A homoleptic octahedral aqueous complex [ M ( H ₂ O )₆]²⁺ , where M is a transition metal from the 3d series, has a spin-only magnetic moment of 3.87 B.M. The sum of the atomic numbers of all possible metals M that satisfy this condition is ______.

Step-by-step solution

The spin-only magnetic moment is given by = n(n+2) B.M. , where n is the number of unpaired electrons. Given = 3.87 B.M. , we have: n(n+2) = 3.87 Squaring both sides gives n(n+2) 15 , which means n = 3 . The complex is [ M ( H ₂ O )₆]²⁺ , so the metal M is in the +2 oxidation state. H ₂ O acts as a weak field ligand for 3d series M ²⁺ ions, leading to high-spin octahedral complexes. We look for 3 d configurations with exactly 3 unpaired electrons in a high-spin octahedral field: The d ^3 configuration has 3 unpaire