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

💡 Solution & Explanation

I need to work backwards from the magnetic moment to identify the complex ion. **Step 1: Convert magnetic moment to number of unpaired electrons** The magnetic moment formula is: $$\mu = \sqrt{n(n+2)} \text{ BM}$$ where $n$ = number of unpaired electrons. Given: $\mu = 2.83$ BM $$2.83 = \sqrt{n(n+2)}$$ Squaring both sides: $$8 = n(n+2)$$ $$n^2 + 2n - 8 = 0$$ $$(n+4)(n-2) = 0$$ Therefore: $n = 2$ unpaired electrons **Step 2: Identify complexes with 2 unpaired electrons** Common complexes with 2 unpaired electrons: - $[Fe(H_2O)_6]^{2+}$ (high spin $d^6$: $t_{2g}^4 e_g^2$) — 4 unpaired ✗ - $[Ni(H_2O)_6]^{2+}$ (high spin $d^8$: $t_{2g}^6 e_g^2$) — 2 unpaired ✓ - $[Cu(H_2O)_6]^{2+}$ (high spin $d^9$: $t_{2g}^6 e_g^3$) — 1 unpaired ✗ - $[Fe(CN)_6]^{4-}$ (low spin $d^6$: $t_{2g}^6 e_g^0$) — 0 unpaired ✗ **Step 3: Verification** For $[Ni(H_2O)_6]^{2+}$ with $n = 2$: $$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \text{ BM}$$ ✓ **Answer:** The complex is most likely **$[Ni(