An octahedral complex with the formula [M(CN)₆]³⁻ has a spin-only magnetic moment of 2.83 BM. The central transition metal M is:

Step-by-step solution

The spin-only magnetic moment is given by = n(n+2) BM, where n is the number of unpaired electrons. Given = 2.83 BM, we have n = 2 unpaired electrons. In the complex [M(CN)₆]³⁻ , the oxidation state of the metal M is +3 . Since CN^- is a strong field ligand, it causes pairing of electrons, forming a low-spin complex. Let us check the electronic configurations of the +3 ions of the given metals in a strong octahedral field: For Cr ³⁺ (3d^3) : t_ 2g ^3 , n = 3 For Mn ³⁺ (3d^4) : low-spin t_ 2g ^4 , n = 2 For Fe ³⁺ (3