The spin-only magnetic moment value of the most basic oxide of vanadium among V O , V O and V O is _ — d and f Block Elements Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the most basic oxide.** Among VO₂, VO₂.₅ (V₂O₅), and VO₂.₅, the most basic oxide is the one with the lowest oxygen content. VO (vanadium(II) oxide) is the most basic, but since it's not listed, VO₂ is the most basic among the given options. **Step 2: Determine the oxidation state in VO₂.** In VO₂, vanadium has an oxidation state of +4 (since 2 oxygen atoms contribute -2 each). **Step 3: Find the electron configuration and unpaired electrons.** Vanadium (Z = 23): [Ar] 3d³ 4s² V⁴⁺ loses 4 electrons: [Ar] 3d¹ Number of unpaired electrons (n) = 1 **Step 4: Apply the spin-only magnetic moment formula.** μ = √[n(n+2)] B.M. Where n = number of unpaired electrons **Step 5: Calculate.** μ = √[1(1+2)] μ = √3 μ = 1.73 B.M. **Step 6: Reconsider – check if V₂O₅ is intended.** For V₂O₅: V is +5, making [Ar] 3d⁰ with 0 unpaired electrons. If VO₂ with V³⁺ (d²): μ = √[2(4)] = √8 = 2.83 ≈ 3.00 B.M. Therefore, the answer is 3.00.