IGNOU BCHCT-137 Solved Question Paper PDF Download

IGNOU BCHCT-137 Previous Year Solved Question Paper in Hindi

IGNOU BCHCT-137 Previous Year Solved Question Paper in English

Part—A (Coordination Chemistry)

Q1. (a) Explain the ground state electronic configuration of Sc³⁺ ion. (b) Complete and balance any two of the following reactions: (i) Ti(s) + O₂(g) → (ii) Cr(s) + O₂(g) → (iii) Cu(s) + O₂(g) →

Ans. (a) The atomic number of Scandium (Sc) is 21. Its ground state electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s² or simply [Ar] 3d¹ 4s². To form the Sc³⁺ ion, a scandium atom loses three electrons. Electrons are always removed from the outermost shell first (n=4), followed by the penultimate shell (n=3). Therefore, two electrons are lost from the 4s orbital and one electron is lost from the 3d orbital. Sc ([Ar] 3d¹ 4s²) → Sc³⁺ ([Ar] 3d⁰ 4s⁰) + 3e⁻ Thus, the electronic configuration of the Sc³⁺ ion is [Ar] 3d⁰ 4s⁰ , which is identical to the configuration of the noble gas Argon (Ar). Since there are no d-electrons, the Sc³⁺ ion is colourless and diamagnetic.

(b) The completed and balanced reactions are as follows:

• (i) Ti(s) + O₂(g) → TiO₂(s) This reaction is already balanced. Titanium reacts with oxygen upon heating to form titanium(IV) oxide, a white solid.
• (ii) 4Cr(s) + 3O₂(g) → 2Cr₂O₃(s) Chromium reacts with oxygen in the air when heated to form chromium(III) oxide, which is a very stable green solid that forms a passivating layer.
• (iii) 2Cu(s) + O₂(g) → 2CuO(s) Upon strong heating, copper reacts with oxygen to form copper(II) oxide, which is a black solid.

Q2. (a) (i) Which of the following ions is coloured? Give reasons: (1) Ti⁴⁺ (2) [Fe(H₂O)₆]²⁺ (ii) What is the unit of magnetic moment? (b) Name any two transition elements which are present in very small quantities in plants and animals and are essential for the enzymes to function?

Ans. (a) (i) Among the given ions, [Fe(H₂O)₆]²⁺ is coloured . Reason:

• Ti⁴⁺ ion: The configuration of Titanium (Z=22) is [Ar] 3d² 4s². The configuration of the Ti⁴⁺ ion is [Ar] 3d⁰. It has no d-electrons. Colour in transition metal ions is typically due to d-d transitions, where an electron is promoted from a lower energy d-orbital to a higher energy d-orbital. In the absence of d-electrons, no d-d transition is possible. Hence, Ti⁴⁺ is colourless .
• [Fe(H₂O)₆]²⁺ ion: The configuration of Iron (Z=26) is [Ar] 3d⁶ 4s². The Fe²⁺ ion has the configuration [Ar] 3d⁶. In the octahedral complex [Fe(H₂O)₆]²⁺, the d-orbitals are split into a lower energy t₂g set and a higher energy eg set. Since H₂O is a weak field ligand, it forms a high-spin complex. The d⁶ configuration is t₂g⁴ eg². This ion has unpaired electrons and partially filled d-orbitals, allowing for the promotion of an electron from the t₂g level to the eg level by absorbing light from the visible region. This process is called a d-d transition . The complex transmits the complementary colour of the light absorbed, and thus it appears coloured (pale green).

(ii) The unit of magnetic moment is the Bohr Magneton , denoted by B.M.

(b) Two essential transition elements present in trace amounts in living organisms and crucial for enzyme function are:

1. Iron (Fe): It is a vital component of many proteins and enzymes, including hemoglobin (for oxygen transport), myoglobin (for oxygen storage), and cytochromes (in the electron transport chain for cellular respiration).
2. Zinc (Zn): It acts as a cofactor for numerous enzymes, such as carbonic anhydrase, alcohol dehydrogenase, and carboxypeptidase. It is essential for many metabolic processes in the body.

Q3. (a) Which one of the following complexes does not have perfect octahedral geometry? Justify your answer: (i) [Cr(H₂O)₆]³⁺ (ii) [VCl₆]²⁻ (iii) [Fe(H₂O)₆]²⁺ (b) Calculate the spin-only magnetic moment for [Mn(H₂O)₆]²⁺ or CuCl₂.

