Defects in crystal Questions in English

(N/A) When the stoichiometry of a crystal is disturbed by the presence of defects,these are called non-stoichiometric defects.
These defects arise due to the presence of either extra cations or extra anions,which disturbs the ratio of cations and anions in the crystal.
$A$ large number of inorganic compounds are known to have constituents in non-stoichiometric ratios due to these defects in the crystal structure.
In compounds with non-stoichiometric defects,electrical neutrality is maintained either by the presence of extra electrons or by the presence of extra positive charges.
There are two types of non-stoichiometric defects:
$(i)$ Metal excess defects.
$(ii)$ Metal deficiency defects.

(N/A) $(i)$ Defects due to anionic vacancies: Alkali halides such as $NaCl$ and $KCl$ show this type of defect.
In this type of defect,an anion is missing from its lattice site,and the site is occupied by an electron to maintain electrical neutrality. These electrons trapped in anionic vacancies are called $F$-centers (from the German word $Farbenzentrum$,meaning color center),which are responsible for the color of the crystals.
$(ii)$ Defects due to the presence of extra cations at interstitial sites: Zinc oxide $(ZnO)$ is white at room temperature. On heating,it loses oxygen and turns yellow due to the following reaction:
$ZnO \xrightarrow{\Delta} Zn^{2+} + \frac{1}{2}O_2 + 2e^-$
Now,there is an excess of zinc in the crystal and its formula becomes $Zn_{1+x}O$. The excess $Zn^{2+}$ ions move to interstitial sites and the electrons to neighboring interstitial sites to maintain electrical neutrality.

(N/A) For an $FCC$ lattice, the number of atoms per unit cell is $Z = 4$. The molar mass of $Ca$ is $M = 40.08 \ g \ mol^{-1}$. The edge length is $a = 556 \ pm = 5.56 \times 10^{-8} \ cm$. The Avogadro constant is $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
$(i)$ In a Schottky defect, some lattice sites are vacant, reducing the number of effective atoms per unit cell. The new number of atoms is $Z' = 4 \times (1 - 0.001) = 3.996$. The density is $\rho = \frac{Z' \times M}{N_A \times a^3} = \frac{3.996 \times 40.08}{6.022 \times 10^{23} \times (5.56 \times 10^{-8})^3} \approx 1.543 \ g \ cm^{-3}$.
$(ii)$ In a Frenkel defect, ions shift to interstitial sites, but the total number of ions in the crystal remains the same. Thus, the number of atoms per unit cell remains $Z = 4$. The density is $\rho = \frac{Z \times M}{N_A \times a^3} = \frac{4 \times 40.08}{6.022 \times 10^{23} \times (5.56 \times 10^{-8})^3} \approx 1.545 \ g \ cm^{-3}$.

(A) $(i)$ Assertion $(A)$ is correct because in point defects like Schottky or metal deficiency defects,the ionic solid maintains electrical neutrality despite the absence of ions.
$(ii)$ Reason $(R)$ is correct because Frenkel defect occurs when a cation leaves its lattice site and occupies an interstitial site,which preserves the overall electrical neutrality of the crystal.
$(iii)$ However,the Reason $(R)$ describes the Frenkel defect,whereas the Assertion $(A)$ refers to a general property of point defects (specifically vacancy-type defects). Therefore,$(R)$ is not the correct explanation for $(A)$.

(A) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $y$.
Total number of $Fe$ ions is $x + y = 0.93$.
Since the crystal is electrically neutral,the total positive charge must equal the total negative charge ($-2$ for $O^{2-}$).
$2x + 3y = 2$.
Substituting $y = 0.93 - x$ into the charge equation:
$2x + 3(0.93 - x) = 2$
$2x + 2.79 - 3x = 2$
$-x = -0.79$
$x = 0.79$.
Thus,the number of $Fe^{2+}$ ions is $0.79$.
The percentage of $Fe^{2+}$ ions is $\frac{0.79}{0.93} \times 100 \approx 84.94\%$.
The nearest integer is $85\%$.

(D) In a vacancy defect,some of the lattice sites are vacant,which leads to a decrease in the number of atoms per unit volume,thereby decreasing the density of the substance. Therefore,the statement that vacancy defect increases the density is incorrect.

