The magnetic moment of a complex ion is 2.83 BM. The complex ion is — Coordination Compounds Chemistry Question

💡 Solution & Explanation

I need to find which complex ion has a magnetic moment of 2.83 BM. **Step 1: Use the magnetic moment formula** $$\mu = \sqrt{n(n+2)} \text{ BM}$$ where $n$ = number of unpaired electrons **Step 2: Solve for n** $$2.83 = \sqrt{n(n+2)}$$ $$8 = n(n+2)$$ $$n^2 + 2n - 8 = 0$$ $$(n+4)(n-2) = 0$$ $$n = 2 \text{ (taking positive value)}$$ **Step 3: Identify the complex with 2 unpaired electrons** Common complexes with $n = 2$: - $[Fe(CN)_6]^{4-}$ → $d^6$ low-spin, 0 unpaired electrons ✗ - $[CoF_6]^{3-}$ → $d^6$ high-spin, 4 unpaired electrons ✗ - $[Fe(H_2O)_6]^{2+}$ → $d^6$ high-spin, 4 unpaired electrons ✗ - $[Ni(H_2O)_6]^{2+}$ → $d^8$, 2 unpaired electrons ✓ - $[Cu(H_2O)_6]^{2+}$ → $d^9$, 1 unpaired electron ✗ **Conclusion:** The complex with 2.83 BM magnetic moment is **$[Ni(H_2O)_6]^{2+}$** (option C), which has $d^8$ configuration with 2 unpaired electrons in octahedral geometry.