The d-and f-Block Elements — Important Questions

Due to d-d transitions where electrons absorb light from the visible region

Step 1: In transition metal ions, d orbitals are split into two energy levels when surrounded by ligands like water (this is explained by Crystal Field Theory). Step 2: An electron from a lower energy d orbital absorbs energy and gets excited to a higher energy d orbital. This is called a d-d transition. Step 3: The energy required for this transition corresponds to the frequency of visible light (wavelength ~400–700 nm). Step 4: The colour we observe is the complementary colour of the light absorbed. For example, Cu²⁺ absorbs red light and appears blue. Step 5: Ions like Sc³⁺ (3d⁰) and Zn²⁺ (

Sodium chromate (Na₂CrO₄)

Step 1: The first step in preparing K₂Cr₂O₇ is the fusion of chromite ore with sodium carbonate in the presence of air (oxygen). Step 2: The balanced reaction is: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂. Step 3: The product is yellow sodium chromate (Na₂CrO₄), not sodium dichromate. Sodium dichromate is formed in the next step when the solution is acidified with H₂SO₄. Step 4: The overall sequence is: Chromite ore → Na₂CrO₄ (by fusion with Na₂CO₃) → Na₂Cr₂O₇ (by acidification) → K₂Cr₂O₇ (by reaction with KCl). Step 5: Potassium chromate and chromic oxide are not the direct products

+2 (Mn forms Mn²⁺)

Step 1: KMnO₄ (potassium permanganate) is a strong oxidising agent. The behaviour of MnO₄⁻ depends on the pH (acidity) of the solution. Step 2: In acidic solution, the half-reaction is: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (E° = +1.52 V). Step 3: Mn goes from +7 oxidation state (in MnO₄⁻) to +2 oxidation state (in Mn²⁺), gaining 5 electrons. Step 4: In neutral or faintly alkaline solution, Mn is reduced to MnO₂ (+4 state). In strongly alkaline solution, Mn is reduced to MnO₄²⁻ (+6 state). Step 5: The change from +7 to +2 is specific to acidic conditions. The solution changes from purple/pink (MnO₄⁻

Lanthanoid contraction

Step 1: As we move along the lanthanoid series (Ce to Lu), the 4f electrons are being added. These 4f electrons shield the outer electrons from the nuclear charge very poorly compared to d electrons. Step 2: This poor shielding causes the nuclear charge experienced by outer electrons to increase gradually, pulling the electron cloud inward. The result is a regular decrease in atomic and ionic radii across the lanthanoid series. Step 3: This cumulative decrease in size across 14 lanthanoid elements is called Lanthanoid Contraction. Step 4: Due to lanthanoid contraction, the 5d elements that com