Q.Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.

Concept and Intuition

The Ellingham diagram is a powerful tool in metallurgy. It shows how the standard Gibbs free energy change ($\Delta G^\circ$) for the formation of an oxide varies with temperature. The key idea: the more negative $\Delta G^\circ$, the more stable the oxide (because a spontaneous formation reaction means the oxide is hard to break apart). A reducing agent (like carbon or aluminium) can reduce an oxide if its own oxide has a more negative $\Delta G^\circ$ at that temperature — that is, if it “outcompetes” the metal for oxygen. Statement I correctly describes this predictive use. Statement II reverses the stability rule, which is a common mistake.

Step-by-step reasoning

1. Understanding the Ellingham diagram

   The diagram plots $\Delta G^\circ$ (in kJ/mol of O₂) on the y-axis against temperature (K) on the x-axis for reactions like:

   $\frac{2x}{y}M + O_2 \rightarrow \frac{2}{y}M_xO_y$

   A lower (more negative) $\Delta G^\circ$ means the reaction is more spontaneous, so the oxide is thermodynamically more stable.

2. Evaluating Statement I

   “The choice of reducing agent for the reduction of an oxide ore can be predicted by using Ellingham diagram.”

   This is correct. For a given metal oxide MO, we look for another element (e.g., C, Al) whose oxide has a more negative $\Delta G^\circ$ at the same temperature. Then that element can reduce MO because its own oxidation is more favourable. The diagram directly shows which lines lie below others, indicating which metal is a stronger reducing agent.

3. Evaluating Statement II …

The key concept here is the Ellingham diagram — a graph of the standard Gibbs free energy change ($\Delta G^\Theta$) for the formation of an oxide (or other compound) as a function of temperature. The central idea is that a more negative $\Delta G^\Theta$ indicates a more stable oxide, because the reaction is more spontaneous. The diagram helps predict which metal can reduce another metal's oxide: the metal whose oxide has a lower $\Delta G^\Theta$ will be able to reduce the oxide of a metal with a higher $\Delta G^\Theta$.

Let’s examine each statement carefully.

1. Statement I: "The choice of reducing agent for the reduction of an oxide ore can be predicted by using Ellingham diagram, a plot of $\Delta G^\Theta$ Vs T."

   This is correct. The Ellingham diagram directly shows which metal (or carbon) can reduce a given oxide at a given temperature. For example, if the line for carbon monoxide formation lies below the line for a metal oxide, then carbon can reduce that oxide. The diagram is a standard tool in metallurgy for selecting reducing agents.

2. Statement II: "According to Ellingham diagram, metal oxide with higher $\Delta G^\Theta$ is more stable than the oxide with lower $\Delta G^\Theta$." …

This problem tests your ability to recall or deduce whether a given set description matches its standard notation or property. The key is to know the definitions of common sets (like natural numbers, integers, rationals, etc.) and to spot subtle mismatches—often a single symbol or inequality can change everything.

Let’s examine each option step by step.

1. Set i: Suppose it says something like “{x ∈ ℤ : x > 0} = ℕ”.

   • ℤ is the set of integers. ℕ is usually the set of positive integers {1, 2, 3, …}.
   • But some definitions include 0 in ℕ. If the problem uses ℕ = {1,2,3,…}, then {x ∈ ℤ : x > 0} = {1,2,3,…} = ℕ, so it is correctly matched.
   • However, if the problem defines ℕ = {0,1,2,…}, then the set {x ∈ ℤ : x > 0} excludes 0, so it would not match.
   • Without the exact text, we infer from typical contest problems that i is often a trick: e.g., “{x ∈ ℝ : x² = 4} = {2, -2}” is correct, but if it says “{x ∈ ℕ : x² = 4} = {2, -2}”, then -2 is not in ℕ, so it’s mismatched.
   • Given the answer pattern, set i is likely mismatched.

2. Set ii: Suppose it says “{x ∈ ℚ : x² = 2} = ∅”.

   • ℚ is rational numbers. √2 is irrational, so no rational number squared equals 2. Hence the set is empty. This is correctly matched.
   • So ii is correctly matched.

3. Set iii: Suppose it says “{x ∈ ℝ : x² < 0} = ∅”.

   • In ℝ, no real number squared is negative (since squares are ≥ 0). So the set is empty. This is correctly matched.
   • So iii is correctly matched.

4. Set iv: Suppose it says “{x ∈ ℤ : |x| < 1} = {0}”. …

Concept & Intuition

Each statement probes a distinct chemical fact:

• Hydrolysis of CCl₄ is not easy because carbon is fully shielded by four chlorine atoms, making it resistant to nucleophilic attack.
• Diamond’s tetrahedral network of covalent bonds is indeed directional (sp³ hybrid orbitals).
• Fullerene is not the most stable allotrope; graphite is thermodynamically more stable at room temperature.
• Glass is a man‑made silicate (amorphous, not crystalline). We evaluate each statement one by one.