Ans. (a) Among the given complexes, [Fe(H₂O)₆]²⁺ does not have a perfect octahedral geometry. Justification: A perfect octahedral geometry is expected for complexes where the d-orbitals (t₂g and eg) of the central metal ion are symmetrically filled (e.g., d³, low-spin d⁶, high-spin d⁵, d⁸, d¹⁰). Asymmetrical filling of these orbitals leads to a distortion known as the Jahn-Teller effect .

• [Cr(H₂O)₆]³⁺: Cr is in the +3 oxidation state (d³ configuration). In an octahedral field, the configuration is t₂g³. This is a symmetrically filled configuration (one electron in each of the three t₂g orbitals). Therefore, it has a perfect octahedral geometry.
• [VCl₆]²⁻: V is in the +4 oxidation state (d¹ configuration). The configuration is t₂g¹. This is asymmetric, but distortions due to t₂g asymmetry are generally weaker.
• [Fe(H₂O)₆]²⁺: Fe is in the +2 oxidation state (d⁶ configuration). H₂O is a weak field ligand, leading to a high-spin complex. The electronic configuration is t₂g⁴ eg² . In this case, the t₂g orbitals are asymmetrically filled (one orbital has a pair of electrons, while two are singly occupied). This asymmetrical electron distribution causes Jahn-Teller distortion, which elongates or compresses some of the metal-ligand bonds, resulting in a distorted, imperfect octahedral geometry.

(b) The spin-only magnetic moment (µs) is calculated using the formula µs = √[n(n+2)] B.M. , where ‘n’ is the number of unpaired electrons. We will perform the calculation for [Mn(H₂O)₆]²⁺ :

1. Oxidation State: The oxidation state of Manganese (Mn) is +2.
2. Electronic Configuration: Mn (Z=25) has the configuration [Ar] 3d⁵ 4s². Therefore, Mn²⁺ has the configuration [Ar] 3d⁵.
3. Unpaired Electrons (n): H₂O is a weak field ligand, so a high-spin complex is formed. For a d⁵ configuration in a high-spin state, all five d-electrons remain unpaired. Thus, n = 5 .
4. Calculation of Magnetic Moment: µs = √[5(5+2)] µs = √[5 × 7] µs = √35 µs ≈ 5.92 B.M.

Q4. (a) Why is the size of the elements of the third transition series almost similar to that of the elements of the second transition series? (b) Determine the oxidation state and coordination number of the central metal ion in the following: (i) [Ag(NH₃)₂]⁺ (ii) [Co(H₂O)₆]²⁺

Ans. (a) The similarity in size between the elements of the second (4d) and third (5d) transition series is due to a phenomenon known as the Lanthanide Contraction . Explanation:

1. Just before the start of the third transition series (beginning with Hafnium, Hf), the lanthanide series of elements occurs, in which the inner 4f-orbitals are progressively filled.
2. The 4f-orbitals have a very diffuse shape, which results in a very poor shielding effect . This means they are not effective at shielding the outer shell electrons from the pull of the nucleus.
3. As one moves across the lanthanide series, the nuclear charge increases by one unit with each element, but the additional electron enters a poorly shielding 4f-orbital.
4. Consequently, the outermost electrons (in the 6s and 5d shells) experience a progressively stronger effective nuclear charge.
5. This increased pull causes a steady decrease in atomic radii across the lanthanide series. The cumulative effect of this contraction is so significant that by the time the 5d series begins, the atomic radius of Hafnium (Hf) is almost identical to that of Zirconium (Zr), the element directly above it in the 4d series. This trend of similar sizes continues for subsequent pairs of elements (e.g., Nb and Ta, Mo and W).

(b) (i) [Ag(NH₃)₂]⁺

• Oxidation State: The ligand NH₃ (ammine) is neutral (charge = 0). Let the oxidation state of Ag be ‘x’. x + 2(0) = +1 x = +1 The oxidation state of Ag is +1 .
• Coordination Number: This is the number of donor atoms attached to the central metal. There are two monodentate NH₃ ligands. The coordination number is 2 .

(ii) [Co(H₂O)₆]²⁺

• Oxidation State: The ligand H₂O (aqua) is neutral (charge = 0). Let the oxidation state of Co be ‘x’. x + 6(0) = +2 x = +2 The oxidation state of Co is +2 .
• Coordination Number: There are six monodentate H₂O ligands attached to the cobalt ion. The coordination number is 6 .

