An aqueous solution of a transition metal ion exhibits a spin-only magnetic moment of 5.92 BM . The ion could be :

Step-by-step solution

The spin-only magnetic moment is given by the formula = n(n+2) BM , where n is the number of unpaired electrons. Given = 5.92 BM , we have n(n+2) = 5.92 . Squaring both sides gives n(n+2) 35 , which yields n = 5 . The electronic configurations of the given ions are: Fe ²⁺ ( Z=26 ): [ Ar ] 3d^6 , which has 4 unpaired electrons. Ni ²⁺ ( Z=28 ): [ Ar ] 3d^8 , which has 2 unpaired electrons. Mn ²⁺ ( Z=25 ): [ Ar ] 3d^5 , which has 5 unpaired electrons. Ti ³⁺ ( Z=22 ): [ Ar ] 3d^1 , which has 1 unpaired electron. Since