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CONCEPT:

Stoichiometry of the Reaction

• Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction.
• In the given reaction:

3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

• From the balanced chemical equation:
• 3 moles of Fe react with 4 moles of H₂O.
• The molar ratio of Fe to H₂O is 3:4.
• Molar masses:
• Molar mass of Fe = 56 g/mol
• Molar mass of H₂O = 18 g/mol

EXPLANATION:

• Given mass of steam (H₂O) = 18 g
• Number of moles of steam:

n_steam = ⅛18 = 1 mol

• From the stoichiometric ratio, moles of Fe required:

n_Fe = ¾

• Mass of Fe required:

Mass of Fe = n_Fe × Molar mass of Fe

= ¾ × 56

= 42 g

Therefore, the mass of iron converted into Fe₃O₄ by the action of 18 g of steam is 42 g.

CONCEPT:

Ionization Energy and Its Dependence on Nuclear Charge (Z)

• The ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state.
• For hydrogen-like ions (single electron systems like $\text{Li}^{2+}$), the ionization energy can be calculated using the formula:

$\text{E} = \text{E}_{\text{H}} \times \text{Z}^2$

• Here:
• $\text{E}_{\text{H}}$ is the ionization energy of a hydrogen atom in its ground state.
• $\text{Z}$ is the atomic number of the element.

EXPLANATION:

• The given reaction is:

$\text{Li}^{2+}\text{(g)} \rightarrow \text{Li}^{3+}\text{(g)} + \text{e}^-$

• Here, $\text{Li}^{2+}$ is a hydrogen-like ion with a single electron in its orbit. Its ionization energy can be calculated using the formula:

$\text{E} = \text{E}_{\text{H}} \times \text{Z}^2$

• The ionization energy of hydrogen, $\text{E}_{\text{H}} = 2.18 \times 10^{-18} \text{ J atom}^{-1}$.
• The atomic number of lithium, $\text{Z} = 3$.
• Substitute the values into the formula:

$\text{E}_{\text{Li}^{2+}} = 2.18 \times 10^{-18} \times (3)^2$

$\text{E}_{\text{Li}^{2+}} = 2.18 \times 10^{-18} \times 9$

$\text{E}_{\text{Li}^{2+}} = 1.962 \times 10^{-17} \text{ J atom}^{-1}$

Therefore, the energy required for the process is $1.962 \times 10^{-17} \text{ J atom}^{-1}$.

CONCEPT:

Bond Order and Unpaired Electrons in Molecular Orbital Theory

• Bond Order:

Bond Order = ^{(Number of bonding electrons - Number of antibonding electrons)}⁄₂

• The bond order is a measure of the number of chemical bonds between a pair of atoms. It is given by the formula:
• A higher bond order corresponds to a shorter and stronger bond, while a lower bond order corresponds to a longer and weaker bond.
• Unpaired Electrons:
• The number of unpaired electrons in a molecule is determined based on the electronic configuration of its molecular orbitals.
• Molecules with unpaired electrons are paramagnetic, while those without unpaired electrons are diamagnetic.

EXPLANATION:

• The electronic configurations of the molecular orbitals for the given oxygen species are as follows:
• O+2O2+

: Bond Order = 2.5, Unpaired Electrons = 1

• O2O2

: Bond Order = 2, Unpaired Electrons = 2

• O−2O2−

: Bond Order = 1.5, Unpaired Electrons = 1

• O2−2O22−

: Bond Order = 1, Unpaired Electrons = 0

• Statement I: The bond length is inversely proportional to the bond order. Therefore, the correct sequence of bond lengths is:

O+2<O2<O−2<O2−2O2+<O2<O2−<O22−

This makes Statement I true.

• Statement II: The number of unpaired electrons decreases in the following order:

O2>O+2>O−2>O2−2O2>O2+>O2−>O22−

This makes Statement II true.

Conclusion: Both Statement I and Statement II are true.

CONCEPT:

Enthalpy of Formation ($\Delta_{\text{f}}H$)

• The enthalpy of formation ($\Delta_{\text{f}}H$) of a compound is the change in enthalpy when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions.
• To calculate the enthalpy of formation for a reaction, the enthalpy changes ($\Delta_{\text{r}}H$) from individual steps involved in the reaction can be summed up using Hess's Law.
• Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps that lead to the reaction.

