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

💡 Solution & Explanation

I notice the question is incomplete—option C isn't provided. However, I can show you **how to solve this type of problem**: ## Step-by-Step Approach **Given:** Magnetic moment ($\mu$) = 2.83 BM **Formula:** $$\mu = \sqrt{n(n+2)} \text{ BM}$$ where $n$ = number of unpaired electrons **Step 1:** Solve for unpaired electrons $$2.83 = \sqrt{n(n+2)}$$ $$8 = n(n+2)$$ $$n^2 + 2n - 8 = 0$$ $$(n+4)(n-2) = 0$$ $$n = 2 \text{ unpaired electrons}$$ **Step 2:** Identify the complex with 2 unpaired electrons - Look for $d^2$, $d^8$ (high spin), or similar configurations - Common examples: $[Fe(CN)_6]^{4-}$ (0 unpaired), $[Fe(H_2O)_6]^{2+}$ (4 unpaired), $[Co(NH_3)_6]^{3+}$ (0 unpaired), $[CoF_6]^{3-}$ (4 unpaired), or $[Ni(en)_3]^{2+}$ (2 unpaired) **Answer:** The correct option is likely a **Ni(II) complex** like $[Ni(en)_3]^{2+}$ or a similar $d^8$ high-spin complex, as these typically have 2 unpaired electrons. *Please provide option C for a definitive answer.*