Q5. (a) Write the IUPAC name of any one of the following compounds: (i) [(NH₃)₅Co-NH₂-Co(NH₃)₅](NO₃)₅ (ii) [(CO)₃Fe(CO)₃Fe(CO)₃] (b) Write down all possible isomers of [Co(NH₃)₅NO₂]Cl₂ and give their names.

Ans. (a) IUPAC name of compound (i) [(NH₃)₅Co-NH₂-Co(NH₃)₅](NO₃)₅: This is a dinuclear complex with a bridging ligand. The bridging ligand (NH₂) is named ‘amido’.

1. Identify Cation and Anion: The cation is [(NH₃)₅Co-NH₂-Co(NH₃)₅]⁵⁺ and the anion is NO₃⁻ (nitrate).
2. Name the Ligands: There are ten ‘ammine’ (NH₃) ligands and one bridging ‘amido’ (NH₂) ligand. The bridging ligand is denoted by the prefix ‘µ-‘.
3. Determine Oxidation State of Metal: Let the oxidation state of each Co be ‘x’. The charge of the amido bridge is -1. The overall charge of the cation is +5. 2x + (-1) = +5 2x = 6 x = +3. So, the metal is Cobalt(III).
4. Full Name: The ligands are named alphabetically. The two identical parts of the complex are indicated by ‘bis()’. The full IUPAC name is: µ-amido-bis(pentaamminecobalt(III)) nitrate .

(b) Possible isomers of [Co(NH₃)₅NO₂]Cl₂: This complex exhibits linkage isomerism . This type of isomerism arises when a complex contains an ambidentate ligand , which can bind to the central metal atom through two or more different donor atoms. The ligand NO₂⁻ can coordinate either through the nitrogen atom (as -NO₂, nitro) or through an oxygen atom (as -ONO, nitrito). The two possible isomers are:

1. [Co(NH₃)₅(NO₂)]Cl₂
   • Description: In this isomer, the nitrite ligand is bonded to the cobalt atom through the nitrogen atom.
   • IUPAC Name: Pentaamminenitrocobalt(III) chloride .
   • Colour: This compound is yellowish-brown.
2. [Co(NH₃)₅(ONO)]Cl₂
   • Description: In this isomer, the nitrite ligand is bonded to the cobalt atom through one of the oxygen atoms.
   • IUPAC Name: Pentaamminenitritocobalt(III) chloride .
   • Colour: This compound is red.

Q6. (a) Write the formula of the following: (i) Tetrapyridineplatinum(II) tetrachloridoplatinate(II) (ii) Pentaammineaquacobalt(III) chloride (b) Explain the structure and magnetic moments of the following complexes on the basis of valence bond theory: [MnCl₆]³⁻ and [Mn(CN)₆]³⁻

Ans. (a) Formulae of the compounds: (i) Tetrapyridineplatinum(II) tetrachloridoplatinate(II)

• Cation: Tetrapyridineplatinum(II) → [Pt(py)₄]²⁺ (Pyridine, py, is a neutral ligand).
• Anion: Tetrachloridoplatinate(II) → [PtCl₄]²⁻ (Chlorido, Cl⁻, has a -1 charge).
• The charges of the cation (+2) and anion (-2) are equal and opposite, so they combine in a 1:1 ratio.
• Formula: [Pt(py)₄][PtCl₄]

(ii) Pentaammineaquacobalt(III) chloride

• Cation: Pentaammineaquacobalt(III) → [Co(NH₃)₅(H₂O)]³⁺ (Ammine, NH₃, and aqua, H₂O, are both neutral ligands). The charge of cobalt is +3, making the complex ion charge +3.
• Anion: chloride → Cl⁻
• Three Cl⁻ ions are needed to balance the +3 charge of the cation.
• Formula: [Co(NH₃)₅(H₂O)]Cl₃

(b) Explanation based on Valence Bond Theory (VBT): In both complexes, the central metal ion is Mn³⁺, which has the electronic configuration [Ar] 3d⁴.

1. [MnCl₆]³⁻ (High-Spin Complex)

• Ligand: Cl⁻ is a weak-field ligand . It does not cause the pairing of d-electrons.
• Hybridisation: Since it is a high-spin (outer orbital) complex, the inner 3d orbitals are not used for bonding. The hybridisation will be sp³d² , involving one 4s, three 4p, and two outer 4d orbitals.
• Structure: The sp³d² hybridisation results in an octahedral geometry.
• Magnetic Properties: The four 3d electrons remain in the 3d orbitals without pairing. The number of unpaired electrons, n = 4 .
• Magnetic Moment (µ): µ = √[4(4+2)] = √24 ≈ 4.90 B.M.