EXPLANATION:

• Given data:
• Step (i): \( 2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}2\text{(g)} \), \( \Delta{\text{r}}H_1 = -1200 , \text{kJ/mol} \)
• Step (ii): \( \text{H}_2\text{(g)} + \text{Cl}2\text{(g)} \rightarrow 2\text{HCl(g)} \), \( \Delta{\text{r}}H_2 = -164 , \text{kJ/mol} \) (multiplied by 3: $-164 \times 3 , \text{kJ/mol}$)
• Step (iii): $\text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)}$, $\Delta_{\text{r}}H_3 = -83 , \text{kJ/mol}$ (multiplied by 6: $-83 \times 6 , \text{kJ/mol}$)
• Step (iv): \( \text{Al}_2\text{Cl}_6\text{(aq)} \rightarrow \text{Al}_2\text{Cl}6\text{(s)} + \text{aq} \), \( \Delta{\text{r}}H_4 = +663 , \text{kJ/mol} \)
• Target reaction: $2\text{Al(s)} + 3\text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6\text{(s)}$
• Using Hess's Law:
• $\Delta_{\text{f}}H = \Delta_{\text{r}}H_1 + \Delta_{\text{r}}H_2 + \Delta_{\text{r}}H_3 + \Delta_{\text{r}}H_4$
• $\Delta_{\text{f}}H = -1200 - (164 \times 3) - (83 \times 6) + 663$
• $\Delta_{\text{f}}H = -1200 - 492 - 498 + 663$
• $\Delta_{\text{f}}H = -1527 , \text{kJ/mol}$

Therefore, the enthalpy of formation ($\Delta_{\text{f}}H$) of anhydrous solid $\text{Al}_2\text{Cl}_6$ is $-1527 , \text{kJ/mol}$.

CONCEPT:

Freezing Point Depression and Degree of Dissociation

• The depression in the freezing point of a solution is given by the equation:

ΔT_f = iK_fm

where:

• ΔT_f = depression in freezing point
• i = van't Hoff factor
• K_f = molal freezing point depression constant
• m = molality of the solution
• The van't Hoff factor, i, is related to the degree of dissociation (α) by the formula:

i = 1 + (n - 1)α

where:

• n = number of ions produced per molecule upon dissociation
• α = degree of dissociation
• The dissociation constant (K_a) is given by the formula:

K_a = (Cα²) / (1 - α)

where:

• C = concentration of the solute
• α = degree of dissociation

EXPLANATION:

• Given data:
• Mass of fluoroacetic acid = 19.5 g
• Molar mass of fluoroacetic acid = 78 g mol^-1
• Mass of water = 500 g = 0.5 kg
• ΔT_f = 1°C
• K_f = 1.86 K kg mol^-1
• Calculate the number of moles of fluoroacetic acid:

= ^19.5 / ₇₈ = 0.25 mol

• Calculate the molality (m):

m = ^{Number of moles of solute} / _{Mass of solvent in kg}

= ^0.25 / _0.5 = 0.5 mol kg^-1

• ΔT_f = iK_fm

1 = i × 1.86 × 0.5

i = ¹ / _{1.86 × 0.5} = ² / _1.86

• i = 1 + α (since fluoroacetic acid dissociates into 2 particles: H⁺ and CH₂FCOO^-)

² / _1.86 = 1 + α

α = ² / _1.86 - 1 = 0.075

• K_a = ^Cα² / _{1 - α}

Here, C = molality = 0.5 mol kg^-1

K_a = ^{0.5 × (0.075)²} / _{1 - 0.075}

K_a = ^{0.5 × 0.005625} / _0.925 = ^0.0028125 / _0.925

K_a ≈ 3 × 10^-3

Therefore, the dissociation constant (K_a) of fluoroacetic acid is approximately 3 × 10^-3.

CONCEPT:

Solubility Product and Molarity Relation

• The solubility product constant ($K_{\text{sp}}$) represents the equilibrium constant for the dissolution of a sparingly soluble salt.
• The molarity of ions in solution can be derived from the solubility product constant using the stoichiometric relationship between the ions and the salt.