(A) Schottky defect is a type of vacancy defect in ionic solids,where an equal number of cations and anions are missing from their lattice sites to maintain electrical neutrality.This defect leads to a decrease in the density of the crystal.It is typically observed in ionic compounds where the cations and anions have similar sizes,such as $NaCl$ and $AgBr$.

(C) For $Fe_{0.93}O \rightarrow Fe_2O_3$:
$A.$ The oxidation state of $Fe$ in $Fe_{0.93}O$ is $x$: $0.93x - 2 = 0 \Rightarrow x = \frac{2}{0.93} \approx 2.15$. In $Fe_2O_3$,$Fe$ is $+3$. The change in oxidation state per $Fe$ atom is $3 - \frac{2}{0.93}$. Total $n$-factor $= 0.93 \times (3 - \frac{2}{0.93}) = 0.93 \times 3 - 2 = 2.79 - 2 = 0.79$. Thus,equivalent weight $= \frac{\text{Molecular weight}}{0.79}$. Statement $A$ is correct.
$B.$ Let moles of $Fe^{2+}$ be $x$ and $Fe^{3+}$ be $y$. $x + y = 0.93$ and $2x + 3y = 2$. Solving gives $x = 0.79$ and $y = 0.14$. Statement $B$ is correct.
$C.$ $Fe_{0.93}O$ is a non-stoichiometric metal-deficient oxide with $O^{2-}$ in $ccp$ arrangement. Statement $C$ is correct.
$D.$ $\% Fe^{2+} = \frac{0.79}{0.93} \times 100 \approx 84.95\% \approx 85\%$. $\% Fe^{3+} = \frac{0.14}{0.93} \times 100 \approx 15.05\% \approx 15\%$. Statement $D$ is correct.
All $4$ statements are correct.

(D) In the metal deficiency defect of $A_{0.95} O$,the total charge must be neutral. Let the number of $A^{3+}$ ions be $x$. Then the number of $A^{2+}$ ions is $(0.95 - x)$.
Equating the total positive charge to the total negative charge (which is $2$ for $O^{2-}$):
$3x + 2(0.95 - x) = 2$
$3x + 1.9 - 2x = 2$
$x = 0.1$
So,there are $0.1$ $A^{3+}$ ions and $0.85$ $A^{2+}$ ions for every $O^{2-}$ ion.
This defect involves the replacement of $A^{2+}$ ions by $A^{3+}$ ions,creating cation vacancies to maintain electrical neutrality. Structure $C$ correctly shows the presence of $A^{2+}$,$A^{3+}$,and a vacant cation site (represented by an empty circle with a dot).

(A) and $C$ are correct statements.
$A$. Incorrect: Frenkel defect is favoured by a large difference in the sizes of cation and anion.
$B$. Correct: Frenkel defect is a dislocation defect where an ion moves from its lattice site to an interstitial site.
$C$. Correct: Trapping of an electron in an anionic vacancy leads to the formation of $F$-centers,which impart color to the crystal.
$D$. Incorrect: Schottky defects decrease the density of the solid,thus affecting its physical properties.

(A) Stainless steel is an alloy of iron $(Fe)$ with other elements like chromium $(Cr)$,nickel $(Ni)$,and carbon $(C)$.
In the crystal lattice of iron,some iron atoms are replaced by atoms of other metals like chromium or nickel.
This type of defect,where foreign atoms replace the host atoms in the lattice,is known as a $Substitutional \ impurity \ defect$.

(B) In a $Frenkel$ defect,an ion (usually the smaller cation) is displaced from its normal lattice site and occupies an interstitial site.
This defect does not change the stoichiometry of the crystal because the number of ions remains the same.
It is commonly observed in ionic compounds where there is a large difference in the size of ions,such as $AgCl$ or $ZnS$.

(C) The correct answer is $C$ (Schottky defect).
$A$ $Schottky$ defect occurs in an ionic solid when equal numbers of cations and anions are missing from their regular positions in the crystal lattice,creating vacancies.
This defect maintains the overall electrical neutrality of the crystal.

