Defects in crystal Questions in English

(B) Doping $NaCl$ with $10^{-4} \ mol \%$ of $SrCl_2$ means that $100 \ mol$ of $NaCl$ contains $10^{-4} \ mol$ of $SrCl_2.$
Therefore,$1 \ mol$ of $NaCl$ contains $10^{-4} / 100 = 10^{-6} \ mol$ of $SrCl_2.$
Each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality,creating one cation vacancy per $Sr^{2+}$ ion introduced.
Concentration of cation vacancies $= (10^{-6} \ mol \ SrCl_2 / mol \ NaCl) \times (N_A \ ions / mol) \times (1 \ \text{vacancy} / Sr^{2+} \ \text{ion}).$
$= 10^{-6} \times 6.02 \times 10^{23} \ \text{vacancies} \ mol^{-1} = 6.02 \times 10^{17} \ mol^{-1}.$

(B) $F$-centers are the sites where anions are missing and instead electrons are present. The appearance of colour in solid alkali metal halides is generally due to these $F$-centers.

(B) Given the formula of the metal oxide is $M_{0.98}O$.
Consider $1 \text{ mole}$ of the oxide.
Moles of $M = 0.98$ and moles of $O^{2-} = 1$.
Let the moles of $M^{3+} = x$.
Then,the moles of $M^{2+} = 0.98 - x$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge:
$(0.98 - x) \times 2 + x \times 3 = 1 \times 2$
$1.96 - 2x + 3x = 2$
$x = 2 - 1.96 = 0.04$.
Therefore,the fraction of metal existing as $M^{3+} = \frac{0.04}{0.98} \times 100 = 4.08 \%$.

(D) In a Schottky defect,an equal number of cations and anions leave their lattice sites to maintain electrical neutrality,creating vacancies.
Thus,both cation and anion vacancies must be present in the crystal lattice.
Option $D$ states that only one type of ion (either cation or anion) is missing,which would violate the principle of electrical neutrality.
Therefore,statement $D$ is incorrect.

(D) The density of a solid is defined as the ratio of mass to volume.
In $SiC$ (Silicon Carbide),the crystal lattice is fixed.
Option $A$ changes the mass of the unit cell.
Option $B$ (Vacancy defect) involves the loss of atoms,which decreases the mass while the volume remains constant,thus changing the density.
Option $C$ involves substituting atoms with different atomic masses ($Ge$ has a different mass than $Si$ or $C$),which changes the mass of the unit cell.
Option $D$ involves interchanging the positions of $Si$ and $C$ atoms within the existing lattice. Since the total number of $Si$ and $C$ atoms remains the same,the total mass of the unit cell remains unchanged. Therefore,the density remains constant.

(D) The number density of silicon atoms is given as $5 \times 10^{28} \ \text{atoms} \ m^{-3}$.
Converting this to $\text{cm}^{-3}$,we get $5 \times 10^{28} \times (10^{-2} \ m)^3 = 5 \times 10^{22} \ \text{atoms} \ cm^{-3}$.
Since $1$ indium atom is doped per $5 \times 10^7$ silicon atoms,the concentration of acceptor atoms $(n_A)$ is calculated as:
$n_A = \frac{5 \times 10^{22}}{5 \times 10^7} = 1 \times 10^{15} \ \text{atoms} \ cm^{-3}$.

(A) Statement $A$ is $WRONG$ because metals can exhibit both stoichiometric and non-stoichiometric defects (such as metal excess or metal deficiency defects).
Statement $B$ is $CORRECT$ because in Schottky defects,equal numbers of cations and anions are missing from the lattice,which decreases the mass while the volume remains effectively constant,thus reducing density.
Statement $C$ is $CORRECT$ because the introduction of foreign atoms (impurities) into the crystal lattice changes the mass of the unit cell,thereby altering the density.
Statement $D$ is $CORRECT$ because $F$-centres are formed when an anion is missing from its lattice site and an electron occupies the vacancy to maintain electrical neutrality,which is a type of metal excess defect.

(B) $1$. Frenkel defect occurs in ionic crystals where there is a large difference in the size of ions. The smaller ion (usually cation) is dislocated from its lattice site to an interstitial site. Thus,statement $B$ is correct,and statement $A$ is incorrect.
$2$. Paramagnetic substances are weakly attracted by a magnetic field,not repelled. Thus,statement $C$ is incorrect.
$3$. Schottky defect involves the missing of equal numbers of cations and anions from the lattice,which decreases the density of the crystal. Thus,it affects the physical properties of solids. Statement $D$ is incorrect.
$4$. Therefore,the only correct statement is $B$.

