How many unpaired electron(s) are present in [NiCl4]2– complex?

To determine the number of unpaired electrons in the [NiCl4]2– complex, we need to follow these steps:

1. Identify the oxidation state of the metal: In this case, Nickel (Ni) is in the +2 oxidation state because the overall charge of the complex is -2 and each Chlorine (Cl) atom contributes -1.

2. Write the electron configuration of the metal ion: The electron configuration of Ni is [Ar] 3d8 4s2. When it loses two electrons to form Ni2+, the electron configuration becomes [Ar] 3d8.

3. Determine the type of complex: [NiCl4]2– is a tetrahedral complex, which means it is a high-spin or spin-free complex. In high-spin complexes, electrons prefer to occupy different orbitals before pairing up in the same orbital.

4. Determine the number of unpaired electrons: In a high-spin complex, the five 3d orbitals are filled in this way: one electron each in the first four orbitals, and the remaining four electrons are paired up in these four orbitals. So, there are no unpaired electrons in [NiCl4]2–.

So, the [NiCl4]2– complex has no unpaired electrons.