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

💡 Solution & Explanation

The spin-only magnetic moment is given by the formula μ = √(n * (n + 2)) B.M., where n is the number of unpaired electrons. A magnetic moment of ~4.9 B.M. corresponds to exactly n = 4 unpaired electrons (since √(4 * 6) = √(24) = 4.899 B.M.). Let's determine n for each ion: (a) Cr^+ = [Ar] 3d^5 => n = 5 unpaired electrons. (b) Ti^4+ = [Ar] 3d^0 => n = 0 unpaired electrons. (c) Fe^2+ = [Ar] 3d^6 => n = 4 unpaired electrons. (d) Cu^2+ = [Ar] 3d^9 => n = 1 unpaired electron. Therefore, the Fe^2+ ion has a magnetic moment of ≈ 4.9 B.M.