Animated Solution for Chemistry - Coordination Compounds: Among $\left[\text{Ni}(\text{CO}{)}_4\right]$, ${\left[{\text{NiCl}}_4\right]}^{2-}$, $\left[\text{Co}({\text{NH}}_3{)}_4{\text{Cl}}_2\right]\text{Cl}$, ${\text{Na}}_3\left[{\text{CoF}}_6\right]$, ${\text{Na}}_2{\text{O}}_2$ and ${\text{CsO}}_2$, the total number of paramagnetic compounds is -

Imagine you have a powerful magnet and you bring it close to a vial containing a chemical compound. Some compounds will be weakly attracted to the magnet, while others will be slightly repelled. This invisible, almost magical interaction is governed entirely by the tiny electrons dancing inside the atoms.

If a molecule has even a single unpaired electron, it acts like a tiny bar magnet and gets attracted to external magnetic fields. We call this property paramagnetism. If all electrons are perfectly paired up, their magnetic fields cancel out, resulting in diamagnetism. In this problem, we are given six different chemical species and tasked with unmasking the paramagnetic ones. Let's dive into the quantum world!

CFT and MOT

To solve this, we need two powerful theoretical tools. For the transition metal coordination complexes, we will use Crystal Field Theory (CFT). CFT tells us that ligands (the molecules surrounding the central metal) can either be "strong field" (forcing electrons to pair up against their will) or "weak field" (allowing electrons to spread out and remain unpaired).

For the oxygen-based compounds, we will rely on Molecular Orbital Theory (MOT). MOT allows us to construct the energy levels of the entire molecule and fill them with valence electrons to see if any are left unpaired at the top.

Let's analyze the four coordination complexes one by one:

1. Nickel Tetracarbonyl, $\left[\text{Ni}(\text{CO}{)}_4\right]$: Here, Nickel is in a $0$ oxidation state, giving it a valence configuration of $3d^{8}4s^2$. Carbon monoxide ($\text{CO}$) is a notoriously strong field ligand. It exerts such a strong push that it forces the two $4s$ electrons to migrate into the $3d$ orbitals, resulting in a $3d^{10}$ configuration. With all 10 d-electrons perfectly paired, this complex is strictly diamagnetic.

2. Tetrachloridonickelate, ${\left[{\text{NiCl}}_4\right]}^{2-}$: In this ion, Nickel is in a $+2$ oxidation state, leaving it with a $3d^8$ configuration. Chloride (${\text{Cl}}^{-}$) is a weak field ligand. It doesn't have the strength to force the electrons to pair up against Hund's rule. As a result, two electrons remain unpaired in the d-orbitals. This makes the complex paramagnetic.

3. Cobalt Ammine Chloride, $\left[\text{Co}({\text{NH}}_3{)}_4{\text{Cl}}_2\right]\text{Cl}$: Cobalt here is in a $+3$ oxidation state, which means it has a $3d^6$ configuration. Ammonia (${\text{NH}}_3$) acts as a strong field ligand for ${\text{Co}}^{3+}$. The strong field splits the d-orbitals significantly, forcing all six electrons to pair up in the lower energy $t_{2g}$ level ($t_{2g}^6{e}_g^0$). Zero unpaired electrons mean it is diamagnetic.

4. Sodium Hexafluoridocobaltate, ${\text{Na}}_3\left[{\text{CoF}}_6\right]$: Again, Cobalt is in a $+3$ state ($3d^6$). However, the ligand this time is Fluoride (${\text{F}}^{-}$), which is a weak field ligand. The crystal field splitting is small, so the electrons prefer to occupy the higher energy $e_g$ orbitals rather than pair up. This results in a high-spin $t_{2g}^4{e}_g^2$ configuration, leaving four unpaired electrons. This complex is highly paramagnetic.

Peroxide vs Superoxide

Now, let's look at the two sodium and cesium compounds. These are not simple oxides; they contain diatomic oxygen anions.

5. Sodium Peroxide, ${\text{Na}}_2{\text{O}}_2$: This compound contains the peroxide ion, ${\text{O}}_2^{2-}$. Each oxygen atom brings 6 valence electrons, and the $-2$ charge adds 2 more, totaling 14 valence electrons. When we fill the molecular orbitals, the last electrons go into the anti-bonding ${\pi }^{\ast }$ orbitals: $\dots {\pi }_{2p}^4{\pi }_{2p}^{\ast 4}$. All orbitals are completely filled and paired. Thus, it is diamagnetic.

6. Cesium Superoxide, ${\text{CsO}}_2$: This compound contains the superoxide ion, ${\text{O}}_2^{-}$. With a $-1$ charge, it has 13 valence electrons. Following the same MOT filling order, we get: $\dots {\pi }_{2p}^4{\pi }_{2p}^{\ast 3}$. Notice that odd number? There is exactly one unpaired electron sitting alone in the ${\pi }^{\ast }$ orbital. This single electron makes the superoxide ion paramagnetic.

We have successfully analyzed all six species. The ones harboring unpaired electrons are: 1. ${\left[{\text{NiCl}}_4\right]}^{2-}$ 2. ${\text{Na}}_3\left[{\text{CoF}}_6\right]$ 3. ${\text{CsO}}_2$

Counting them up, we find exactly 3 paramagnetic compounds. The correct option is (B).