CBSE Class 12 Chemistry Solid State Assignment Set 03

Very Short Answer Type Questions

Question. Which point defect in crystals does not alter the density of the relevant solid?
Answer: This is known as the Frenkel defect.
In simple words: The Frenkel defect does not change the density of the crystal at all.

Exam Tip: Remember that in a Frenkel defect, ions only shift positions within the lattice, so the total mass and volume remain unchanged.

Question. Which point defect in its crystal units alters the density of a solid?
Answer: It is called the Schottky defect.
In simple words: The Schottky defect changes the density of the crystal.

Exam Tip: Remember that Schottky defects involve missing ions, which directly reduces the density of the solid.

Question. Which point defect in its crystal units increases the density of a solid?
Answer: A metal excess defect raises the density of a solid. This happens because of the occurrence of extra cations in the interstitial locations.
In simple words: When extra metal ions squeeze into the empty spaces of the crystal, the density of the solid goes up.

Exam Tip: Clearly state that the addition of extra cations in the interstitial sites increases the overall mass without changing the volume.

Question. How do metallic and ionic substances differ in conducting electricity?
Answer: Electrical conduction in metallic materials occurs because of free electrons, whereas in ionic substances, conductivity arises due to the occurrence of ions.
In simple words: Metals conduct electricity using moving electrons, but ionic solids need moving ions to conduct electricity.

Exam Tip: Differentiate clearly by stating that metals conduct in the solid state via electrons, while ionic compounds require a molten or dissolved state to release mobile ions.

Question. Which point defect of its crystals decreases the density of a solid?
Answer: This is the Schottky defect.
In simple words: The Schottky defect lowers the density of the crystal.

Exam Tip: Associate Schottky defect with missing pairs of cations and anions, which leads to a decrease in mass and density.

Question. What is the number of atoms in a unit cell of a face-centred cubic crystal?
Answer: There are a total of 4 atoms present within one unit cell of an fcc crystal lattice.
In simple words: A face-centered cubic unit cell has a total of four atoms in it.

Exam Tip: Show the calculation if required: \( 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4 \) to ensure full marks.

Question. Write a feature which will distinguish a metallic solid from an ionic solid.
Answer: Electrical conduction in metallic solids happens due to mobile electrons, whereas in ionic solids, conductivity is caused by the existence of ions.
In simple words: Metals carry electricity using electrons, whereas ionic solids do so using charged ions.

Exam Tip: You can also mention state-wise conductivity: metals conduct in both solid and liquid states, whereas ionic solids only conduct when molten or dissolved.

Question. Which point defect in crystals of a solid does not change the density of the solid?
Answer: It is the Frenkel defect.
In simple words: The Frenkel defect keeps the density of the solid exactly the same.

Exam Tip: Explain that no ions leave the lattice in a Frenkel defect; they merely relocate to interstitial voids.

Question. Which point defect in crystals of a solid decreases the density of the solid?
Answer: This is known as the Schottky defect.
In simple words: The Schottky defect makes the density of the solid go down.

Exam Tip: Remember that mass decreases because ions are completely missing from their normal lattice sites.

Question. What type of interactions hold the molecules together in a polar molecular solid?
Answer: Dipole-dipole attractive forces bind the molecules together in polar molecular solids.
In simple words: Polar molecules are held together by the attraction between their positive and negative ends.

Exam Tip: Specifically use the term "dipole-dipole interactions" as it is the key term examiners look for.

Question. What type of semiconductor is obtained when silicon is doped with arsenic?
Answer: An n-type semiconductor is created.
In simple words: Doping silicon with arsenic gives us an n-type semiconductor.

Exam Tip: Mention that arsenic is a pentavalent impurity (Group 15), which provides extra conducting electrons.

Question. Write a distinguishing feature of metallic solids.
Answer: Metallic solids exhibit great electrical and thermal conductivities because of the existence of mobile electrons.
In simple words: Metals conduct heat and electricity very well because they have free-moving electrons.

Exam Tip: Mention high malleability, ductility, or the presence of a "sea of mobile electrons" as other defining properties.

