Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question A(g) → B(g) + C(g) is a first-order reaction . The reaction is followed using pressure method . Let: P t P_t  = total pressure at time t t t P ∞ P_\infty  = total pressure at infinite time Reaction started with only A Which of the following expressions gives the correct rate constant k k ? 🖼️ Question Image ✍️ Short Solution To derive the expression for k k , use pressure as a measure of extent of reaction . 🔹 Step 1 — Initial condition Reaction: A → B + C A \rightarrow B + C Initial pressure of A: P A = P 0 P_A = P_0 ​ 🔹 Step 2 — Pressure at time t t Let x x x be the amount (in pressure units) of A decomposed at time t t . Then: Pressure of A remaining = P 0 − x P_0 - x Pressure of B formed = x x Pressure of C formed = x x Total pressure at time t t t : P t = ( P 0 − x ) + x + x = P 0 + x P_t = (P_0 - x) + x + x = P_0 + x So: x = P t − P 0 x = P_t - P_0 ​ 🔹 Step 3 — Total pressure at infinite time When rea...

  ❓ Question The number of unpaired electrons responsible for the paramagnetic nature of the following complex ions are, respectively: [ Fe(CN) 6 ] 3 − [ FeF 6 ] 3 − [ \text{FeF}_6 ]^{3-} [ CoF 6 ] 3 − [ \text{CoF}_6 ]^{3-} [ Mn(CN) 6 ] 3 − [ \text{Mn(CN)}_6 ]^{3-} 🖼️ Question Image ✍️ Short Solution We analyze each complex using: Oxidation state Electronic configuration Nature of ligand (strong field or weak field) High-spin vs low-spin Count unpaired electrons ✅ (1) [ Fe(CN) 6 ] 3 − [ \text{Fe(CN)}_6 ]^{3-} Step-1: CN⁻ is a strong-field ligand → large Δ₀ → low-spin . Oxidation state: Fe + 6 ( − 1 ) = − 3 ⇒ Fe 3 + \text{Fe} + 6(-1) = -3 \Rightarrow \text{Fe}^{3+} Fe 3 + : 3 d 5 \text{Fe}^{3+} : 3d^5 In low-spin d⁵ , electrons pair as much as possible: Configuration: t 2 g 5 e g 0 t_{2g}^5 e_g^0 ​ Unpaired electrons = 1 ✅ (2) [ FeF 6 ] 3 − [ \text{FeF}_6 ]^{3-} F⁻ is a weak-field ligand → small Δ₀ → high-spin . Oxidation state o...

❓ Question The electric field in a region is given by E = ( 2 i ^ + 4 j ^ + 6 k ^ ) × 10 3   N/C . The flux of the field through a rectangular surface parallel to the x–z plane is 6.0 N·m²·C⁻¹ . Find the area of the surface . 🖼️ Question Image ✍️ Short Solution We’ll use the electric flux formula : Φ = E ⃗ ⋅ A ⃗ = E A cos ⁡ θ Where: Φ \Phi  = Electric flux E ⃗ \vec{E}  = Electric field A ⃗ \vec{A}  = Area vector (perpendicular to the surface) θ \theta  = angle between E ⃗ \vec{E}  and A ⃗ \vec{A} 🔹 Step 1 — Determine direction of the area vector The surface is parallel to the x–z plane , ⇒ its normal vector is along the y-axis (positive or negative ĵ ). Thus, A ⃗ = A j ^​ 🔹 Step 2 — Find component of electric field perpendicular to surface Electric field: E ⃗ = ( 2 i ^ + 4 j ^ + 6 k ^ ) × 10 3 Only the ĵ-component (i.e., E y E_y E y ​ ) contributes to the flux through the x–z surface. So, E y = 4 × 10 3   N/C 🔹 Step 3...