EXPLANATION:

• For $\text{Ag}_2\text{CrO}_4$:
• The dissociation equation is:

$\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$

• The solubility product is given by:

$K_{\text{sp}} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]$

• If the molarity of $\text{Ag}_2\text{CrO}_4$ is $S_1$, then:
• $[\text{Ag}^+] = 2S_1$
• $[\text{CrO}_4^{2-}] = S_1$
• Substitute these values into the solubility product equation:

$K_{\text{sp}} = (2S_1)^2(S_1)$

$K_{\text{sp}} = 4S_1^3$

• Given $K_{\text{sp}} = 32x$, solve for $S_1$:

$4S_1^3 = 32x$

$S_1 = 2x^{1/3}$

• For $\text{AgBr}$:
• The dissociation equation is:

$\text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^-$

• The solubility product is given by:

$K_{\text{sp}} = [\text{Ag}^+][\text{Br}^-]$

• If the molarity of $\text{AgBr}$ is $S_2$, then:
• $[\text{Ag}^+] = S_2$
• $[\text{Br}^-] = S_2$
• Substitute these values into the solubility product equation:

$K_{\text{sp}} = S_2^2$

• Given $K_{\text{sp}} = 4y$, solve for $S_2$:

$S_2^2 = 4y$

$S_2 = 2y^{1/2}$

• Calculate the ratio $\left( \frac{\text{S}_1}{\text{S}_2} \right)$:

$\frac{\text{S}_1}{\text{S}_2} = \frac{2x^{1/3}}{2y^{1/2}}$

$\frac{\text{S}_1}{\text{S}_2} = \frac{\sqrt[3]{x}}{\sqrt{y}}$

Therefore, the ratio $\left( \frac{\text{molarity of Ag}_2\text{CrO}_4}{\text{molarity of AgBr}} \right)$ is $\frac{\sqrt[3]{x}}{\sqrt{y}}$.

CONCEPT:

Standard Cell Potential (E⁰_cell)

• The standard cell potential (E⁰_cell) is the difference in the standard reduction potentials of the cathode and the anode in an electrochemical cell.
• The standard cell potential can be calculated using the equation:

E⁰_cell = E⁰_cathode - E⁰_anode

EXPLANATION:

• In the given reaction:

Fe(s) + 2OH^-(aq) + 2AgBr(s) → Fe(OH)₂(s) + 2Ag(s) + 2Br^-(aq)

• Iron (Fe) is oxidized to Fe(OH)₂, so Fe is the anode.
• Silver bromide (AgBr) is reduced to Ag, so AgBr is the cathode.
• Given standard reduction potentials:
• E⁰_AgBr = +0.07 V (for AgBr(s) + e^- → Ag(s) + Br^-(aq))
• E⁰_Fe = -0.88 V (for Fe(OH)₂(s) + 2e^- → Fe(s) + 2OH^-(aq))
• Using the standard cell potential equation:
• E⁰_cell = E⁰_cathode - E⁰_anode
• = 0.07 V - (-0.88 V)
• = 0.07 V + 0.88 V
• = 0.95 V
• Key points to note:
• Fe is oxidized in the reaction, not reduced.
• E⁰_cell is not an extensive property; it is an intensive property, as it does not depend on the amount of material.

Therefore, the correct overall reaction is:

Fe(s) + 2OH^-(aq) + 2AgBr(s) → Fe(OH)₂(s) + 2Ag(s) + 2Br^-(aq)

The standard cell potential (E⁰_cell) is 0.95 V.

CONCEPT:

Zero-order and First-order Reaction Time Relationships

• For a zero-order reaction, the rate of reaction is constant and independent of the concentration of the reactant.
• For a first-order reaction, the rate of reaction is proportional to the concentration of the reactant.
• The times required for 50% completion (\( \text{t}{1/2} \)) and 100% completion (\( \text{t}{100%} \)) of the reaction differ for zero-order and first-order reactions.