Question. ‘Crystalline solids are anisotropic in nature.’ What does this statement mean?
Answer: This indicates that crystalline materials display varying values for physical properties like electrical conductivity, refractive index, and thermal expansion along different directions.
In simple words: Crystalline solids have different physical properties depending on the direction you measure them from.

Exam Tip: State that anisotropy arises due to the different arrangement of particles along different directions in a crystalline lattice.

Question. Which stoichiometric defect in crystals increases the density of a solid?
Answer: An interstitial defect in crystals raises the density of a solid.
In simple words: An interstitial defect increases how heavy a solid is for its size because extra particles fit into the empty spaces.

Exam Tip: Remember that extra constituent particles occupy interstitial sites, which adds mass to the same volume.

Question. What is meant by ‘doping’ in a semiconductor?
Answer: The process of adding a proper impurity to a semiconductor to enhance its electrical conductivity is called doping.
In simple words: Doping means adding a tiny bit of another element to a semiconductor to help it conduct electricity better.

Exam Tip: Always specify that the added impurity must be in a tiny, controlled quantity to modify conductivity.

Short Answer Type Questions-I (SA-I) (2 Marks)

Question. Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of unit cell of its crystal.
Answer:
Let us assume:
- Edge length of the unit cell = \( a \text{ pm} \)
- Number of atoms present per unit cell = \( Z \)
Therefore, the volume of the unit cell is:
\[ \text{Volume} = (a \text{ pm})^3 = (a \times 10^{-10} \text{ cm})^3 = a^3 \times 10^{-30} \text{ cm}^3 \]
The density of the unit cell is given by:
\[ d = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} \quad \text{--- (i)} \]
The mass of the unit cell is:
\[ \text{Mass of unit cell} = \text{Number of atoms inside the unit cell} \times \text{mass of each single atom} = Z \times m \]
The mass of a single atom can be calculated using its molar mass (\( M \)) and Avogadro's number (\( N_0 \)):
\[ m = \frac{\text{Molar mass}}{\text{Avogadro's number}} = \frac{M}{N_0} \]
Placing these expressions into equation (i), we obtain:
\[ d = \frac{Z \times M}{a^3 \times 10^{-30} \times N_0} \]
If the edge length \( a \) is expressed directly in centimeters (\( \text{cm} \)), then the formula becomes:
\[ d = \frac{Z \times M}{a^3 \times N_0} \text{ g/cm}^3 \]
Rearranging this equation, the molar mass \( M \) of the unknown metal can be calculated as:
\[ M = \frac{d \times a^3 \times N_0}{Z} \]
In simple words: By finding the volume of the unit cell and combining it with the density, we can calculate the total mass. Dividing this by the number of atoms gives us the mass of a single atom, which helps us find the molar mass of the metal.

Exam Tip: Make sure to clearly state the units used, especially converting the edge length from picometers to centimeters using the factor \( 10^{-10} \text{ cm/pm} \).

Question. Calculate the packing efficiency of a metal crystal for a simple cubic lattice.
Answer:
Let the edge length of the simple cubic unit cell be \( a \) and the radius of the sphere (atom) be \( r \).
In a simple cubic lattice, the atoms at the corners touch each other along the edge, so:
\[ a = 2r \]
The total volume of the cubic unit cell is:
\[ \text{Volume of unit cell} = a^3 = (2r)^3 = 8r^3 \]
A simple cubic unit cell contains exactly 1 atom.
The volume of 1 spherical atom is:
\[ \text{Volume of one sphere} = \frac{4}{3}\pi r^3 \]
The packing efficiency is defined as:
\[ \text{Packing Efficiency} = \frac{\text{Volume of one atom}}{\text{Total volume of unit cell}} \times 100 \]
\[ \text{Packing Efficiency} = \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100 = \frac{\pi}{6} \times 100 \approx 52.4\% \]
Therefore, the packing efficiency of a simple cubic lattice is \( 52.4\% \).
In simple words: In a simple cubic lattice, the atoms touch along the edges, meaning about \( 52.4\% \) of the total space is occupied by atoms, while the rest is empty.

Exam Tip: Clearly show the relation \( a = 2r \) and the volume formula of a sphere to score full marks in this derivation.