  ❓ Question Match List - I with List - II List - I List - II (A) Solution of chloroform and acetone (I) Minimum boiling azeotrope (B) Solution of ethanol and water (II) Dimerizes (C) Solution of benzene and toluene (III) Maximum boiling azeotrope (D) Solution of acetic acid in benzene (IV) ΔVₘᵢₓ = 0 🖼️ Question Image ✍️ Short Solution Let’s match each solution with its correct behavior using the concept of ideal and non-ideal liquid mixtures . 🔹 Step 1 — Recall: Types of Solutions Based on Raoult’s Law Ideal Solutions: Obey Raoult’s Law completely. No enthalpy or volume change on mixing. Example: Benzene + Toluene. ⇒ ΔHₘᵢₓ = 0, ΔVₘᵢₓ = 0. Non-Ideal Solutions: Deviate from Raoult’s Law due to differences in molecular forces. Can show positive or negative deviation. 🔹 Step 2 — Identify the Type of Deviation Type Interaction Strength Vapor Pressure Boiling Point Example Positive Deviation A–B < A–A or B–B Higher than expected Minimu...

  ❓ Question Which of the following statements is correct ? 1️⃣ Δ f H 298 ∘ \Delta_f H^\circ_{298} Δ f ​ H 298 ∘ ​ is zero for O(g) 2️⃣ The standard state of a pure gas is the pure gas at a pressure of 1 bar and temperature 273 K 3️⃣ The term standard state implies that the temperature is 0 °C 4️⃣ Δ f H 500 ∘ \Delta_f H^\circ_{500} Δ f ​ H 500 ∘ ​ is zero for O₂(g) 🖼️ Question Image ✍️ Short Solution Let’s analyze each statement carefully using thermodynamic definitions. 🔹 Step 1 — Recall definition of standard enthalpy of formation (ΔfH°) The standard enthalpy of formation is the enthalpy change when 1 mol of a substance is formed from its constituent elements in their most stable states at 1 bar pressure and a specified temperature (usually 298 K ). By convention: Δ f H ∘ = 0 for all elements in their standard states at 298 K. 🔹 Step 2 — Check each statement (1) Δ f H 298 ∘ \Delta_f H^\circ_{298} is zero fo...

  ❓ Question The extra stability of half-filled subshells is due to which of the following factors? Options: (A) Symmetrical distribution of electrons (B) Smaller Coulombic repulsion energy (C) Presence of electrons with same spin in non-degenerate orbitals (D) Larger exchange energy (E) Relatively smaller shielding of electrons by one another 🖼️ Question Image ✍️ Short Solution We often see exceptions in electronic configurations, like: Cr:  [ Ar ]   3 d 5   4 s 1 instead of  [ Ar ]   3 d 4   4 s 2 \text{Cr: } [\text{Ar}]\,3d^5\,4s^1 \quad \text{instead of } [\text{Ar}]\,3d^4\,4s^2 Cu:  [ Ar ]   3 d 10   4 s 1 instead of  [ Ar ]   3 d 9   4 s 2 \text{Cu: } [\text{Ar}]\,3d^{10}\,4s^1 \quad \text{instead of } [\text{Ar}]\,3d^9\,4s^2 This happens because half-filled (d⁵) and fully-filled (d¹⁰) subshells are more stable than uneven ones. Let’s understand why . 🔹 Step 1 — Symmetrical distribution of electrons In a half-filled subsh...

  ❓ Question In SO 2 \text{SO}_2 ​ , NO 2 − \text{NO}_2^- ​ , and N 3 − \text{N}_3^- ​ , the hybridizations at the central atom are respectively: 🖼️ Question Image ✍️ Short Solution We’ll find hybridization at each central atom using the Steric Number (SN) method: Steric number = Number of σ-bonds + Number of lone pairs on central atom and use: Steric Number Hybridization Example 2 sp BeCl₂ 3 sp² BF₃ 4 sp³ CH₄ 5 sp³d PCl₅ 6 sp³d² SF₆ 🔹 (A) For SO 2 \text{SO}_2 ​ Step 1: Central atom = S Valence electrons of S = 6 Each O atom forms a bond with S; one double bond and one coordinate bond (resonance). → S forms 2 σ-bonds + has 1 lone pair Steric number = 3 → sp² hybridization ✅ Hybridization of S in SO₂ = sp² Geometry: Bent / V-shaped (bond angle ≈ 119°) 🔹 (B) For NO 2 − \text{NO}_2^- ​ Step 1: Central atom = N Valence electrons of N = 5 , plus 1 extra electron (for negative charge) → total 6 N forms 2 σ-bon...