Animated Solution for Chemistry - Coordination Compounds: Nickel ($Z=28$) combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex $[Ni{X}_4{]}^{2-}$. The number of unpaired electron (s) in the nickel and geometry of this complex ion are, respectively

Imagine you are a molecular architect tasked with building a coordination complex. You are given a central Nickel atom ($Z=28$) and four uninegative monodentate ligands, denoted as $X^{-}$. The blueprint explicitly states that the final structure must be paramagnetic.

Our first crucial step is to determine the oxidation state of the central Nickel atom. We know the overall charge of the complex $[Ni{X}_4{]}^{2-}$ is $-2$, and each of the four ligands carries a $-1$ charge. Setting up a simple algebraic equation, we get:

Solving for $x$, we find that the oxidation state of Nickel is $+2$. This means our central atom is an $N{i}^{2+}$ ion.

Now, let's look at the electronic configuration. A neutral Nickel atom has an atomic number of $28$, which gives it an outer configuration of $4s^{2}3d^8$. When it loses two electrons to form the $N{i}^{2+}$ ion, it loses them from the outermost $4s$ orbital first. Thus, the configuration becomes:

Let's visualize these eight electrons in the five $3d$ orbitals. Following Hund's rule of maximum multiplicity, we first place one electron in each of the five orbitals, and then we pair up the remaining three. This leaves us with exactly two unpaired electrons in the $3d$ subshell.

Here is where the problem gives us a massive hint: the complex is paramagnetic. Paramagnetism arises from the presence of unpaired electrons. If the ligand $X^{-}$ were a strong field ligand, it would force those two unpaired electrons to pair up against Hund's rule, making the complex diamagnetic.

Because the complex remains paramagnetic, we know for a fact that the electrons do not pair up. The two unpaired electrons stay exactly where they are.

To build the complex, the $N{i}^{2+}$ ion needs to accept four pairs of electrons from the four $X^{-}$ ligands. This requires four empty orbitals of equivalent energy.

Since the inner $3d$ orbitals are fully occupied or contain unpaired electrons that refuse to move, the metal ion must look outward. It utilizes the empty $4s$ orbital and the three empty $4p$ orbitals. By mixing these together, it creates four equivalent $s{p}^3$ hybrid orbitals.

In the world of molecular geometry, $s{p}^3$ hybridization perfectly corresponds to a tetrahedral shape. The four hybrid orbitals point towards the corners of a regular tetrahedron to minimize electron repulsion.

Therefore, our complex has two unpaired electrons and a tetrahedral geometry. It is a beautiful demonstration of how magnetic properties directly dictate the physical shape of a molecule!