Question. Define the following terms in relation to crystalline solids : (i) Unit cell (ii) Coordination number. Give one example in each case.
Answer:
(i) Unit cell This is the smallest three-dimensional repeating unit of a space lattice which, when repeated in different directions, generates the entire crystal lattice.
Example: Cubic unit cell, hexagonal unit cell.
(ii) Coordination number This is the total number of immediate neighboring spheres that are in direct contact with any given sphere in a crystal lattice.
Example: The coordination number of an atom in a hexagonal close-packed (hcp) structure is 12.
In simple words: A unit cell is the basic building block of a crystal that repeats to make the whole lattice. The coordination number is simply the number of touching neighbor atoms.

Exam Tip: Underline "smallest repeating unit" for unit cell and "number of nearest neighbors" for coordination number to catch the examiner's eye.

Question. The unit cell of an element of atomic mass 108 u and density 10.5 g cm–3 is a cube with edge length, 409 pm. Find the type of unit cell of the crystal.[Given : Avogadro's constant = 6.023 × 1023 mol–1]
Answer:
The given values are:
- Atomic mass (\( M \)) = \( 108 \text{ g/mol} \)
- Density (\( \rho \)) = \( 10.5 \text{ g cm}^{-3} \)
- Edge length (\( a \)) = \( 409 \text{ pm} = 409 \times 10^{-10} \text{ cm} \)
- Avogadro's constant (\( N_A \)) = \( 6.023 \times 10^{23} \text{ mol}^{-1} \)
Using the relation for density:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \]
We can solve for \( Z \) (number of atoms per unit cell):
\[ Z = \frac{\rho \times a^3 \times N_A}{M} \]
Substituting the given values:
\[ Z = \frac{10.5 \times (409 \times 10^{-10})^3 \times 6.023 \times 10^{23}}{108} \]
\[ Z = \frac{10.5 \times 68.42 \times 10^{-24} \times 6.023 \times 10^{23}}{108} \]
\[ Z = \frac{432.65}{108} \approx 4 \]
Since \( Z = 4 \), the crystal has a face-centered cubic (fcc) or cubic close-packed (ccp) unit cell structure.
In simple words: By using the crystal density formula and solving for the number of atoms, we find that there are exactly 4 atoms per unit cell, which means it is an fcc (or ccp) structure.

Exam Tip: Always double-check your calculation of \( a^3 \) and keep track of negative exponents to get the correct integer value of \( Z \).

Question. Explain the following terms with suitable examples : Ferromagnetism and Ferrimagnetism
Answer:
Ferromagnetic solids:These are substances that experience strong attraction to an external magnetic field and retain their alignment even when the magnetic field is taken away. This behavior is called ferromagnetism.
Examples include Iron (\( \text{Fe} \)), Cobalt (\( \text{Co} \)), and Nickel (\( \text{Ni} \)).
Ferrimagnetic solids These materials are predicted to exhibit high magnetism because of unpaired electrons, but actually show a very weak net magnetic moment because their domains are aligned in opposite directions in unequal numbers.
Examples include Magnetite (\( \text{Fe}_3\text{O}_4 \)) and ferrites.
In simple words: Ferromagnetic materials become permanent magnets and stay strongly magnetic. Ferrimagnetic materials have a much weaker magnetic force because some of their internal magnetic parts point in opposite directions and cancel each other out.

Exam Tip: Draw domain arrow alignments to get extra marks: all pointing up for ferromagnetism, and unequal opposite pointing arrows for ferrimagnetism.