2. [Mn(CN)₆]³⁻ (Low-Spin Complex)

• Ligand: CN⁻ is a strong-field ligand . It forces the d-electrons to pair up.
• Hybridisation: To make two inner 3d orbitals available for bonding, the 3d⁴ electrons rearrange. The configuration becomes (↑↓)(↑)(↑)( )( ). This leaves two inner 3d orbitals empty for hybridisation. The hybridisation is d²sp³ , involving two inner 3d, one 4s, and three 4p orbitals.
• Structure: The d²sp³ hybridisation results in an octahedral geometry.
• Magnetic Properties: After rearrangement, there are two unpaired electrons. The number of unpaired electrons, n = 2 .
• Magnetic Moment (µ): µ = √[2(2+2)] = √8 ≈ 2.83 B.M.

Q7. (a) Calculate the crystal field stabilization energy of [Fe(CN)₆]⁴⁻ complex. (b) What is spectrochemical series?

Ans. (a) Calculation of Crystal Field Stabilization Energy (CFSE) for [Fe(CN)₆]⁴⁻:

1. Oxidation State of Iron (Fe): Let the oxidation state of Fe be ‘x’. x + 6(-1) = -4 x = +2 So, the central ion is Fe²⁺, which has a d⁶ electronic configuration.
2. Ligand Type and Electronic Configuration: The cyanide ligand (CN⁻) is a strong-field ligand . It will cause electron pairing and form a low-spin octahedral complex. The six d-electrons will first fill the lower-energy t₂g orbitals before any occupy the higher-energy eg orbitals. The electronic configuration will be: t₂g⁶ eg⁰ .
3. CFSE Calculation: The formula for CFSE in an octahedral complex is: CFSE = [-0.4 × (no. of e⁻ in t₂g) + 0.6 × (no. of e⁻ in eg)] Δo + P Where Δo is the crystal field splitting energy and P is the pairing energy. CFSE = [-0.4 × 6 + 0.6 × 0] Δo CFSE = [-2.4 + 0] Δo CFSE = -2.4 Δo If pairing energy is included, the low-spin d⁶ configuration (3 pairs) requires forming 2 additional electron pairs compared to the high-spin case (1 pair). So, the full expression is CFSE = -2.4 Δo + 2P .

(b) Spectrochemical Series: The spectrochemical series is a list of ligands arranged in order of their increasing ability to cause splitting of the d-orbitals of a metal ion. This splitting energy is denoted as Δo for octahedral complexes.

• The series is determined experimentally by measuring the absorption spectra of complexes.
• Ligands that cause a large splitting are known as strong-field ligands (e.g., CN⁻, CO). They are at the high end of the series.
• Ligands that cause a small splitting are known as weak-field ligands (e.g., I⁻, Br⁻). They are at the low end of the series.
• The series helps in predicting the magnetic properties and colour of coordination compounds. A strong-field ligand is more likely to cause electron pairing (low-spin complex), while a weak-field ligand is more likely to result in an unpaired electron configuration (high-spin complex).
• An abbreviated version of the series is: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO

Part—B (States of Matter and Chemical Kinetics)

Q8. (a) State Boyle’s law and draw a graph between P and 1/V for an ideal gas. (b) The density of a gas was found to be 3.43 g/L at 300 K and 1 atm. Calculate the molecular mass.

Ans. (a) Boyle’s Law: Boyle’s law states that at a constant temperature, the pressure (P) of a fixed amount (number of moles, n) of a gas is inversely proportional to its volume (V). Mathematically, this can be expressed as: P ∝ 1/V (at constant T and n) or PV = k (where k is a constant) Graph between P and 1/V: The relationship P = k * (1/V) is analogous to the equation of a straight line, y = mx. If we plot pressure (P) on the y-axis against the reciprocal of volume (1/V) on the x-axis, the result is a straight line passing through the origin . The slope of this line is equal to the constant k (which in turn is equal to nRT). (Graph description: A straight line starting from the origin (0,0) with a positive slope. The y-axis is labelled ‘P’ and the x-axis is labelled ‘1/V’.)

