sqrt 2 }} \approx 0.74048048\end{array}$$
hence option (D) is correct
Question 9
1 / -0

Aluminium crystallizes in a cubic close packed structure. Its metallic radius is $$125pm$$.
i) Calculate the edge length of unit cell
ii) How many unit cells are there in $$1.00cm^3$$ of aluminium?

A
$$354 pm, 2.41\times 10^{22}$$

B
$$355 pm, 3.41\times 10^{20}$$

C
$$356 pm, 4.41\times 10^{21}$$

D
$$357 pm, 5.41\times 10^{23}$$

Solution

For a $$FCC$$ or $$CCP$$ unit cell,
$$ 4r = \sqrt2 \times a$$ or $$a = \dfrac{4r}{\sqrt2} = 354pm$$
$$\because$$ Volume of one unit crystal $$= a^3 = 44361864\times 10^{-30}cm^3$$
$$\therefore$$ No. of unit cells in $$1cm^3$$ aluminium = $$\dfrac{1\times 10^{30}}{44361864}= 2.41\times 10^{22}$$
Question 10
1 / -0

In a solid $$AB$$ having $$NaCl$$ structure atoms, occupy the corners of the unit cell. If all the face centred atoms along one of the axis are removed, then the resultant stoichiometry of the solid is:

A
$$AB_2$$

B
$$A_2B$$

C
$$A_4B_3$$

D
$$A_3B_4$$

Solution

A occupies $$8$$ corners which contributes $$\cfrac{1}{8}\Rightarrow \cfrac{1}{8}\times 8=1$$
and $$6$$ face centered contributes $$\cfrac{1}{2}\Rightarrow \cfrac{1}{2}\times 6=3$$
But face centered atom along any one axis is removed which will be $$2$$.
Now, face centre have $$4\times \cfrac{1}{2}=2$$
$$Z_{effective}$$ of $$A$$ after removing face centres at any axis $$=4\times \cfrac{1}{2}+\cfrac{1}{8}\times 8=3$$
$$B$$ occupies octahedral voids which contributes $$\cfrac{1}{4}$$
Therefore, $$1+12\times \cfrac{1}{4}=4=Z_{effective}$$ of $$B$$
Formula $$=A_3B_4$$

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Re-Attempt Weekly Quiz Competition

Solid State Test - 66

Result

Solid State Test - 66

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SHARING IS CARING

Question 1

cubic

orthorhombic

tetragonal

trigonal

Solution

Question 2

frenkel defect

metal excess defect

metal deficiency defect

impurity defect

Solution

Question 3

$$SiC$$

$$TiC$$

$$Li_{2}C_{2}$$

$$Al_{4}C_{3}$$

Solution

Question 4

3 corners and 1 face centre

3 face centres and 1 corner

2 face centres and 2 corners

2 face centres, 1 corner and 1 body centre

Solution

Question 5

NaCl structure

ZnS structure

fluorite structure

antifluorite structure

Solution

Question 6

$$2.86$$

$$2.68$$

$$3.16$$

$$3.68$$

Solution

Question 7

$$123.2\times10^{-36} m^3, 2160\times10^{-36} m^3$$

$$12.32\times10^{-36} m^3, 216\times10^{-36} m^3$$

$$22.62\times10^{-36} m^3, 76.36\times10^{-36} m^3$$

$$93.23\times10^{-36} m^3, 142\times10^{-36} m^3$$

Solution

Question 8

$$\frac{{\sqrt 3 }}{8}\pi $$

$$\frac{\pi }{6}$$

$$\frac{{\sqrt 2 }}{6}\pi $$

$$\frac{1}{{3\sqrt 2 }}\pi $$

Solution

Question 9

$$354 pm, 2.41\times 10^{22}$$

$$355 pm, 3.41\times 10^{20}$$

$$356 pm, 4.41\times 10^{21}$$

$$357 pm, 5.41\times 10^{23}$$

Solution

Question 10

$$AB_2$$

$$A_2B$$

$$A_4B_3$$

$$A_3B_4$$

Solution

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