(A) Frenkel defect occurs in ionic compounds where there is a large difference in the size of ions,specifically where the cation is much smaller than the anion. Option $A$ is incorrect because Frenkel defect is characteristic of compounds with a large difference in ionic sizes,whereas Schottky defect occurs in compounds where cation and anion sizes are almost equal.

(A) Schottky defect is a type of point defect in ionic crystals where an equal number of cations and anions are missing from their lattice sites.
$1$. Electrical neutrality is maintained because the number of missing cations equals the number of missing anions.
$2$. It typically occurs in ionic compounds where the cation and anion sizes are similar.
$3$. It is favored in compounds with high coordination numbers.
$4$. Due to the loss of ions from the lattice,the density of the crystal decreases.
Therefore,the statement that electrical neutrality is not preserved is incorrect.

(A) In Schottky defect,cations and anions are missing from their lattice sites in stoichiometric proportion.
It is a type of vacancy defect in ionic solids.
To maintain electrical neutrality,the number of missing cations and anions must be equal.
This defect decreases the density of the crystal.

(D) Frenkel defect is observed in ionic solids where there is a large difference in the size of the ions.
Specifically,it occurs in crystals where the smaller ion (usually the cation) can leave its lattice site and occupy an interstitial site.
$AgCl$ shows Frenkel defect because the $Ag^+$ ion is significantly smaller than the $Cl^-$ ion,allowing it to move into interstitial positions.
In contrast,$NaCl$,$CsCl$,and $KCl$ typically exhibit Schottky defects due to the comparable sizes of their constituent ions.

(A) $ZnS$ exhibits Frenkel defect because the $Zn^{2+}$ ion is small enough to occupy interstitial sites,whereas Schottky defect is typically observed in ionic solids with similar-sized cations and anions,such as $NaCl$,$KCl$,and $CsCl$.

(D) defect is called non-stoichiometric if the ratio of the number of cations to anions in the crystal becomes different from that indicated by its chemical formula.
This occurs due to the presence of metal excess or metal deficiency in the crystal lattice.

(C) Brass is an alloy of $Cu$ and $Zn$. In brass,$Zn$ atoms replace some of the $Cu$ atoms in the crystal lattice of copper. Since the host atoms are replaced by foreign atoms of similar size,this is classified as a substitutional impurity defect.

(B) Option $(B)$ is correct.
$1$. Frenkel defect occurs in compounds where there is a large difference in the size of ions,specifically where anions are much larger than cations.
$2$. Frenkel defect is known as a dislocation defect because an ion is dislocated from its normal lattice site to an interstitial site.
$3$. $F$-centers are formed by the trapping of electrons in anion vacancies,not protons.
$4$. Schottky defect leads to a decrease in the density of the crystal due to the presence of vacancies.

(C) The correct statement is that the trapping of an electron in the lattice leads to the formation of an $F$-centre.
Frenkel defect is a dislocation defect where an ion leaves its lattice site and occupies an interstitial site.
Schottky defect is a vacancy defect where equal numbers of cations and anions are missing from the lattice,which leads to a decrease in the density of the solid.

(C) When $SrCl_{2}$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^{+}$ ions to maintain electrical neutrality.
One $Sr^{2+}$ ion creates one cationic vacancy.
Therefore,$1 \ \text{mole}$ of $SrCl_{2}$ creates $1 \ \text{mole}$ of cationic vacancies.
$10^{-5} \ \text{mole}$ of $SrCl_{2}$ will create $10^{-5} \ \text{mole}$ of cationic vacancies.
Number of cationic vacancies $= 10^{-5} \times 6.022 \times 10^{23} = 6.022 \times 10^{18}$.

(B) Schottky defect arises when an equal number of cations and anions are missing from their normal lattice sites to maintain electrical neutrality. This defect leads to a decrease in the density of the crystal.

(B) Dislocation defect (Frenkel defect) occurs in ionic crystals where there is a large difference in the size of the ions,allowing the smaller ion (usually the cation) to occupy an interstitial site. In alkali halides,the sizes of the cations and anions are almost equal,which prevents the smaller ion from fitting into the interstitial spaces. Therefore,they do not show dislocation defect.

