A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of 3.86 BM. The atomic number of the metal is

Solution

Chemistry ma’am · 3 steps

Turn 3.86 BM into unpaired electrons first, then find which +2 ion has that many.

1. μ = √n(n+2) BM

2. Find n. Try small values of n: n = 3 gives √15 ≈ 3.87 BM, which matches 3.86. So the ion has 3 unpaired electrons.
3. Form the ion. For first row metals, the two electrons of the +2 ion leave from 4s, not 3d. So each M²⁺ ion is [Ar] 3d^(Z-20).

   Z	Ion	Configuration	Unpaired n	μ (BM)
   22	Ti²⁺	3d²	2	√8 ≈ 2.83
   23	V²⁺	3d³	3	√15 ≈ 3.87
   26	Fe²⁺	3d⁶	4	√24 ≈ 4.90
   25	Mn²⁺	3d⁵	5	√35 ≈ 5.92

4. So. Only V²⁺ (3d³) has 3 unpaired electrons, so the metal is vanadium with atomic number 23.

The idea to keep. Learn the spin-only ladder by heart: 1.73, 2.83, 3.87, 4.90, 5.92 BM for 1 to 5 unpaired electrons. For a +2 ion, always remove the 4s electrons first.

Answer from the NTA final answer key. Solution by PrepWiser.