To calculate the magnetic moment of the compound Ni CO 4, we first need to determine the electronic configuration of the central metal ion, which is nickel in this case.Nickel Ni has an atomic number of 28, and its ground state electronic configuration is [Ar] 3d^8 4s^2. In Ni CO 4, nickel is in the +0 oxidation state, as CO is a neutral ligand. Therefore, the electronic configuration of Ni in Ni CO 4 remains the same as its ground state, which is [Ar] 3d^8 4s^2.Now, we need to consider the crystal field theory. In Ni CO 4, the coordination geometry around the Ni atom is tetrahedral. In a tetrahedral crystal field, the d-orbitals split into two sets: the lower energy set dxy, dxz, dyz and the higher energy set dz^2, dx^2-y^2 . However, CO is a strong field ligand and causes the pairing of electrons in the d-orbitals.In the case of Ni CO 4, the 3d^8 electronic configuration in a tetrahedral crystal field would result in the following distribution of electrons:Lower energy set dxy, dxz, dyz : Higher energy set dz^2, dx^2-y^2 : As we can see, all the d-orbitals are filled with paired electrons, and there are no unpaired electrons in the 4s orbital either.The magnetic moment can be calculated using the formula: = n n+2 Bwhere n is the number of unpaired electrons, and B is the Bohr magneton.Since there are no unpaired electrons in Ni CO 4 n = 0 , the magnetic moment will be: = 0 0+2 B = 0 BA magnetic moment of 0 B indicates that the compound is diamagnetic, as there are no unpaired electrons to contribute to a net magnetic moment. Therefore, Ni CO 4 is a diamagnetic compound.