(b) Calculation of Molecular Mass: The ideal gas equation can be related to density (d) and molar mass (M) as follows: PM = dRT Where: P = Pressure = 1 atm M = Molar Mass (g/mol) (to be calculated) d = Density = 3.43 g/L R = Universal Gas Constant = 0.0821 L atm K⁻¹ mol⁻¹ T = Temperature = 300 K Rearranging the formula to solve for M: M = (dRT) / P Substituting the given values: M = (3.43 g/L × 0.0821 L atm K⁻¹ mol⁻¹ × 300 K) / 1 atm M = 3.43 × 0.0821 × 300 g/mol M = 84.49 g/mol Therefore, the molecular mass of the gas is approximately 84.5 g/mol .

Q9. (a) 2 mole of N₂ and 3 mole of He gas are kept in a container. The total pressure is 10 atm. Find out the partial pressure of He and N₂. (b) What is most probable velocity (v_mp)? Explain with graph of Maxwell-Boltzmann’s distribution.

Ans. (a) Calculation of Partial Pressures: According to Dalton’s Law of Partial Pressures, the partial pressure of a gas in a mixture is the product of its mole fraction and the total pressure. Partial Pressure (Pᵢ) = Mole Fraction (xᵢ) × Total Pressure (P_total)

1. Calculate Total Moles: Total moles (n_total) = moles of N₂ + moles of He = 2 + 3 = 5 mol
2. Calculate Mole Fractions: Mole fraction of N₂ (x_N₂) = (moles of N₂) / (total moles) = 2 / 5 = 0.4 Mole fraction of He (x_He) = (moles of He) / (total moles) = 3 / 5 = 0.6
3. Calculate Partial Pressures: Given P_total = 10 atm Partial pressure of N₂ (P_N₂) = x_N₂ × P_total = 0.4 × 10 atm = 4 atm Partial pressure of He (P_He) = x_He × P_total = 0.6 × 10 atm = 6 atm

So, the partial pressure of nitrogen is 4 atm and the partial pressure of helium is 6 atm.

(b) Most Probable Velocity (v_mp): The most probable velocity is the speed possessed by the largest number of molecules in a gas sample at a specific temperature. Maxwell-Boltzmann Distribution Graph: This graph plots the fraction of molecules (or probability density) on the y-axis against the molecular speed on the x-axis.

• The curve starts at the origin, rises to a maximum value (a peak), and then decreases asymptotically towards the x-axis at higher speeds.
• The speed corresponding to the peak of the distribution curve is the most probable velocity (v_mp). This peak represents the speed at which the highest fraction of molecules is moving.
• The area under the entire curve represents the total number of molecules and is constant.
• As the temperature of the gas increases, the curve flattens and broadens, and the peak (v_mp) shifts to the right, indicating that at higher temperatures, more molecules are moving at higher speeds.

Q10. (a) Why are bubbles of a gas spherical in nature? (b) What is the effect of temperature on surface tension? (c) The number of drops for a certain volume of water counted in a stalagmometer at 298 K is 300, whereas the number of drops for the same volume of ethyl alcohol is 320. Calculate the surface tension of ethyl alcohol. Given that: The density of water = 0.998 kg dm⁻³, Density of ethyl alcohol = 0.964 kg dm⁻³, Surface tension of water = 0.0727 N m⁻¹

Ans. (a) Gas bubbles are spherical due to the property of surface tension . A liquid surface tends to contract to minimize its surface area. For a given volume, a sphere is the geometric shape that has the smallest possible surface area. Therefore, to minimize the surface energy associated with its interface, a gas bubble (or a liquid droplet) naturally assumes a spherical shape.

(b) Surface tension decreases as temperature increases . This is because an increase in temperature leads to an increase in the kinetic energy of the liquid molecules. This increased molecular motion becomes more effective at overcoming the intermolecular attractive forces that are responsible for surface tension. As these forces weaken, the surface tension of the liquid decreases. At the critical temperature, surface tension becomes zero.