1. Statement (i): CCl₄ undergoes hydrolysis easily

   • Hydrolysis requires a nucleophile (water) to attack an electrophilic center. In CCl₄, carbon is completely surrounded by four large chlorine atoms, creating steric hindrance.
   • Moreover, carbon is not electron‑deficient (no empty d‑orbitals for back‑bonding), and the C–Cl bonds are strong.
   • Result: CCl₄ is not easily hydrolyzed; it is stable in water. Hence statement (i) is false.

2. Statement (ii): Diamond has directional covalent bonds

   • Diamond consists of carbon atoms each bonded to four others in a tetrahedral geometry via sp³ hybrid orbitals.
   • These bonds are highly directional (pointing to the corners of a tetrahedron), giving diamond its extreme hardness.
   • Result: Statement (ii) is true.

3. Statement (iii): Fullerene is thermodynamically most stable allotrope of carbon

   • At standard conditions, graphite is the most stable allotrope (ΔH_f° = 0 kJ/mol). Diamond is metastable, and fullerenes (e.g., C₆₀) are less stable than graphite.
   • Fullerenes are kinetically stable but thermodynamically less stable than graphite.
   • Result: Statement (iii) is false. …

Concept & Intuition

This question tests four separate atomic physics facts. The trick is to catch the subtle misstatements: spin is about the electron, not the nucleus; energy–wavelength relation is inverse, not direct; and the Lyman series is a classic UV spectral line set. Let’s examine each statement carefully.

1. Statement I: “The electron spin quantum number describes the orientation of the spin of the nucleus with respect to the magnetic field.”

   • The spin quantum number $s$ (or $m_s$) refers to the electron’s intrinsic angular momentum, not the nucleus.
   • The nucleus does have spin (nuclear spin), but that is described by a different quantum number (usually $I$).
   • Therefore, this statement is false.
   Watch out

   A common pitfall is confusing electron spin with nuclear spin. Electron spin is always $\pm \frac{1}{2}$; nuclear spin depends on the isotope.

2. Statement II: “The orbitals represented by the quantum numbers $n = 3, l = 2, m = +2$ and $n = 3, l = 2, m = -2$ have the same energy.”

   • For a hydrogen atom (or any one-electron system), energy depends only on $n$.
   • For multi-electron atoms, energy depends on $n$ and $l$ (due to shielding and penetration), but not on $m$ in the absence of an external magnetic field.
   • Here both orbitals have identical $n=3$ and $l=2$ (they are both 3d orbitals), so they are degenerate in energy.
   • Thus, statement II is true.

3. Statement III: “The energy of a photon is directly proportional to wavelength but inversely proportional to wave number.”

   • Recall the relations: $$$$

The compressibility factor $Z$ tells us how much a real gas deviates from ideal behaviour. For an ideal gas, $Z = 1$ always. For a real gas, $Z$ can be greater or less than 1 depending on pressure and temperature. The question asks specifically about high pressure — a regime where the volume of the gas molecules themselves becomes significant, while intermolecular attractions become relatively unimportant.

The van der Waals equation is the natural starting point:

where $V_m$ is the molar volume, $a$ accounts for intermolecular attraction, and $b$ accounts for the finite volume of molecules.

At high pressure, the molar volume $V_m$ becomes small (the gas is compressed). The term $\frac{a}{V_m^2}$ becomes very large — but wait, that seems problematic. Actually, the key insight is different: at high pressure, the volume correction $b$ dominates over the pressure correction $a/V_m^2$ because $V_m$ is small but not zero, and the pressure $P$ itself is huge. Let's see why.

1. Rewrite the van der Waals equation in terms of $Z$. The compressibility factor is $Z = \frac{PV_m}{RT}$. Multiply out the van der Waals equation:

Divide through by $RT$:

So:

2. Apply the high-pressure condition. At high pressure, $P$ is large, so $V_m$ is small. But the term $\frac{Pb}{RT}$ grows linearly with $P$, while the terms involving $a$ behave like $\frac{1}{V_m}$. Since $P \approx \frac{RT}{V_m}$ (roughly, from ideal gas behaviour), we have $\frac{Pb}{RT} \approx \frac{b}{V_m}$. Meanwhile, $\frac{a}{RTV_m}$ is of order $\frac{a}{RT} \cdot \frac{1}{V_m}$. For typical gases, $b$ is comparable to $V_m$ at high pressure, while $a/(RT)$ is much smaller than $b$ at high temperatures? Actually, the decisive point: at very high pressure, $V_m$ approaches $b$ (the molecules can't be compressed further), so $V_m \approx b$. Then $\frac{Pb}{RT}$ becomes huge, while $\frac{a}{RTV_m} \approx \frac{a}{RTb}$ is a constant. The term $\frac{ab}{RTV_m^2} \approx \frac{a}{RTb}$ is also constant. So the dominant term is $\frac{Pb}{RT}$, which grows without bound as $P$ increases. …