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Step 1: Determine the oxidation state of Nickel in the complex. In the complex $[NiX_4]^{2-}$, the ligand $X^-$ is uninegative. Let the oxidation state of Nickel be $x$. The overall charge of the complex is $-2$. Therefore, we can set up the equation: $x + 4(-1) = -2$ Solving this gives: $x - 4 = -2$ $x = +2$. Thus, Nickel is in the +2 oxidation state.Step 2: Write the electronic configuration of Nickel. The atomic number of Nickel (Ni) is 28. The electronic configuration of neutral Nickel is: $[Ar] 3d^8 4s^2$. When Nickel is in the +2 oxidation state, it loses two electrons, typically from the 4s orbital first. Therefore, the electronic configuration of $Ni^{2+}$ is: $[Ar] 3d^8$.Step 3: Determine the number of unpaired electrons. In the $3d$ subshell, the distribution of electrons for $Ni^{2+}$ ($3d^8$) is as follows: The $3d$ subshell can hold a maximum of 10 electrons. In the case of $Ni^{2+}$, there are 8 electrons in the $3d$ subshell. The distribution of these electrons is: $↑↓ ↑↓ ↑ ↑ ↑$. There are 2 unpaired electrons in the $3d$ subshell.Step 4: Determine the geometry of the complex. The coordination number of Nickel in this complex is 4 (due to 4 ligands). For a coordination number of 4, the possible geometries are tetrahedral and square planar. Since Nickel in the +2 oxidation state prefers to adopt a tetrahedral geometry when coordinated to four monodentate ligands, the geometry of the complex $[NiX_4]^{2-}$ is tetrahedral.Conclusion: Therefore, the number of unpaired electrons in the Nickel and the geometry of the complex ion $[NiX_4]^{2-}$ are, respectively, two and tetrahedral.