$(A)$ For an $FCC$ lattice,the number of atoms per unit cell $(Z)$ is $4$.
Due to $4\%$ Schottky defects,the effective number of atoms per unit cell $(Z_{\text{effective}})$ is:
$Z_{\text{effective}} = 4 \times (1 - 0.04) = 4 \times 0.96 = 3.84$.
The edge length $a = 400 \, pm = 400 \times 10^{-10} \, cm = 4 \times 10^{-8} \, cm$.
The observed density $(d)$ is given by the formula:
$d = \frac{Z_{\text{effective}} \times M}{N_A \times a^3}$.
Substituting the values:
$d = \frac{3.84 \times 24}{6 \times 10^{23} \times (4 \times 10^{-8})^3} = \frac{92.16}{6 \times 10^{23} \times 64 \times 10^{-24}} = \frac{92.16}{38.4} = 2.4 \, g/cm^3$.

(D) The doping concentration is $10^{-5} \, mol \, \%$,which means $10^{-5} \, mol$ of $AlCl_3$ per $100 \, mol$ of $NaCl$.
Number of moles of $AlCl_3$ per mole of $NaCl = \frac{10^{-5}}{100} = 10^{-7} \, mol$.
Since each $Al^{3+}$ ion replaces one $Na^+$ ion and creates vacancies to maintain electrical neutrality,one $Al^{3+}$ ion replaces three $Na^+$ ions. This results in $2$ cation vacancies per $Al^{3+}$ ion.
Total cation vacancies per mole of $NaCl = 2 \times (\text{moles of } Al^{3+}) \times N_A$.
$= 2 \times 10^{-7} \times 6.022 \times 10^{23} \, mol^{-1}$.
$= 12.044 \times 10^{16} \, mol^{-1}$.

(D) In $Schottky$ defect,equal number of cations and anions are missing from the lattice,which leads to a decrease in the density of the crystal.
In $Frenkel$ defect,an ion leaves its lattice site and occupies an interstitial site,so the number of ions remains the same,and the density of the crystal remains unchanged.
Therefore,the statement that the density of the crystal increases in $Frenkel$ defect is incorrect.

(D) In Schottky defect,equal number of cations and anions are missing from the lattice,so the stoichiometry remains unchanged.
Frenkel defect is observed in ionic compounds where there is a large difference in size between cation and anion,typically associated with low coordination numbers.
$F$-centres are anionic vacancies occupied by electrons,which absorb visible light and impart colour to the crystal.
Doping does not always increase the density of a crystal; it depends on the nature of the dopant (whether it is an interstitial impurity or a substitutional impurity). Therefore,the statement that density always increases is incorrect.

(D) When $Ru^{3+}$ ions are doped into $AgCl$,each $Ru^{3+}$ ion replaces $3 \ Ag^+$ ions to maintain electrical neutrality.
This creates $2$ cationic vacancies for every $Ru^{3+}$ ion added.
Given doping is $10^{-4} \, \text{mole percent}$,which is $\frac{10^{-4}}{100} = 10^{-6} \, \text{moles of } RuCl_3 \text{ per mole of } AgCl$.
Number of vacancies $= 2 \times 10^{-6} \, \text{moles per mole of } AgCl$.
Concentration of vacancies $= 2 \times 10^{-6} \times 6.023 \times 10^{23} \, \text{vacancies/mole}$.
$= 12.046 \times 10^{17} \, \text{cationic vacancies per mole of } AgCl$.

(A) $AgBr$ is a unique compound that exhibits both Schottky and Frenkel defects.
This occurs because the $Ag^+$ ion is small enough to occupy interstitial sites (Frenkel defect) and the crystal lattice structure allows for the formation of vacancies (Schottky defect) due to the specific radius ratio and coordination number.

(D) In a $Schottky$ defect,an equal number of cations and anions are missing from their lattice sites.
Since the number of ions decreases while the volume of the crystal remains the same,the overall mass of the crystal decreases.
Consequently,the density of the crystal decreases.
In contrast,$Frenkel$ defects do not change the density because ions merely shift to interstitial sites.
$Interstitial$ defects increase density by adding extra atoms to interstitial spaces.

(D) The appearance of colour in solid alkali metal halides is generally due to $F$-centres (Farbe centres).
These are anionic vacancies occupied by unpaired electrons.
When white light falls on these crystals,the electrons absorb energy from the visible region to get excited,which results in the crystal exhibiting a complementary colour.
For example,$NaCl$ with $F$-centres appears yellow,$KCl$ appears magenta,and $KBr$ appears blue.

