Tag: imperfections in solids

Questions Related to imperfections in solids

In a crystal, at $827^{\circ}C$, one out of $10^{10}$ lattice site is found to be vacant, while in the same solid, one out of $2 \times 10^9$ lattice site is found to be vacant at $927^{\circ}C$. What is the enthalpy of vacancy formation in kJ/mol unit?

  1. $76.8$

  2. $176.8$

  3. $33$

  4. $23$


Correct Option: B
Explanation:

$\cfrac {^nV}{N}=Ae^{-\Delta H _V/RT} \longrightarrow (x)$

$\left(\cfrac {^nV}{N}\right) _{827^oC}=\cfrac {1}{10^{10}} \longrightarrow (1)$
$\left(\cfrac {^nV}{N}\right) _{927^oC}= \cfrac {1}{2 \times 10^9} \longrightarrow (2)$                                   $\therefore T _1= 827+273=1100K ; T _2=927+273=1200K$

$\Rightarrow \cfrac {(2)}{(1)}= \cfrac {1/2\times 10^9}{1/10^{10}}= \cfrac {10^ {10}}{2 \times 10^9}= 0.5 \times 10=5$
$x \Rightarrow 2.303 \log 5= - \cfrac {\Delta H _V}{R} \left(\cfrac {1}{T _1}-\cfrac {1}{T _2}\right)$
$\therefore \Delta H _V= 2.303 \times \log 5 \times 8.314 \times \left(\cfrac {1100\times 1200}{100}\right)$
               $=1.766 \times 10^5 J$
               $=176.6kJ/mol$ $unit$ .

Calcium crystallizes in a face centred cubic unit cell with a $=0.556$ nm. Calculate the density if it contains:

(i) $0.1\%$ Frenkel defect
(ii) $0.1\%$ Schottky defect

  1. $1.546, 1.546$

  2. $1.546, 1.544$

  3. $15.46, 15.46$

  4. $21.76, 41.66$


Correct Option: B
Explanation:

$1.$ In frenkel defect there is no change in original density

$\rho=\cfrac{Z\times M _A}{N _A\times a^3}$
$\implies \rho=\cfrac{4\times 40}{6.023\times 10^{23}\times 171.8\times 10^{-24}}$
$\implies \rho=1.546$ $g/cm^3$

$2.$ In schottky defect $Z _{eff}$ changes so density changes
$Z _{new}=4-\cfrac{4\times 0.1}{100}=3.996$
$\rho =\cfrac { Z\times M _{ A } }{ N _{ A }\times a^{ 3 } } $
$\implies \rho =\cfrac { 3.996\times 40 }{ 6.023\times 10^{ 23 }\times 171.8\times 10^{ -24 } } $
$\implies \rho =1.544$$g/cm^{ 3 }$

Statement 1: KCl is more likely to show schottky defect while Lil is more likely to show Frenkel defect.
Statement 2: Schottky defect is more likely in ionic solids in which cations and anions are of comparable size while Frenkel defect is more likely is which cations and anions have large differences in their ionic sizes.

  1. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation for Statement-1

  2. Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation for Statement-1

  3. Statement-1 is true but Statement-2 is false

  4. Statement-1 is false but Statement-2 is true


Correct Option: A
Explanation:

In case of KCl, the cation and the anion are of comparable size while in case of LiI, the cation Li+ is very small as compared to the iodide ion, I-. Hence, KCl is more likely to show Schottky defect while LiI tends to show Frenkel defect. Thus, the statements 1 and 2 are true and 2 is correct explanation of 1.

Statement-I:  Solids having more $F-$centres possess intense colours.

Statement-II:  Excess of $Na^{+}$ in $NaCl$ solid having F-centres makes it appear to pink.

  1. Statement-I is correct but Statement-II is wrong

  2. Statement-I is wrong but Statement-II is correct

  3. Both Statement-I and Statement-II are correct and Statement-II is correct explanation of Statement-I

  4. Both Statement-I and Statement-II are correct but Statement-II is not correct explanation of Statement-I


Correct Option: A

Which of the following statements are correct?

  1. Dislocation of ion from lattice site to interstitial site is called Frenkel defect.

  2. Missing of +ve and -ve ions from their respective position producing a pair of holes is called Schottky defect.

  3. The presence of ions in the vacant interstital sites along with lattice point is called interstital defect.

  4. Non-stoichiometric $NaCl$ is yellow solid.


Correct Option: A,B,C,D
Explanation:

Frenkel defect: This type of defect is created when an ion leaves its appropriate site in the lattice and occupies an interstitial site. A hole or vacancy is thus produced in the lattice.

Schottky defect: This type of defect is produced when one cation and anion are missing from their respective positions leaving behind a pair of holes.

The presence of ions in the vacant interstital sites along with lattice point is called interstital defect.

 When alkali metal halides are heated in an atmosphere of vapours of the alkali metal, anion vacancies are created. 


