Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ 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...

  ❓ Question The hydration energies of K + \text{K}^+  and Cl − \text{Cl}^-  are − x -x  and − y  kJ/mol -y~\text{kJ/mol}  respectively. If the lattice energy of KCl is − z  kJ/mol -z~\text{kJ/mol} , then what is the heat of solution of KCl? 🧪 Concept Summary When an ionic solid (like KCl) dissolves in water: Lattice breaks down — ions separate. (Energy absorbed ) Ions get hydrated — water molecules surround ions. (Energy released ) These two processes combine to give the enthalpy (heat) of solution ( Δ H sol ) (\Delta H_{\text{sol}}) . 🖼️ Question Image ✍️ Step-by-Step Solution Step 1 — Write the process KCl(s) → dissolve K + ( a q ) + Cl − ( a q ) This happens in two stages : 1️⃣ Lattice dissociation (endothermic): KCl(s) → K + ( g ) + Cl − ( g ) Energy absorbed = + z  kJ/mol \text{KCl(s)} \rightarrow \text{K}^+ (g) + \text{Cl}^- (g) \quad \text{Energy absorbed} = +z~\text{kJ/mol} 2️⃣ Hydration (exothermic):...

  ❓ Question An inductor of reactance X L = 100   Ω X_L = 100~\Omega , a capacitor of reactance X C = 50   Ω X_C = 50~\Omega , and a resistor of resistance R = 50   Ω R = 50~\Omega  are connected in series with an AC source of 10 V , f = 50  Hz f = 50~\text{Hz} . The average power dissipated by the circuit is W W . Find the value of W W . 🖼️ Question Image ✍️ Step-by-Step Solution Step 1 — Given data R = 50   Ω , X L = 100   Ω , X C = 50   Ω , V rms = 10  V . Step 2 — Find net reactance The total reactance in a series RLC circuit is the difference between inductive and capacitive reactances: X = X L − X C = 100 − 50 = 50   Ω . Step 3 — Find impedance The total impedance Z Z  is given by Z = R 2 + X 2 = 50 2 + 50 2 = 5000 = 70.71   Ω . Step 4 — Find current (rms value) I rms = V rms Z = 10 70.71 = 0.1414  A . Step 5 — Find power factor (cos φ) Power factor for a series RLC circuit: cos ⁡ ϕ ...

❓ Question FOR: Let M M  and R R  be the mass and radius of a (uniform) solid disc. A small disc of radius R / 3 R/3  is removed from the bigger disc (hole cut out) as shown in the figure. The moment of inertia of the remaining part about an axis A B AB  passing through the centre O O  and perpendicular to the plane of the disc is given as I = x 4   M R 2 . Find the value of x x . (We assume the small disc is concentric with the big disc — i.e. the hole is at the center — this is the standard interpretation unless the figure indicates an off-centre cut.) 🖼️ Question Image ✍️ Short Solution Step 1 — Moment of inertia of the original (full) disc For a uniform solid disc of mass M M  and radius R R , I full = 1 2 M R 2 . Step 2 — Mass and moment of inertia of the removed small disc Area (hence mass) scales with R 2 R^2 . The small disc has radius r = R 3 r=\dfrac{R}{3} , so its area is ( r / R ) 2 = ( 1 / 3 ) 2 = 1 / 9 (r/R)^2=(1/3)^2=1/9  ...

  ❓ Question Two cylindrical rods A and B made of different materials are joined end-to-end in a straight line. The ratios are: L A L B = 1 2 , r A r B = 2 , K A K B = 1 2 . The free ends of rods A and B are held at 400 K and 200 K respectively. Find the interface temperature (temperature at the junction) when steady-state equilibrium is established. 🖼️ Question Image ✍️ Short Solution In steady state the heat current I I  (rate of heat flow) is the same through both rods. Thermal resistance of a rod of length L L , cross-sectional area A A , and thermal conductivity K K  is: R = L K A . For series rods, temperature drops are proportional to their resistances. If T i T_i ​ is the interface temperature then: 400 − T i R A    =    T i − 200 R B . So compute R A R_A  and R B R_B  up to a common factor using given ratios. Step 1 — Express areas via radii Area A = π r 2 A=\pi r^2 . Given r A / r B = 2 r_A/r_B=2  → A A / A B = ( r A / r B )...

  ❓ Question A parallel plate capacitor has a charge of 5 × 10 − 6  C 5 \times 10^{-6}\ \text{C} . A dielectric slab is inserted between the plates and almost fills the space between them. If the induced charge on one face of the slab is 4 × 10 − 6  C 4 \times 10^{-6}\ \text{C} , then find the dielectric constant ( K K ) of the slab. 🖼️ Question Image ✍️ Short Solution Step 1 — Recall the concept When a dielectric slab is inserted into a charged capacitor: The free charge Q Q  on the plates remains the same (if disconnected from battery). The dielectric becomes polarized , producing induced charge Q ′ Q'  on its surfaces. Relation between induced charge and free charge: Q ′ = Q ( 1 − 1 K ) where K K  = dielectric constant. Step 2 — Substitute given values Given: Q = 5 × 10 − 6  C , Q ′ = 4 × 10 − 6  C . So, 4 × 10 − 6 = 5 × 10 − 6 ( 1 − 1 K ) Step 3 — Simplify the equation Cancel 10 − 6 10^{-6}  from both s...