Question. An element X crystallizes in f.c.c structure. 208 g of it has 4.2832 × 1024 atoms. Calculate the edge of the unit cell, if density of X is 7.2 g cm–3.
Answer:
We are given the following information:
- Crystal system = fcc, so \( Z = 4 \)
- Density (\( d \)) = \( 7.2 \text{ g cm}^{-3} \)
- Mass of sample = \( 208 \text{ g} \)
- Number of atoms in sample = \( 4.2832 \times 10^{24} \text{ atoms} \)
First, we find the molar mass (\( M \)) of element X, which is the mass of Avogadro's number (\( 6.022 \times 10^{23} \)) of atoms:
\[ M = \frac{208 \text{ g}}{4.2832 \times 10^{24}} \times 6.022 \times 10^{23} \approx 29.24 \text{ g mol}^{-1} \]
Next, we use the density formula to find the volume of the unit cell (\( a^3 \)):
\[ a^3 = \frac{Z \times M}{d \times N_A} \]
Substituting the values:
\[ a^3 = \frac{4 \times 29.24}{7.2 \times 6.022 \times 10^{23}} \]
\[ a^3 = \frac{116.96}{4.336 \times 10^{24}} \approx 269.6 \times 10^{-24} \text{ cm}^3 \]
Taking the cube root of both sides to get the edge length (\( a \)):
\[ a = \sqrt[3]{269.6 \times 10^{-24} \text{ cm}^3} \approx 6.46 \times 10^{-8} \text{ cm} = 6.46\text{ \(\text{Å}\)} \]
In simple words: First, we calculate the mass of one mole of atoms to get the molar mass. Then, using the density of the crystal, we find the volume of a single unit cell and take its cube root to find the edge length, which is \( 6.46 \times 10^{-8} \text{ cm} \).

Exam Tip: Remember to convert the final answer to Angstroms (\( \text{Å} \)) or picometers (\( \text{pm} \)) if required, since edge lengths are very small.

Question. What is a semiconductor? Describe the two main types of semiconductors.
Answer:
Semiconductor: These are solids that have electrical conductivity levels falling between those of highly conducting metals and non-conducting insulators. Typically, their conductivity is in the range of \( 10^2 \) to \( 10^{-9} \ \Omega^{-1}\text{cm}^{-1} \).
The two main types of semiconductors are:
(a) n-type semiconductors: These are formed by doping a semiconductor with a higher-group (electron-rich) impurity. For example, when Germanium is doped with Arsenic.
(b) p-type semiconductors: These are created by doping a semiconductor with a lower-group (electron-deficient) impurity, which creates vacant electron sites known as holes.
In simple words: Semiconductors are materials that conduct electricity better than insulators but not as well as metals. Adding electron-rich elements makes them n-type, while adding electron-poor elements makes them p-type.

Exam Tip: State the conductivity range accurately and give standard examples like Germanium doped with Silicon or Arsenic to score high.

Question. Account for the following:
(i) Schottky defects lower the density of related solids
(ii) Conductivity of silicon increases on doping it with phosphorus.
Answer:
(i) A Schottky defect arises when an equal number of cations and anions go missing from their normal lattice sites. Since mass is lost from the lattice while the volume stays the same, the density of the solid decreases.
(ii) Silicon's conductivity increases because doping it with pentavalent phosphorus introduces extra, weakly-bound electrons that can move freely through the lattice.
In simple words: (i) Schottky defects leave empty spaces in the crystal by losing ions, which makes the solid lighter and less dense. (ii) Phosphorus has more electrons than silicon, so adding it provides free-moving electrons that help carry electricity.

Exam Tip: Highlight "missing equal numbers of cations and anions" for (i) and "extra free electrons from pentavalent impurity" for (ii).

Question. Aluminium crystallizes in an fcc structure. Atomic radius of the metal is 125 pm. What is the length of the side of the unit cell of the metal?
Answer:
For a face-centered cubic (fcc) lattice, the relation between atomic radius (\( r \)) and edge length (\( a \)) is:
\[ r = \frac{a}{2\sqrt{2}} \ ]
We are given:
- Atomic radius (\( r \)) = \( 125 \text{ pm} \)
Rearranging the formula to find the edge length \( a \):
\[ a = 2\sqrt{2} \times r \]
Substituting the given values:
\[ a = 2 \times 1.414 \times 125 \text{ pm} = 353.5 \text{ pm} \]
Therefore, the side length of the unit cell is \( 353.5 \text{ pm} \).
In simple words: In an fcc crystal, the atoms touch along the face diagonal. Using this relationship, we find that the edge length is about \( 353.5 \text{ pm} \).

Exam Tip: Make sure to memorize the different relations between \( a \) and \( r \) for simple cubic, bcc, and fcc systems, as these are frequently tested.

