To calculate the magnetic moment of the Ni CN 4 2- complex ion, we first need to determine the oxidation state of the nickel Ni atom and the number of unpaired electrons in its d-orbitals.The complex ion is Ni CN 4 2-. The cyanide ion CN- has a charge of -1, so the total charge contributed by the four cyanide ions is -4. To balance the overall charge of the complex ion -2 , the nickel atom must have an oxidation state of +2.Now, let's find the electron configuration of Ni2+. The atomic number of nickel is 28, and its ground state electron configuration is [Ar] 3d8 4s2. When nickel loses two electrons to form Ni2+, the electron configuration becomes [Ar] 3d8.In a tetrahedral complex, the d-orbitals split into two groups: a lower-energy group dxy, dxz, dyz and a higher-energy group dz2, dx2-y2 . However, the energy difference between these groups is relatively small in a tetrahedral complex compared to an octahedral complex. Therefore, the electrons will fill the orbitals according to Hund's rule, which states that electrons will occupy orbitals singly before pairing up.The electron configuration of Ni2+ in the Ni CN 4 2- complex is as follows:Lower-energy group t2 : dxy , dxz , dyz Higher-energy group e : dz2 , dx2-y2 There are no unpaired electrons in the Ni CN 4 2- complex ion.The magnetic moment can be calculated using the formula: = n n+2 ,where n is the number of unpaired electrons.Since there are no unpaired electrons n = 0 , the magnetic moment of the Ni CN 4 2- complex ion is: = 0 0+2 = 0The magnetic moment of the Ni CN 4 2- complex ion is 0, indicating that the complex is diamagnetic.