(C) Analysis of the statements:
$(1)$ Incorrect: Frenkel defect is favored by a large difference in the sizes of cation and anion.
$(2)$ Incorrect: Frenkel defect is a dislocation defect,not a metal excess defect.
$(3)$ Correct: Trapping of electrons in anionic vacancies in the crystal lattice leads to the formation of $F$-centres (Farbenzenter),which are responsible for the color of the crystal.
$(4)$ Incorrect: Schottky defect decreases the density of the solid,thus affecting its physical properties.

(C) Let there be $100$ formula units of $Ni_{0.98}O_{1.00}$. Thus,for $100$ $O^{2-}$ ions,there are $98$ nickel ions ($Ni^{2+}$ or $Ni^{3+}$).
Let $x$ be the number of $Ni^{2+}$ ions.
Then,the number of $Ni^{3+}$ ions is $(98 - x)$.
For the compound to be electrically neutral,the total positive charge must equal the total negative charge:
$2x + 3(98 - x) + 100(-2) = 0$
$2x + 294 - 3x - 200 = 0$
$-x + 94 = 0$
$x = 94$
So,the number of $Ni^{2+}$ ions is $94$.
The percentage of $Ni^{2+}$ ions is $\frac{94}{98} \times 100 \approx 95.92\%$,which is approximately $96\%$.

(A) Among the given statements,$(B)$ and $(C)$ are incorrect.
$(B)$ Silicon $(Si)$ and germanium $(Ge)$ are group $14$ elements and are intrinsic semiconductors,not impurities.
$(C)$ Phosphorus $(P)$ and arsenic $(As)$ are group $15$ elements. They have $5$ valence electrons. When doped into group $14$ elements,$4$ electrons form covalent bonds,and the $5$th electron remains free. Thus,they are electron-rich impurities,not electron-deficient.

(B) When molten $NaCl$ containing a small amount of $SrCl_2$ is crystallized,some $Na^{+}$ ion sites are occupied by $Sr^{2+}$ ions.
Each $Sr^{2+}$ ion replaces two $Na^{+}$ ions to maintain electrical neutrality.
One $Sr^{2+}$ ion occupies the site of one $Na^{+}$ ion,while the other site remains vacant.
Thus,the number of cationic vacancies produced is equal to the number of $Sr^{2+}$ ions incorporated into the crystal lattice.

(D) Given the formula of nickel oxide is $Ni_{0.98}O_{1.00}$.
Let the number of $Ni^{2+}$ ions be $x$ and the number of $Ni^{3+}$ ions be $(0.98 - x)$.
Since the total charge of the compound must be zero,the sum of positive charges equals the sum of negative charges.
$2x + 3(0.98 - x) + 1(-2) = 0$
$2x + 2.94 - 3x - 2 = 0$
$-x + 0.94 = 0$
$x = 0.94$
So,the fraction of $Ni^{2+}$ ions $= \frac{0.94}{0.98} \times 100 \approx 96 \%$.
The fraction of $Ni^{3+}$ ions $= \frac{0.98 - 0.94}{0.98} \times 100 = \frac{0.04}{0.98} \times 100 \approx 4 \%$.
Therefore,the fractions are $Ni^{2+}=96 \%$ and $Ni^{3+}=4 \%$.

(B) . Frenkel defect: Shown by ionic substances with a large difference in ion sizes,e.g.,$AgCl$.
$B$. Schottky defect: Shown by ionic substances with similar cation and anion sizes,e.g.,$NaCl$.
$C$. Vacancy defect: Occurs when some lattice sites are vacant,leading to decreased density.
$D$. Metal deficiency defect: Occurs in non-stoichiometric solids like $FeO$ (often $Fe_{0.95}O$).

(C) The concentration of Schottky defects in a crystal is given by the formula $n = N \exp(-E/2kT)$,where $n$ is the number of Schottky pairs per $cm^3$,$N$ is the total number of lattice sites per $cm^3$,$E$ is the energy required to form a defect,$k$ is the Boltzmann constant,and $T$ is the temperature.
For $NaCl$ at room temperature $(298 \ K)$,the calculated value for the number of Schottky pairs is approximately $10^6$ per $cm^3$.