The anions (halide ions) diffuse to the surface of the crystal from their appropriate lattice sites to combine with the newly generated metal cations. The electron lost by the metal atom diffuse through the crystal is known as F-centres. 


The main consequence of metal excess defect in the development of colour in the crystal. For example, when $NaCl$ crystal is heated in an atmosphere of $Na$ vapours, it becomes yellow.


Therefore, all statements are correct.

Which of the following statements is/are correct ?

  1. If three $Fe^{2+ }$ ions are missing from their lattice in FeO,  there must be two $Fe^{3+ }$ ions somewhere in the lattice.

  2. Crystals with metal deficiency defects are called super conductors.

  3. Crystals with metal deficiency are called semiconductors.

  4. Bohr Magneton = $9.27 \times 10^{-24} A :m^2$.


Correct Option: A,C,D
Explanation:

The statements A, C and D are correct.
(A) If three $Fe^{2+}$ ions are missing from their lattice in FeO,  there must be two $Fe^{3+}$ ions somewhere in the lattice.
The charge on three $Fe^{2+}$ ions is $3 \times (+2) = +6$
The charge on two $Fe^{3+}$ ions is $2 \times{3+} = +6$
Thus, the electrical neutrality is maintained.

Two type of vacancies of iron are present here. This is called crystal with metal deficiency defect.

(B) Crystals with metal deficiency defects are not called super conductors. The idea of Superconductors is that it possesses zero resistance at a certain temperature. 

(C) Crystals with metal deficiency are called semiconductors.

(D) Bohr Magneton = $9.27 \times 10^{-24} A :m^2$.
The orbital or spin magnetic momentum of an electron is expressed in terms of Bohr magneton. Its symbol is $\mu _B$.

Except option B all are correct.

Assertion: Solids containing F centers are paramagnetic.
Reason: Solids containing F centers conduct electricity and it is an n-type semiconductor. 
  1. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion

  3. Assertion is correct but Reason is not correct

  4. Assertion is not correct but Reason is correct

  5. Both Assertion and Reason are not correct


Correct Option: A
Explanation:

F centers are color centres consisting of an electron trapped by a negative ion vacancy in an ionic crystal such as an alkali halide or an alkaline-earth fluoride or oxide.
As F-center have free unpaired electron, they are paramagnetic in nature and because of this free electron, they conduct electricity.
Also, because of this extra electron, they form n-type semiconductor.

A strong current of trivalent gaseous boron passed through a germanium crystal decreases the density of the crystal due to part replacement of germanium by boron and due to interstitial vacancies created by missing Ge atoms. In one such experiment, one gram of germanium is taken and the boron atoms are found to be $150$ ppm by weight when the density of the Ge crystal decreases by $4\%$. Calculate the percentage of missing vacancies due to germanium which are filled up by boron atoms. Atomic weight of Ge $= 72.6$ amu and $B = 11$ amu.

  1. $2.4\, \%$

  2. $1.2\, \%$

  3. $6.6\, \%$

  4. None of the above


Correct Option: A
Explanation:

As given,
$1$ g Germanium $\equiv  150$ ppm of Boron $=1.5 \times 10^{-4}$ g of Boron
$4\%$ decrease in density of Ge occurs in one experiment due to replacement of Ge by B.

The weight becomes $0.96$ g (Ge + B).
Hence, $0.04$ g is missing.
$0.04$ missing means $(0.04-0.00015)$ g Ge missing $= \dfrac{(0.04-0.00015)}{72.61}$ moles
Ge missing $=$ $5.488\times 10^{-4}$ moles
Boron replaced $=$ $1.5\times 10^{-4} g = \dfrac{(1.5\times 10^{-4})}{10}$ moles
Equivalently replaced $= \dfrac 34\times $ $\dfrac{1.5\times 10^{-4}}{10}$ moles of Ge ( Valency of B is $3$ and Ge is $4$.)
$=0.1125\times 10^{-4}$ moles of Ge
Percentage of missing vacancies filled by B atoms $=\dfrac {0.1125\times 100}{5.488}$ $=2.05\%$

Crystals have 'vacant sites' or 'defects' in them. When light strikes a photographic silver bromide paper, silver atoms move in through these defects to:

  1. develop the film

  2. form tiny clumps of silver atoms

  3. form negative images

  4. form a colour image


Correct Option: C
Explanation:

When light strikes a photographer $(AgBr)$ paper, it gives energy to the electrons present in the film. These energetic electrons when strike silver ions turn them to silver atoms. So eventually,  ions leave their lattice site and occupy interstitial sites. Since silver atoms are black in color so whenever light strikes a silver ion the photographic film will turn black.

The pink colour of lithium chloride crystal is due to:

  1. frenkel defect

  2. metal excess defect

  3. metal deficiency defect

  4. impurity defect


Correct Option: B
Explanation:

$LiCl$ 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.