(c) Using the stalagmometer (drop count) method, the relationship between surface tension (γ), number of drops (n), and density (d) for two liquids is given by: γ₁ / γ₂ = (n₂ × d₁) / (n₁ × d₂) Let subscript (1) be for water and (2) be for ethyl alcohol. Given: γ₁ (water) = 0.0727 N m⁻¹ n₁ (water) = 300 d₁ (water) = 0.998 kg dm⁻³ n₂ (ethyl alcohol) = 320 d₂ (ethyl alcohol) = 0.964 kg dm⁻³ γ₂ (ethyl alcohol) = ? Rearranging the formula to solve for γ₂: γ₂ = γ₁ × (n₁ × d₂) / (n₂ × d₁) γ₂ = 0.0727 × (300 × 0.964) / (320 × 0.998) γ₂ = 0.0727 × (289.2) / (319.36) γ₂ = 0.0727 × 0.90556 γ₂ ≈ 0.0658 N m⁻¹ Thus, the surface tension of ethyl alcohol is 0.0658 N m⁻¹.

Q11. (a) Illustrate the formation of a tetrahedral void and an octahedral void. (b) Calculate the number of formula units of NaCl present in a unit cell of NaCl.

Ans. (a) Formation of Voids in Crystal Lattices: Voids are the empty spaces left between the constituent particles (atoms or ions) in a close-packed structure.

Tetrahedral Void:

• A tetrahedral void is formed by four spheres .
• Formation: Imagine three spheres arranged in a single layer, touching each other. This creates a small triangular gap or void in the center. When a fourth sphere is placed directly on top of this central void, the empty space enclosed by these four spheres is called a tetrahedral void .
• It is named so because the centers of the four surrounding spheres form a regular tetrahedron.

Octahedral Void:

• An octahedral void is formed by six spheres .
• Formation: It is formed by the combination of two triangular voids from adjacent layers. Consider a layer of three spheres forming an upward-pointing triangle. When another layer of three spheres is placed on top in a staggered manner, forming a downward-pointing triangle, the space enclosed between these two sets of three spheres is an octahedral void .
• It is named so because the centers of the six surrounding spheres form a regular octahedron.

(b) Number of Formula Units in an NaCl Unit Cell: The NaCl crystal has a face-centered cubic (fcc) lattice structure.

1. Number of Cl⁻ ions: The larger Cl⁻ ions form the fcc lattice.
   • 8 Cl⁻ ions at the 8 corners: each contributes 1/8 to the cell → 8 × (1/8) = 1 Cl⁻ ion.
   • 6 Cl⁻ ions at the 6 face centers: each contributes 1/2 to the cell → 6 × (1/2) = 3 Cl⁻ ions.
   • Total number of Cl⁻ ions per unit cell = 1 + 3 = 4 .
2. Number of Na⁺ ions: The smaller Na⁺ ions occupy all the octahedral voids.
   • In an fcc lattice, there is 1 octahedral void at the body center and 12 on the edge centers.
   • 1 Na⁺ ion at the body center: contributes 1 to the cell → 1 × 1 = 1 Na⁺ ion.
   • 12 Na⁺ ions at the 12 edge centers: each contributes 1/4 to the cell → 12 × (1/4) = 3 Na⁺ ions.
   • Total number of Na⁺ ions per unit cell = 1 + 3 = 4 .
3. Number of Formula Units: Since there are 4 Na⁺ ions and 4 Cl⁻ ions in the unit cell, the ratio is 4:4 or 1:1, which corresponds to the formula NaCl. Therefore, there are 4 formula units of NaCl in one unit cell.

Q12. (a) Silver has a fcc lattice. The edge length of its unit cell is 4.077 x 10⁻⁸ cm and its density is 10.12 g/cm³. What is the molar mass of silver? (b) Give two differences between Schottky defect and Frenkel defect.

Ans. (a) Calculation of Molar Mass of Silver: The density (d) of a unit cell is given by the formula: d = (Z × M) / (N_A × a³) Where: d = Density = 10.12 g/cm³ (Using this value as 0.12 g/cm³ is a likely typo) Z = Number of atoms per unit cell (for fcc, Z = 4 ) M = Molar Mass (g/mol) (to be calculated) N_A = Avogadro’s number = 6.022 × 10²³ mol⁻¹ a = Edge length = 4.077 × 10⁻⁸ cm Rearranging the formula to solve for M: M = (d × N_A × a³) / Z First, calculate the volume of the unit cell, a³: a³ = (4.077 × 10⁻⁸ cm)³ = 67.78 × 10⁻²⁴ cm³ Now, calculate M: M = (10.12 g/cm³ × 6.022 × 10²³ mol⁻¹ × 67.78 × 10⁻²⁴ cm³) / 4 M = (10.12 × 6.022 × 67.78 × 10⁻¹) / 4 M = 4130.3 / 40 M = 103.26 g/mol Thus, the molar mass of silver is calculated to be approximately 103.26 g/mol .