(B) $Frenkel$ defect occurs when an ion (usually a cation) is dislocated from its normal lattice site to an interstitial site.
Because the ions remain within the crystal lattice,the total mass and volume of the crystal do not change.
Therefore,the density of the crystal remains constant,not less than that of a pure perfect crystal.
This defect is typically observed in ionic solids where there is a large difference in the size of the ions.
$Schottky$ and $Frenkel$ defects can coexist in the same crystal.
Alkali halides (except $AgBr$) generally do not show $Frenkel$ defect due to the similar sizes of their ions.

(B) When $CaCl_2$ is added to $AgCl$ crystal,each $Ca^{2+}$ ion replaces two $Ag^{+}$ ions to maintain electrical neutrality.
Since one $Ca^{2+}$ ion replaces two $Ag^{+}$ ions,one $Ag^{+}$ site remains vacant,creating a cation vacancy.
This type of defect is a form of impurity defect,which is related to the Schottky defect mechanism where vacancies are created in the lattice.
Therefore,the introduction of $CaCl_2$ into $AgCl$ leads to the formation of cation vacancies,which is characteristic of Schottky-type defects.
Thus,the correct option is $B$.

(A) In a Frenkel defect,an ion leaves its lattice site and occupies an interstitial site within the same crystal lattice.
Since no atoms or ions are lost from or added to the crystal,the total mass $(M)$ remains constant $(M = M_0)$.
The volume $(V)$ of the crystal remains unchanged because the lattice structure does not expand or contract significantly $(V = V_0)$.
Since density $(D)$ is defined as mass divided by volume $(D = M/V)$,and both mass and volume remain constant,the density remains unchanged $(D = D_0)$.
Therefore,the correct option is $(A)$.

(C) When $CdCl_2$ is added to $AgCl$,each $Cd^{2+}$ ion replaces two $Ag^+$ ions to maintain electrical neutrality.
One $Cd^{2+}$ ion occupies the site of one $Ag^+$ ion,while the other $Ag^+$ site remains vacant.
Thus,the number of cation vacancies created is equal to the number of $Cd^{2+}$ ions added.
Given that $1 \times 10^{-4}$ mole percent of $CdCl_2$ is added,this means $1 \times 10^{-4} \text{ moles of } CdCl_2$ are present in $100 \text{ moles of } AgCl$.
Therefore,the number of moles of $Cd^{2+}$ ions per mole of $AgCl$ is $\frac{1 \times 10^{-4}}{100} = 1 \times 10^{-6} \text{ mol}$.
The number of cation vacancies per mole is $1 \times 10^{-6} \times N_A$,where $N_A = 6.023 \times 10^{23} \text{ mol}^{-1}$.
Number of vacancies $= 1 \times 10^{-6} \times 6.023 \times 10^{23} = 6.023 \times 10^{17} \text{ mol}^{-1}$.

(B) Zinc oxide is white in colour at room temperature. On heating,it loses oxygen and turns yellow.
$ZnO \xrightarrow{\text{heating}} Zn^{2+} + \frac{1}{2} O_2 + 2e^{-}$
The excess $Zn^{2+}$ ions move to interstitial sites and electrons move to neighbouring interstitial sites.
This is a 'metal excess' defect due to the presence of extra cations at interstitial sites.
Therefore,statements $(I)$ and $(III)$ are correct.

(C) When $NaCl$ crystals are heated in an atmosphere of $Na$ vapour,the $Na$ atoms deposit on the surface of the crystal.
$Cl^-$ ions diffuse to the surface and combine with $Na$ atoms to form $NaCl$,which releases electrons into the crystal lattice.
These electrons occupy the vacant anionic sites,known as $F$-centres ($F$ stands for the German word 'Farbe',meaning colour).
This phenomenon is a type of metal excess defect caused by anionic vacancies.

(C) The crystal of $FeO$ exhibits a metal deficiency defect because some $Fe^{2+}$ ions are missing from their lattice sites.
To maintain electrical neutrality,the loss of positive charge is compensated by the presence of $Fe^{3+}$ ions.
Since the number of metal ions is less than the stoichiometric ratio,it is a metal deficiency defect.
Typically,the formula is represented as $Fe_{0.95} O$.

(A) When a crystal lattice contains extra cations in the interstitial sites to maintain electrical neutrality,it is known as a metal excess defect.
This is a type of non-stoichiometric defect.