(A) The density of a crystal is given by the formula $d = \frac{Z \times M}{N_A \times a^3}$.
For a face-centred cubic $(FCC)$ unit cell,the number of atoms per unit cell $Z = 4$.
The molar mass of Calcium $(Ca)$ is $M = 40 \ g \ mol^{-1}$.
The edge length $a = 0.556 \ nm = 0.556 \times 10^{-7} \ cm$.
Frenkel defects involve the migration of ions into interstitial sites,which does not change the mass or the volume of the unit cell; therefore,the theoretical density remains unchanged.
Substituting the values: $d = \frac{4 \times 40}{6.022 \times 10^{23} \times (0.556 \times 10^{-7})^3}$.
$d = \frac{160}{6.022 \times 10^{23} \times 1.719 \times 10^{-22}} \approx 1.55 \ g \ cm^{-3}$.

(B) Let the amount of $Fe^{+3}$ be $x$ and $Fe^{+2}$ be $(0.93 - x)$.
The total charge must be zero for the neutral compound $Fe_{0.93}O_{1.00}$.
$3x + 2(0.93 - x) - 2 = 0$
$3x + 1.86 - 2x - 2 = 0$
$x = 0.14$
The fraction of $Fe^{+3}$ ions among total iron cations is $\frac{0.14}{0.93}$.
Therefore,the percentage of $Fe^{+3}$ is $\frac{0.14}{0.93} \times 100 = \frac{14}{93} \times 100$.

(D) In a $Schottky$ defect,equal numbers of cations and anions are missing from their lattice sites.
Because the number of missing cations and anions is equal,the stoichiometry of the crystal remains unaffected.
However,because atoms/ions are missing from the lattice,the total mass of the crystal decreases while the volume remains constant,which leads to a decrease in the density of the crystal.
Therefore,the statement that the density remains unaffected is incorrect.

(D) $ZnO$ is white at room temperature. On heating,it loses oxygen and turns yellow due to the formation of metal excess defect. The reaction is: $ZnO \xrightarrow{\Delta} Zn^{2+} + \frac{1}{2}O_2 + 2e^-$. The excess $Zn^{2+}$ ions move to interstitial sites and the electrons move to neighboring interstitial sites to maintain electrical neutrality. These trapped electrons in the interstitial sites behave like $F$-centres,which absorb light from the visible region and impart a yellow colour to the crystal.

(C) $1$. Schottky defect is observed in ionic solids where the size of the cation and anion are almost similar.
$2$. Frenkel defect is observed in ionic solids where there is a large difference in the size of the ions,specifically when the cation is much smaller than the anion,allowing it to occupy interstitial sites.
$3$. As temperature increases,the thermal energy of the particles increases,which leads to an increase in the number of defects in a crystal lattice.
$4$. Therefore,statement $C$ is correct.

(B) The Schottky defect is a type of point defect in ionic crystals where an equal number of cations and anions are missing from their respective lattice sites to maintain electrical neutrality.

(A) $Schottky$ defect occurs when equal numbers of cations and anions are missing from their lattice sites.
Since the number of ions decreases while the volume of the crystal remains the same,the mass of the crystal decreases.
Therefore,the density of the solid decreases.

(A) In $F$-centers,electrons are trapped in anionic vacancies. These centers are responsible for the color of the crystal.

(C) The $Schottky$ defect arises due to the missing of an equal number of cations and anions from the crystal lattice,which maintains electrical neutrality.

(B) The decrease in density is due to the presence of vacancies.
Percentage of vacancies = Percentage decrease in density
$= \frac{\text{Density}_{\text{ideal}} - \text{Density}_{\text{observed}}}{\text{Density}_{\text{ideal}}} \times 100$
$= \frac{2.178 \times 10^3 - 2.165 \times 10^3}{2.178 \times 10^3} \times 100$
$= \frac{0.013}{2.178} \times 100 = 0.5969 \% \approx 5.96 \times 10^{-1} \%$

(D) Stoichiometric defects do not disturb the stoichiometry of the solid.
Since these defects arise due to the absorption of thermal energy,they are also known as intrinsic defects or thermodynamic defects.

(B) Vacancy defect occurs when some of the lattice sites are vacant in a crystal.
Since some atoms or ions are missing from the lattice,the total mass of the crystal decreases while the volume remains the same.
Therefore,the density of the solid substance decreases.

(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,maintaining electrical neutrality.
This defect is typically observed in ionic compounds with high coordination numbers and where the size difference between cations and anions is small.
$NaCl$ is a classic example of an ionic crystal that exhibits Schottky defect.
$ZnS$ typically shows Frenkel defect due to the small size of $Zn^{2+}$ ions.
$SrCl_2$ is often doped with $NaCl$ to create Schottky-type vacancies,but $NaCl$ itself is the primary example.

