Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

 Question: The equilibrium constant for the decomposition of H 2 O ( g )    ⇌    H 2 ( g ) + 1 2 O 2 ( g ) is K = 8.0 × 10⁻³ at 2300 K . If the total equilibrium pressure = 1 bar, the degree of dissociation (α) of water is ____ × 10⁻² (nearest integer).

  Question: 20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is ______ M. (Nearest Integer value).

 Question: The number of species from the following that are involved in sp³d² hybridization is: [Co(NH₃)₆]³⁺, SF₆, [CrF₆]³⁻, [CoF₆]³⁻, [Mn(CN)₆]³⁻ and [MnCl₆]³⁻ 📷 Question Image: Short Solution (Text): Key idea (VBT quick check) Octahedral complexes can be: d²sp³ (inner-orbital) when strong-field ligands pair electrons in (n−1)d (low spin). sp³d² (outer-orbital) when weak-field ligands don’t pair (high spin), so central atom uses the outer nd orbitals. SF₆ (molecular, not a complex): classic sp³d² around S. Evaluate each species [Co(NH₃)₆]³⁺ Co³⁺: d⁶; NH₃ ~ intermediate/weak-to-borderline but for Co³⁺ it gives low spin (pairing favored) → inner (d²sp³). ❌ SF₆ S uses 3s, three 3p, two 3d → sp³d² (octahedral). ✅ [CrF₆]³⁻ Cr³⁺: d³; F⁻ weak field, but d³ doesn’t need pairing; inner (n−1)d are available → d²sp³ (inner). ❌ [CoF₆]³⁻ Co³⁺: d⁶; F⁻ weak field → high spin , no pairing in (n−1)d → uses outer nd → sp³d² . ✅ [Mn(CN)₆]³⁻ ...

  Question: The correct decreasing order of spin-only magnetic moment values (BM) of Cu⁺, Cu²⁺, Cr²⁺ and Cr³⁺ ions is : Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺ Cu²⁺ > Cu⁺ > Cr²⁺ > Cr³⁺ Cu⁺ > Cu²⁺ > Cr³⁺ > Cr²⁺ Cr³⁺ > Cr²⁺ > Cu⁺ > Cu²⁺

  Question: Correct statements for an element with atomic number 9 are: A. There can be 5 electrons for which mₛ = +1/2 and 4 electrons for which mₛ = −1/2. B. There is only one electron in the p₂ orbital. C. The last electron goes to the orbital with n = 2 and l = 1. D. The sum of angular nodes of all the atomic orbitals is 1. 📷 Question Image:

  Question: Given below are two statements: Statement I: H₂Se is more acidic than H₂Te Statement II: H₂Se has higher bond enthalpy for dissociation than H₂Te 📷 Question Image:

In a first order decomposition reaction, the time taken for the decomposition of reactant to one-fourth and one-eighth of its initial concentration are t₁ and t₂(s) , respectively. The ratio t₁/t₂ will be: 📷 Question Image:

  Atomic Number of the Element with Lowest First Ionisation Enthalpy | JEE Chemistry 📌 Question The atomic number of the element from the following with the lowest first ionisation enthalpy is: 📚 Concept Used – First Ionisation Enthalpy Definition: First ionisation enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Periodic Trend: Across a period → IE increases (nuclear charge ↑, atomic size ↓) Down a group → IE decreases (atomic size ↑, shielding effect ↑) Lowest IE Element: The element at the bottom-left of the periodic table has the lowest IE, because it has large size and weak nuclear attraction for the outermost electron. 🖋 Solution The element with the lowest first ionisation enthalpy is Francium (Fr) , but in practical chemistry, Cesium (Cs) is often considered because Francium is radioactive and rare . Francium’s atomic number = 87 Therefore, the answer ...

  Match the List-I with List-II: Coordination Complexes, Shapes, and Magnetic Moments Introduction Coordination chemistry is a fascinating branch of inorganic chemistry that deals with the structure, bonding, and properties of complexes. In competitive exams like JEE, NEET, and GATE , questions often test your ability to relate complex formulas with their shapes and magnetic moments .