Question. (a) Why does presence of excess of lithium makes LiCl crystals pink?
(b) A solid with cubic crystal is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound?
Answer:
(a) The pink coloration is due to a metal excess defect caused by anionic vacancies. In this defect, chloride ions leave the lattice, and their vacant sites are occupied by unpaired electrons, forming F-centres that absorb light.
(b) For the compound:
- Atoms of \( Q \) are located at the 8 corners of the cube, so their contribution is:
\[ 8 \times \frac{1}{8} = 1 \text{ atom of } Q \]
- Atom of \( P \) is located at the body center, contributing:
\[ 1 \times 1 = 1 \text{ atom of } P \]
The ratio of atoms \( P : Q \) is \( 1 : 1 \), making the empirical formula \( \text{PQ} \).
In simple words: (a) Heating lithium chloride traps extra electrons in empty spaces of the crystal, creating F-centres that make it look pink. (b) Since Q atoms at the corners add up to 1 and the P atom in the middle is 1, the formula is PQ.

Exam Tip: Explicitly mention the name "F-centre" for part (a) to secure full marks.

Question. (a) What change occurs when AgCl is doped with CdCl₂?
(b) What type of semiconductor is produced when silicon is doped with boron?
Answer:
(a) Doping \( \text{AgCl} \) with \( \text{CdCl}_2 \) creates an impurity defect. For every divalent \( \text{Cd}^{2+} \) ion that enters, it replaces two \( \text{Ag}^+ \) ions, leaving one site vacant. These cation vacancies significantly increase the electrical conductivity of the solid.
(b) A p-type semiconductor is created when silicon is doped with trivalent boron.
In simple words: (a) When cadmium chloride is added, cadmium ions replace silver ions but leave behind empty spaces, which helps electricity flow better. (b) Boron has fewer electrons than silicon, leaving "holes" that turn the material into a p-type semiconductor.

Exam Tip: For part (a), explain that one \( \text{Cd}^{2+} \) ion replaces two \( \text{Ag}^+ \) ions to maintain electrical neutrality, leaving one cation vacancy.

Question. If NaCl is doped with 10–3 mole percent SrCl2, what will be the concentration of cation vacancies? (NA = 6.02 × 1023 mol–1)
Answer:
A doping level of \( 10^{-3} \text{ mole percent} \) means that \( 100 \text{ moles} \) of \( \text{NaCl} \) are doped with \( 10^{-3} \text{ moles} \) of \( \text{SrCl}_2 \).
Therefore, the amount of \( \text{SrCl}_2 \) added per mole of \( \text{NaCl} \) is:
\[ \text{Moles of } \text{Sr}^{2+} \text{ per mole of NaCl} = \frac{10^{-3}}{100} = 10^{-5} \text{ mole} \]
Because each divalent \( \text{Sr}^{2+} \) ion replaces two \( \text{Na}^+ \) ions to maintain neutrality, it creates exactly one cation vacancy.
Thus, the concentration of cation vacancies is:
\[ \text{Vacancies} = 10^{-5} \text{ mole} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 6.02 \times 10^{18} \text{ vacancies per mole} \]
In simple words: Adding strontium chloride replaces sodium ions and leaves behind vacancies. Since each strontium ion creates one vacant spot, multiplying the moles of strontium by Avogadro's number gives \( 6.02 \times 10^{18} \) vacant spots per mole.

Exam Tip: Always explain the relationship that 1 \( \text{Sr}^{2+} \) ion introduces exactly 1 cation vacancy to show clear conceptual understanding.

Question. What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer: Semiconductor These are solid materials with electrical conductivities that are intermediate between those of metals and insulators, typically ranging from \( 10^{-6} \) to \( 10^4 \ \Omega^{-1}\text{m}^{-1} \). Examples include Silicon and Germanium.
Semiconductors are divided into two main categories:
1. Intrinsic Semiconductors Pure elements that act as insulators at absolute zero but exhibit low conductivity at room temperature as thermal energy excites electrons across the band gap.
2. Extrinsic Semiconductors:Created by introducing impurities through doping. These are further classified based on their conduction mechanism:
- n-type Semiconductor: Formed by doping with a Group 15 element (like phosphorus or arsenic). Four valence electrons bond with neighboring silicon atoms, leaving a fifth unbonded electron that is free to conduct electricity.
- p-type Semiconductor: Formed by doping with a Group 13 element (like boron or gallium). The impurity atom can only form three bonds, leaving an electron vacancy or "positive hole" that allows charge to flow.
In simple words: Semiconductors are materials whose ability to conduct electricity lies between metals and non-metals. Intrinsic ones are pure, while extrinsic ones are doped with impurities. n-type semiconductors use extra electrons to conduct, while p-type semiconductors use missing electron spaces, called holes, to move charge.