(A) Schottky defect is a type of point defect in ionic crystals.
It is generally observed in ionic compounds with high coordination numbers where the cations and anions are of similar sizes.
$Alkali$ $halides$ like $NaCl$,$KCl$,and $CsCl$ are classic examples of compounds exhibiting Schottky defects.

(B) In a crystal lattice,when a smaller ion (usually a cation) leaves its normal lattice site and occupies an interstitial site,it creates a vacancy at its original position and an interstitial defect at its new position. This type of defect is known as the $Frenkel$ defect. It is commonly observed in ionic compounds where there is a large difference in the size of ions,such as $AgCl$,$AgBr$,and $ZnS$.

(A) Schottky defect is a type of point defect in ionic crystals.
It is observed in ionic compounds where the size of the cation and the anion are almost similar.
This defect occurs due to the missing of an equal number of cations and anions from their lattice sites,maintaining electrical neutrality.

(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 to maintain electrical neutrality.
Therefore,it involves both cation and anion vacancies,not just cation vacancies.
Compounds exhibiting Schottky defect typically have high coordination numbers (e.g.,$NaCl$,$KCl$,$CsCl$).
Since the statement that it contains only cation vacancies is incorrect,the correct option is $A$.

(A) The 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.
This defect is temperature-dependent because the number of vacancies increases with an increase in temperature.
As the temperature increases,the thermal energy of the ions increases,making it easier for them to leave their lattice sites.
Therefore,the Schottky defect is maximum at the highest temperature among the given options.
Comparing the temperatures: $18\,^oC = 291\, K$,$273\, K$,$283\, K$,and $-273\,^oC = 0\, K$.
The highest temperature is $291\, K$ $(18\,^oC)$.

(B) The Frenkel defect is a type of point defect in crystals where an ion (usually a cation) leaves its lattice site and occupies an interstitial site.
Since no ions leave the crystal lattice entirely,the total mass and volume of the crystal remain unchanged.
Therefore,the density of the solid remains constant.

(A) Frenkel defect occurs when an ion is missing from its normal lattice site and occupies an interstitial site. Because the ion is displaced from its original position to an interstitial position,it is also known as $Dislocation \ defect$.

(B) Schottky defect is shown by ionic compounds with similar cation and anion sizes,such as $NaCl$.
Frenkel defect is shown by ionic compounds with a large difference in size between the cation and anion,such as $AgCl$ and $AgI$.
$AgBr$ is a unique case that exhibits both Schottky and Frenkel defects due to the intermediate size of the $Ag^+$ ion.

(D) In an $Interstitial \ defect$,some constituent particles (atoms or molecules) occupy an interstitial site.
This increases the number of particles per unit volume,thereby increasing the density of the solid.

(C) Non-stoichiometric defects are those in which the ratio of the constituent elements does not follow the law of constant proportions. These include $Metal \ excess \ defect$ and $Metal \ deficiency \ defect$. $Dislocation \ defect$ (also known as $Frenkel \ defect$) and $Impurity \ defect$ are types of stoichiometric defects. Therefore,$Dislocation \ defect$ is not a non-stoichiometric defect.

(A) When $NaCl$ crystals are heated in an atmosphere of sodium vapor,the sodium atoms deposit on the surface of the crystal.
$Cl^-$ ions diffuse to the surface and combine with $Na$ atoms to form $NaCl$.
This happens by the loss of electrons by sodium atoms to form $Na^+$ ions.
The released electrons diffuse into the crystal and occupy anionic sites.
These anionic sites occupied by unpaired electrons are called $F$-centers.
This results in an excess of sodium metal in the crystal,hence it is a metal excess defect.

(A) In ionic crystals like $NaCl$ and $KCl$,the cations and anions are of similar size.
Such compounds typically exhibit Schottky defects,which involve the missing of an equal number of cations and anions from their lattice sites to maintain electrical neutrality.
Therefore,the correct answer is $A$.

(B) When $LiCl$ crystals are heated in an atmosphere of $Li$ vapor,excess $Li$ atoms deposit on the surface of the crystal.
$Cl^-$ ions diffuse to the surface and combine with $Li$ atoms to form $LiCl$.
This process releases electrons which diffuse into the crystal and occupy the anionic sites (vacancies) previously held by $Cl^-$ ions.
These anionic sites occupied by unpaired electrons are called $F$-centers (from the German word $Farbenzenter$,meaning color centers).
These $F$-centers absorb energy from visible light and impart a characteristic color to the crystal.
For $LiCl$,the presence of $F$-centers results in a $Pale pink$ color.

