Which of the following will have magnetic moment, about 4.9 B.M.? — Atomic Structure Chemistry Question

💡 Solution & Explanation

Magnetic moment $\mu = \sqrt{n(n+2)}$ B.M. For $\mu \approx 4.9$, $n=4$ unpaired electrons. (a) Cr$^+$ ($3d^5$) has 5. (b) Ti$^{4+}$ ($3d^0$) has 0. (c) Fe$^{2+}$ ($3d^6$) has 4. (d) Cu$^{2+}$ ($3d^9$) has 1. Fe$^{2+}$ has 4 unpaired electrons. Therefore, correct answer is C.