Transition and Inner Transition Elements

+7

Step 1: To find the oxidation state of Mn in KMnO₄, apply the rule that the sum of all oxidation states equals zero for a neutral compound. Step 2: Assign known oxidation states: K = +1, each O = -2. Step 3: Let oxidation state of Mn = x. Then: (+1) + x + 4(-2) = 0. Step 4: 1 + x - 8 = 0, so x = +7. Step 5: This is the highest oxidation state of Mn. It is because Mn can lose all 7 valence electrons (2 from 4s and 5 from 3d). This high oxidation state makes KMnO₄ a strong oxidizing agent. The +4 state corresponds to MnO₂, +6 corresponds to MnO₄²⁻ (manganate ion).

5.91 BM

Step 1: Find the electronic configuration of Fe³⁺. Fe has configuration [Ar] 3d⁶ 4s². Step 2: Fe³⁺ loses 3 electrons (2 from 4s and 1 from 3d): Fe³⁺ = [Ar] 3d⁵. Step 3: In the 3d⁵ configuration, all 5 d-electrons are unpaired (half-filled, one in each orbital by Hund's rule). So n = 5. Step 4: Apply formula: μ = √n(n+2) = √5(5+2) = √35 = 5.91 BM. Step 5: Options A (1.73 BM) corresponds to n=1 (d¹), B (3.87 BM) to n=3 (d³), and D (4.89 BM) to n=4 (d⁴). Fe³⁺ has d⁵ configuration (same as Mn²⁺), giving the highest magnetic moment in the 3d series.

Cr³⁺

Step 1: In K₂Cr₂O₇, chromium is in the +6 oxidation state. When it acts as an oxidizing agent, it gets reduced. Step 2: The half-reaction in acidic medium is: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Step 3: Chromium is reduced from +6 to +3 (gains 3 electrons per Cr atom, total 6 electrons for 2 Cr atoms). Step 4: The complete reaction is: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. Step 5: CrO₄²⁻ is chromate (Cr is +6, not a reduction product). Cr²⁺ would require reduction to +2, which needs even stronger reducing conditions. CrO₃ is an oxidized form, not a product of reduction.

Cold dilute alkaline KMnO₄; detects unsaturation (C=C bonds)

Step 1: Baeyer's reagent is specifically cold dilute alkaline potassium permanganate (KMnO₄) solution. Step 2: When an alkene (compound with C=C double bond) is treated with Baeyer's reagent, the purple/violet color of KMnO₄ is decolorized. Step 3: The reaction converts alkenes to diols (vicinal diols): C=C + KMnO₄ (alk.) → CHOH-CHOH (diol). Step 4: The decolorization of the purple KMnO₄ solution confirms the presence of unsaturation (C=C or C≡C). Step 5: Hot concentrated KMnO₄ cleaves the double bond entirely (different reaction), so 'hot' and 'cold' conditions give different products. K₂Cr₂O