(B) When $ZnO$ is heated,it loses oxygen and becomes non-stoichiometric: $ZnO \xrightarrow{\Delta} Zn^{2 } \frac{1}{2} O_2 2e^-$.
The excess $Zn^{2 }$ ions move to interstitial sites,and the electrons move to neighboring interstitial sites to maintain electrical neutrality.
This creates an $F$-center-like defect,which absorbs light in the visible region,causing the substance to appear yellow.

(C) Metal deficiency defect occurs when the metal is present in a ratio less than the stoichiometric proportion.
In the compound $Fe_{0.95}O$,the ratio of $Fe$ to $O$ is $0.95:1$,which is less than the ideal $1:1$ ratio.
This indicates that some $Fe^{2+}$ ions are missing from the lattice,and to maintain electrical neutrality,some $Fe^{2+}$ ions are replaced by $Fe^{3+}$ ions.
Therefore,$Fe_{0.95}O$ exhibits metal deficiency defect.

(B) Let the total number of $Fe$ atoms be $0.95$ and $O$ atoms be $1$.
Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.95 - x)$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge.
$2(x) + 3(0.95 - x) = 2(1)$
$2x + 2.85 - 3x = 2$
$-x = 2 - 2.85$
$x = 0.85$ (number of $Fe^{2+}$ ions).
Number of $Fe^{3+}$ ions $= 0.95 - 0.85 = 0.10$.
Percentage of $Fe^{3+} = (0.10 / 0.95) \times 100 = 10.53\%$.

(D) Let the total number of $Fe$ ions be $0.95$ and $O$ ions be $1$.
Let the number of $Fe^{+2}$ ions be $x$.
Then the number of $Fe^{+3}$ ions is $(0.95 - x)$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge.
$2x + 3(0.95 - x) = 2$
$2x + 2.85 - 3x = 2$
$-x = 2 - 2.85$
$-x = -0.85$
$x = 0.85$.
Percentage of $Fe^{+2} = (\frac{0.85}{0.95}) \times 100 = 89.47\%$.

(A) In $Fe_{0.94}O$,the total positive charge must balance the total negative charge to maintain electrical neutrality.
Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.94 - x)$.
The charge on $O^{2-}$ is $-2$.
So,$2x + 3(0.94 - x) = 2$.
$2x + 2.82 - 3x = 2$.
$-x = -0.82$,so $x = 0.82$.
The number of $Fe^{3+}$ ions is $0.94 - 0.82 = 0.12$.
The total iron ions are $0.94$.
The fraction of $Fe^{3+}$ ions is $(0.12 / 0.94) \times 100 \approx 12.77\%$.
However,the question asks for the metal deficiency percentage.
The total iron expected is $1.00$,but we have $0.94$.
Deficiency = $1.00 - 0.94 = 0.06$.
Percentage deficiency = $(0.06 / 1.00) \times 100 = 6\%$.

(C) When $SrCl_2$ is added to molten $NaCl$,$Sr^{2+}$ ions occupy some of the sites of $Na^+$ ions.
Since each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality,one $Na^+$ site remains vacant.
This type of defect is known as an impurity defect.

(A) Impurity defects occur when foreign atoms are present in the lattice sites of the host crystal.
When $CdCl_2$ is added to $AgCl$,$Cd^{2+}$ ions replace some $Ag^+$ ions.
To maintain electrical neutrality,two $Ag^+$ ions are replaced by one $Cd^{2+}$ ion,creating a cation vacancy.
This is a classic example of an impurity defect in ionic solids.

(A) When $NaCl$ is doped with $SrCl_2$,each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality.
One $Sr^{2+}$ ion creates one cation vacancy.
Given doping is $10^{-3} \, mol \% \, SrCl_2$,which means $10^{-3} \, mol$ of $SrCl_2$ in $100 \, mol$ of $NaCl$.
So,$Sr^{2+}$ ions per mole of $NaCl = \frac{10^{-3}}{100} = 10^{-5} \, mol$.
Since each $Sr^{2+}$ ion creates one cation vacancy,the concentration of cation vacancies is $10^{-5} \, mol \, per \, mole$ of $NaCl$.
To find the number of vacancies per mole,we multiply by Avogadro's number $(N_A = 6.022 \times 10^{23})$:
Number of vacancies $= 10^{-5} \times 6.022 \times 10^{23} = 6.022 \times 10^{18} \, mol^{-1}$.