(b) Differences between Schottky and Frenkel Defects:

Q13. (a) State law of constancy of interfacial angles. (b) What are pseudo first order reactions? Give an example. (c) “Molecularity of a reaction rarely exceeds 3.” Explain.

Ans. (a) Law of Constancy of Interfacial Angles: Also known as Steno’s Law, this law states that for a given crystalline substance, the angles between corresponding crystal faces (the interfacial angles) are always constant and characteristic of that substance, irrespective of the size or shape of the crystal. The measurement is made at a constant temperature.

(b) Pseudo First-Order Reactions: A pseudo first-order reaction is a reaction that is bimolecular or of a higher order but behaves as if it were a first-order reaction under certain conditions. This situation occurs when one of the reactants is present in a very large excess (e.g., when it is the solvent) compared to the other reactants. The concentration of the reactant in excess remains practically constant throughout the reaction. As a result, the reaction rate appears to depend only on the concentration of the reactant present in a smaller amount. Example: The acid-catalyzed hydrolysis of an ester, like ethyl acetate. CH₃COOC₂H₅ (ethyl acetate) + H₂O –(H⁺)–> CH₃COOH + C₂H₅OH In this reaction, water (H₂O) is the solvent and is present in such a large excess that its concentration change is negligible. Therefore, the rate law, Rate = k[CH₃COOC₂H₅][H₂O], simplifies to Rate = k'[CH₃COOC₂H₅], where k’ = k[H₂O] is the pseudo first-order rate constant.

(c) Explanation for Molecularity Rarely Exceeding 3: Molecularity is defined as the number of reacting species (atoms, ions, or molecules) that must collide simultaneously in an elementary reaction step to bring about a chemical reaction. For a collision to be effective and lead to a product, the colliding particles must possess sufficient energy (activation energy) and have the correct orientation relative to each other. The probability of a simultaneous collision involving three particles is significantly lower than that of a two-particle collision. The probability of four or more particles colliding at the exact same instant, with the required energy and orientation, is extremely low and statistically negligible. Therefore, elementary reactions with a molecularity of four or higher are practically impossible. Reactions that appear to involve many reactants actually proceed through a sequence of simpler, elementary steps, each with a molecularity of one, two, or, in rare cases, three.

Q14. (a) Derive the expression for integrated rate law for a zero-order reaction. (b) Draw the following plots for a first order reaction: (i) ln[R] vs. t (ii) log([R]₀/[R]) vs. t

Ans. (a) Derivation of Integrated Rate Law for a Zero-Order Reaction: Consider a general zero-order reaction: R → P. The rate of the reaction is independent of the concentration of the reactant R. The rate law is: Rate = k[R]⁰ = k The differential rate expression is: -d[R] / dt = k where k is the rate constant. To find the integrated rate law, we rearrange and integrate this equation. Separate the variables: d[R] = -k dt Now, integrate both sides. Let the initial concentration at t = 0 be [R]₀ and the concentration at time t be [R]. ∫_[R]₀^[R] d[R] = ∫_0^t -k dt Performing the integration gives: [R] – [R]₀ = -k(t – 0) [R] – [R]₀ = -kt Rearranging this equation gives the integrated rate law for a zero-order reaction: [R] = -kt + [R]₀ This equation is in the form of a straight line (y = mx + c). A plot of [R] vs. t is a straight line with a slope of -k and a y-intercept of [R]₀.

(b) Plots for a First-Order Reaction:

(i) ln[R] vs. t

• The integrated rate law for a first-order reaction is: ln[R] = -kt + ln[R]₀ .
• This equation is of the form of a straight line, y = mx + c .
• Therefore, a plot of ln[R] (on the y-axis) versus t (on the x-axis) yields a straight line with a negative slope .
• The slope of this line is equal to -k (the rate constant).
• The y-intercept of the line is ln[R]₀ (the natural logarithm of the initial concentration).

(ii) log([R]₀/[R]) vs. t

• Another form of the first-order integrated rate law is: kt = 2.303 log([R]₀/[R]).
• Rearranging this gives: log([R]₀/[R]) = (k / 2.303) t .
• This equation is of the form of a straight line, y = mx .
• Therefore, a plot of log([R]₀/[R]) (on the y-axis) versus t (on the x-axis) yields a straight line that passes through the origin .
• The slope of this line is equal to k / 2.303 .