Exam Tip: Draw the band gap diagrams showing the donor level (just below conduction band) for n-type, and acceptor level (just above valence band) for p-type if possible.

Question. A compound forms hcp structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Answer: First, we find the number of constituent atoms in \( 0.5 \text{ mol} \) of the hcp lattice:
\[ \text{Number of atoms } (N) = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \text{ atoms} \]
In any close-packed lattice:
- The number of octahedral voids is equal to the number of atoms (\( N \)):
\[ \text{Octahedral voids} = 3.011 \times 10^{23} \]
- The number of tetrahedral voids is twice the number of atoms (\( 2N \)):
\[ \text{Tetrahedral voids} = 2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23} \]
The total number of voids is the sum of both:
\[ \text{Total Voids} = 3.011 \times 10^{23} + 6.022 \times 10^{23} = 9.033 \times 10^{23} \]
In simple words: Half a mole of a substance contains \( 3.011 \times 10^{23} \) atoms. This creates the same number of octahedral voids and twice as many tetrahedral voids (\( 6.022 \times 10^{23} \ )), adding up to \( 9.033 \times 10^{23} \) total voids.

Exam Tip: Always state the relationship between the number of lattice points and the two types of voids clearly before doing calculations.

Question. An element crystallizes in a structure having fcc unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 × 1023 atoms.
Answer: We are given:
- Edge length (\( a \)) = \( 200 \text{ pm} = 200 \times 10^{-10} \text{ cm} = 2 \times 10^{-8} \text{ cm} \)
- Number of atoms in sample = \( 24 \times 10^{23} \text{ atoms} \)
- Mass of sample = \( 200 \text{ g} \)
- Crystal structure is fcc, so \( Z = 4 \)
First, let's calculate the molar mass (\( M \)) of the element:
\[ M = \frac{\text{Mass of sample}}{\text{Total atoms}} \times N_A = \frac{200 \text{ g}}{24 \times 10^{23}} \times 6.022 \times 10^{23} \approx 50.18 \text{ g mol}^{-1} \]
Now, we apply the density formula:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \]
Substituting the values:
\[ \rho = \frac{4 \times 50.18}{(200 \times 10^{-10})^3 \times 6.022 \times 10^{23}} \]
\[ \rho = \frac{200.72}{8 \times 10^{-24} \times 6.022 \times 10^{23}} \]
\[ \rho = \frac{200.72}{8 \times 6.022 \times 10^{-1}} = \frac{200.72}{4.8176} \approx 41.66 \text{ g cm}^{-3} \]
Thus, the density of the element is \( 41.66 \text{ g cm}^{-3} \).
In simple words: We first calculate the molar mass of the element by finding the weight of a mole of its atoms. Then, we use the density equation to find that the density of the crystal is \( 41.66 \text{ g/cm}^3 \).

Exam Tip: Carefully simplify the denominator by rewriting \( 200 \times 10^{-10} \) as \( 2 \times 10^{-8} \) before cubing to prevent arithmetic errors.

Question. An element with density 11.2 g cm–3 forms a f.c.c. lattice with edge length of 4 × 10–8 cm. Calculate the atomic mass of the element. (Given : NA = 6.022 × 1023 mol–1)
Answer: We are given:
- Density (\( \rho \)) = \( 11.2 \text{ g cm}^{-3} \)
- Edge length (\( a \)) = \( 4 \times 10^{-8} \text{ cm} \)
- Avogadro's number (\( N_A \)) = \( 6.022 \times 10^{23} \text{ mol}^{-1} \)
- For an fcc lattice, \( Z = 4 \)
Using the density equation:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \]
Rearranging to solve for molar mass (\( M \)):
\[ M = \frac{\rho \times a^3 \times N_A}{Z} \]
Substituting the given values:
\[ M = \frac{11.2 \times (4 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{4} \]
\[ M = \frac{11.2 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{4} \]
\[ M = \frac{4316.56 \times 10^{-1}}{4} = \frac{431.656}{4} \approx 107.91 \text{ g mol}^{-1} \]
Thus, the atomic mass of the element is \( 107.91 \text{ u} \) (or \( \text{g mol}^{-1} \)).
In simple words: By rearranging the density formula, we multiply the density, cell volume, and Avogadro's number, then divide by 4 to get the atomic mass of \( 107.91 \text{ g/mol} \).