(D) $Schottky$ defect is a type of point defect in ionic crystals where an equal number of cations and anions are missing from their lattice sites to maintain electrical neutrality.
Because atoms/ions are missing from the lattice,the total mass of the crystal decreases while the volume remains constant.
Therefore,the density of the crystal decreases,not increases.
Thus,the statement in option $D$ is incorrect.

(D) The correct statement is that the trapping of electrons in the lattice leads to the formation of $F$-centres.
$1-$ Frenkel defect is favoured when the difference between the size of the cation and anion is large.
$2-$ Frenkel defect is also known as a dislocation defect.
$3-$ Schottky defect affects the physical properties of solids,specifically by decreasing the density of the substance.

(A) In a solid with Schottky defects, atoms are missing from their lattice positions, which decreases the effective number of atoms per unit cell.
For an $FCC$ lattice, the number of atoms per unit cell is $Z = 4$.
With $0.1 \%$ Schottky defects, the effective number of atoms per unit cell is:
$Z_{effective} = 4 \times (1 - \frac{0.1}{100}) = 4 \times 0.999 = 3.996$.
Using the density formula $d = \frac{Z_{effective} \times M}{N_A \times a^3}$:
Given $M = 56 \ g/mol$, $N_A = 6.022 \times 10^{23} \ mol^{-1}$, and $a = 400 \ pm = 4 \times 10^{-8} \ cm$.
$d = \frac{3.996 \times 56}{(6.022 \times 10^{23}) \times (4 \times 10^{-8})^3} \approx \frac{223.776}{6.022 \times 10^{23} \times 64 \times 10^{-24}} \approx \frac{223.776}{38.54} \approx 5.81 \ g/cm^3$.

(B) In a Schottky defect,an equal number of cations and anions are missing from the lattice,which reduces the effective number of atoms per unit cell $(Z_{eff})$.
For an $FCC$ lattice,the number of atoms per unit cell is $Z = 4$.
With $0.1 \%$ Schottky defects,the new effective number of atoms is $Z_{new} = 4 - (4 \times 0.001) = 3.996 \approx 4$.
However,using the provided calculation logic: $Z_{new} = 4 - 0.004 = 3.996$.
Given the atomic mass of Potassium $(K)$ is $39 \ g \ mol^{-1}$.
The edge length $a = 0.5 \ nm = 5 \times 10^{-8} \ cm$.
The density formula is $\rho = \frac{Z \times M}{N_A \times a^3}$.
$\rho = \frac{3.996 \times 39}{6.022 \times 10^{23} \times (5 \times 10^{-8})^3} = \frac{155.844}{6.022 \times 10^{23} \times 125 \times 10^{-24}} = \frac{155.844}{75.275} \approx 2.07 \ g \ cm^{-3}$.
Rounding to the nearest value,we get $2.1 \ g \ cm^{-3}$.
Thus,Option $B$ is correct.

(C) Anion vacancies in alkali halides are produced by heating the alkali halide crystals in an atmosphere of the alkali metal vapour.
When the metal atoms deposit on the surface,they diffuse into the crystal.
After ionization,the alkali metal ion occupies cationic vacancies,whereas the electron occupies the anionic vacancy.
Electrons trapped in anion vacancies are referred to as $F$-centers,which give rise to the characteristic yellow colour in $NaCl$ crystals.

(A) Let $x$ be the number of $M^{3+}$ ions and $y$ be the number of $M^{2+}$ ions.
Total number of metal ions is $x + y = 0.96$.
Therefore,$y = 0.96 - x$.
Since the oxide is electrically neutral,the total positive charge must equal the total negative charge ($-2$ for $O^{2-}$).
$3x + 2y = 2$.
Substituting $y = 0.96 - x$ into the equation:
$3x + 2(0.96 - x) = 2$.
$3x + 1.92 - 2x = 2$.
$x = 0.08$.
Thus,$y = 0.96 - 0.08 = 0.88$.
The fraction of $M^{3+} = \frac{0.08}{0.96} = 0.083$.
The fraction of $M^{2+} = \frac{0.88}{0.96} = 0.917$.

(C) In $Schottky$ defect,an equal number of cations and anions are missing from the crystal lattice,which creates an equal number of cation and anion vacancies to maintain electrical neutrality.