An unknown 3d transition metal M forms an octahedral complex ion [M(H₂O)₆]²⁺ having a spin-only magnetic moment of 4.90 BM . The same metal ion forms another octahedral complex ion [M(CN)₆]⁴⁻ which has a spin-only magnetic moment of 0 BM . The atomic number of the metal M is ________.

Step-by-step solution

The spin-only magnetic moment of [M(H₂O)₆]²⁺ is 4.90 BM . Using the formula = n(n+2) , we find the number of unpaired electrons n = 4 . Since H₂O is a weak field ligand, the complex is high-spin. An octahedral high-spin complex with 4 unpaired electrons corresponds to either a d^4 ( t_ 2g ^3 e_g^1 ) or a d^6 ( t_ 2g ^4 e_g^2 ) configuration for the M²⁺ ion. The spin-only magnetic moment of [M(CN)₆]⁴⁻ is 0 BM , meaning n = 0 . Since CN^- is a strong field ligand, the complex is low-spin. If M²⁺ were d^4 , the low-sp