Exam Tip: Pay attention to units. Since the density is in \( \text{g/cm}^3 \) and edge length is in \( \text{cm} \), the calculated mass will naturally be in \( \text{g/mol} \).

Question. Examine the given defective crystal:

Answer the following questions:
(i) What type of stoichiometric defect is shown by the crystal?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances show such defect?
Answer: (i) This is a Schottky defect because there are missing cations and anions in equal numbers from the lattice.
(ii) The density of the crystal decreases because ions leave the crystal, reducing its mass while the volume remains the same.
(iii) This defect is shown by highly ionic compounds where the cations and anions are of similar size (e.g., \( \text{NaCl} \ ), \( \text{KCl} \ )).
In simple words: (i) This is a Schottky defect because there are missing ions. (ii) It makes the crystal less dense because mass is lost. (iii) Strongly ionic crystals with positive and negative ions of similar sizes show this defect.

Exam Tip: Clearly identify that an equal number of positive and negative charges are missing to prove it is a Schottky defect.

Question. An element with density 2.8 g cm–3 forms a f.c.c. unit cell with edge length 4 × 10–8 cm. Calculate the molar mass of the element. (Given : NA = 6.022 × 1023 mol–1)
Answer: We are given:
- Density (\( \rho \)) = \( 2.8 \text{ g cm}^{-3} \)
- Edge length (\( a \)) = \( 4 \times 10^{-8} \text{ cm} \)
- Avogadro's number (\( N_A \)) = \( 6.022 \times 10^{23} \text{ mol}^{-1} \)
- For fcc, \( Z = 4 \)
Using the formula:
\[ M = \frac{\rho \times a^3 \times N_A}{Z} \]
Substituting the given values:
\[ M = \frac{2.8 \times (4 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{4} \]
\[ M = \frac{2.8 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{4} \]
\[ M = \frac{107.91}{4} \approx 26.98 \text{ g mol}^{-1} \]
Therefore, the molar mass of the element is \( 26.98 \text{ g mol}^{-1} \).
In simple words: We find the molar mass by putting the given density and volume into the rearranged density formula, giving a final molar mass of approximately \( 26.98 \text{ g/mol} \).

Exam Tip: Recognizing that the molar mass corresponds to a real element (Aluminium, 27 g/mol) can give you confidence in your calculated result.

Question. (i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
(ii) What type of stoichiometric defect is shown by NaCl?
Answer: (i) This is caused by a metal excess defect due to anionic vacancies where free electrons are trapped in the empty anionic sites, forming colored F-centres.
(ii) The Schottky defect is exhibited by NaCl.
In simple words: (i) Trapped electrons in vacant spots, called F-centres, make the LiCl crystal look pink. (ii) NaCl shows a Schottky defect where positive and negative ions are missing.

Exam Tip: Associate F-centres with metal excess defects to get maximum marks for color-related questions.

Question. How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell
Answer: (i) Tetrahedral and octahedral voids: A tetrahedral void is formed when four spheres are in contact with each other in a tetrahedral shape, making its coordination number 4. On the other hand, an octahedral void is created by six surrounding spheres, which gives it a coordination number of 6. Also, a tetrahedral void is smaller in size than an octahedral void.
(ii) Crystal lattice and unit cell: A crystal lattice is the complete three-dimensional pattern showing how particles are arranged in space. In contrast, a unit cell is the smallest portion of this lattice that, when repeated in different directions, builds up the whole crystal structure.
In simple words: A tetrahedral void is surrounded by 4 atoms, while an octahedral void is surrounded by 6 atoms. A crystal lattice is the complete grid of particles, while a unit cell is just one repeating block of that grid.

Exam Tip: Draw small diagrams or indicate the coordination numbers (4 for tetrahedral, 6 for octahedral) to clearly display the structural differences.