Animated Solution for Chemistry - Coordination Compounds: The magnetic moment (spin only) of $[NiC{l}_4{]}^{2-}$ is

Setting the Stage

To unravel the magnetic properties of the $[NiC{l}_4{]}^{2-}$ complex, our first mission is to determine the oxidation state of the central metal atom, Nickel. Let's assign the oxidation state of Nickel as $x$.

We know that each chloride ($C{l}^{-}$) ligand carries a charge of $-1$. Since there are four chloride ligands and the overall charge of the complex ion is $-2$, we can set up a simple algebraic equation:

Solving for $x$, we find that $x=+2$. This tells us that Nickel exists as a $N{i}^{2+}$ ion in this coordination sphere.

The Nickel Ion

Nickel has an atomic number of 28. In its neutral ground state, its electronic configuration is $[Ar]3d^{8}4s^2$. When it loses two electrons to form the $N{i}^{2+}$ ion, it sheds the outermost $4s$ electrons first.

This leaves us with the configuration:

If we visualize the $3d$ subshell, which consists of five degenerate orbitals, we distribute the eight electrons according to Hund's Rule. This results in three fully paired orbitals and two unpaired electrons.

Weak vs Strong

Now, we must evaluate the nature of the incoming ligands. According to the spectrochemical series, the chloride ion ($C{l}^{-}$) is a weak field ligand.

Because it is a weak field ligand, the crystal field splitting energy (${\Delta }_o$) it produces is relatively small—less than the pairing energy ($P$). Consequently, it does not possess the energetic leverage required to force the two unpaired $3d$ electrons to pair up against their natural electrostatic repulsion. The electrons remain happily unpaired.

Making Room for Ligands

To accommodate the four electron pairs donated by the four chloride ligands, the $N{i}^{2+}$ ion needs four empty orbitals. Since the inner $3d$ orbitals are occupied, it utilizes its outer empty orbitals: one $4s$ and three $4p$ orbitals.

These orbitals mix to form four equivalent $s{p}^3$ hybrid orbitals. The geometry associated with $s{p}^3$ hybridization is tetrahedral. The chloride ions donate their lone pairs into these empty hybrid orbitals to form coordinate covalent bonds.

Spin-Only Magnetic Moment

The magnetic character of a complex is directly tied to the number of unpaired electrons ($n$). In our $[NiC{l}_4{]}^{2-}$ complex, we established that $n=2$.

We calculate the spin-only magnetic moment ($\mu$) using the formula:

Substituting $n=2$ into the equation:

Evaluating the square root of 8 gives us approximately $2.82\text{ BM}$. Because the complex has a non-zero magnetic moment, it is classified as paramagnetic. Thus, the correct option is (c).