EXPLANATION:

• Zero-order Reaction:
• For a zero-order reaction, the time required for the reaction to complete 100% is given by:

$\text{t}_{100\%} = \frac{\text{a}}{\text{k}}$

• The time required for 50% completion is:

$\text{t}_{1/2} = \frac{\text{a}}{2\text{k}}$

• By dividing \( \text{t}{100%} \) by \( \text{t}{1/2} \), we get:

\( \text{t}{100%} = 2\text{t}{1/2} \)

• First-order Reaction:
• For a first-order reaction, the time required for 100% completion is theoretically infinite, because the reactant concentration never truly reaches zero.
• The time required for 50% completion is:

$\text{t}_{1/2} = \frac{\ell\text{n } 2}{\text{k}}$

• Since $\text{t}_{100\%}$ is infinite, we represent the relationship as:

\( \text{t}{100%} = (\text{t}{1/2})^\infty \)

So, For a zero-order reaction, \( \text{t}{100%} = 2\text{t}{1/2} \).

• For a first-order reaction, \( \text{t}{100%} = (\text{t}{1/2})^\infty \).

Therefore, the correct relationship is:

\( \text{t}{100%} = 2\text{t}{1/2} \) for zero-order reactions and \( \text{t}{100%} = (\text{t}{1/2})^\infty \) for first-order reactions.

CONCEPT:

Ionization Enthalpy

• The ionization enthalpy is the energy required to remove an electron from the outermost shell of an isolated gaseous atom.
• The first ionization enthalpy corresponds to the removal of the first electron, the second ionization enthalpy corresponds to the removal of the second electron, and so on.
• Ionization enthalpy is influenced by factors such as atomic size, nuclear charge, shielding effect, and electron configuration.

EXPLANATION:

• Statement (I): The first ionization enthalpy of the elements Na, Mg, Cl, and Ar follows the order Na > Mg > Cl > Ar.
• This statement is false because the correct order of first ionization enthalpy is Na < Mg < Cl < Ar.
• The trend in ionization energy across a period increases from left to right due to increasing nuclear charge and decreasing atomic size.
• Na has the lowest ionization enthalpy, followed by Mg, Cl, and Ar, which has the highest due to its stable noble gas configuration.
• Statement (II): Among Ca, Al, Fe, and B, the third ionization enthalpy is very high for Ca.
• This statement is true.
• Calcium (Ca) has a very high third ionization enthalpy because after losing two electrons, it attains a stable noble gas configuration (similar to Argon: [Ar]).
• Removing a third electron from this stable configuration requires a significantly higher amount of energy.

So, Statement I is false but Statement II is true.

CONCEPT:

Oxidizing Power of Halogens

• Oxidizing power refers to the ability of a substance to oxidize another substance, i.e., to accept electrons.
• Among the halogens, the oxidizing power decreases down the group in the periodic table in the following order: F₂ > Cl₂ > Br₂ > I₂.
• This trend is due to the decreasing standard reduction potentials of the halogens as we move down the group, making them less effective oxidizing agents.

Layer Test

• The "Layer Test" is a chemical test used to identify halogens in aqueous solutions based on their relative oxidizing powers.
• In this test, a more powerful oxidizing agent (e.g., Cl₂) is used to oxidize halide ions (e.g., Br^- or I^-) to their molecular forms (Br₂ or I₂), which can then be detected by their characteristic colors.
• For example:
• Cl₂ oxidizes bromide ions (Br^-) to bromine (Br₂), which appears as an orange-brown color in aqueous solution:

NaBr + Cl₂-water → Br₂ + NaCl

• Cl₂ also oxidizes iodide ions (I^-) to iodine (I₂), which appears as a violet color in organic solvents or a brown color in aqueous solution:

NaI + Cl₂-water → I₂ + NaCl

EXPLANATION:

• Statement (I): "Oxidizing power of halogens decreases in the order F₂ > Cl₂ > Br₂ > I₂, which is the basis of 'Layer test'" is true. This is because fluorine has the highest oxidizing power, followed by chlorine, bromine, and iodine, consistent with their standard reduction potentials.
• Statement (II): "'Layer test' to identify Br₂ and I₂ in aqueous solution involves the oxidation of bromide or iodide into Br₂ or I₂ respectively with Cl₂, which is a type of displacement redox reaction" is also true. This statement correctly describes the chemical basis of the layer test, where Cl₂, being a stronger oxidizing agent, displaces Br₂ and I₂ from their respective halide ions.
• Since both statements are correct and Statement (II) explains the process mentioned in Statement (I), the correct interpretation is that both Statement I and Statement II are true.

Therefore, the answer is that both Statement I and Statement II are true.