Animated Solution for Chemistry - Coordination Compounds: The crystal field stabilisation energy (CFSE) of $[Fe(H_{2}O{)}_6]C{l}_2$ and $K_2[NiC{l}_4]$, respectively, are

Welcome to a fascinating journey into the heart of coordination chemistry! Today, we are going to unravel the mysteries of Crystal Field Stabilization Energy (CFSE) by comparing two very different complexes: the octahedral $[Fe(H_{2}O{)}_6]C{l}_2$ and the tetrahedral $K_2[NiC{l}_4]$.

To master this concept, we must first understand the identity of our central metal ions. Let's break down their oxidation states. In the iron complex, water ($H_{2}O$) is a neutral ligand. Since there are two chloride ions ($C{l}^{-}$) outside the coordination sphere, the iron must carry a $+2$ charge to balance them out. Similarly, in the nickel complex, the two potassium ions ($K^{+}$) provide a $+2$ charge, and the four chloride ligands inside provide a $-4$ charge. To balance this to an overall $-2$ charge for the complex ion, nickel must also be in a $+2$ state.

Before we dive into the electron configurations, we need our mathematical tools. The CFSE formulas are the keys to solving this problem.

For an octahedral field, the $d$-orbitals split into a lower energy $t_{2g}$ set and a higher energy $e_g$ set. The formula is: Here, $x$ is the number of electrons in the $t_{2g}$ orbitals, and $y$ is the number of electrons in the $e_g$ orbitals.

For a tetrahedral field, the splitting is inverted! The $e$ orbitals are lower in energy, and the $t_2$ orbitals are higher. The formula becomes: Here, $x$ is the number of electrons in the $e$ orbitals, and $y$ is the number of electrons in the $t_2$ orbitals.

Let's focus on $[Fe(H_{2}O{)}_6{]}^{2+}$. With six ligands, it is undeniably octahedral. The crucial detail here is the nature of the water ligand. Water is a weak field ligand, which means it produces a small crystal field splitting (${\Delta }_o$). Because the splitting is small, it is energetically easier for electrons to jump to the higher $e_g$ orbitals than to pair up in the lower $t_{2g}$ orbitals. This results in a high spin complex.

Iron(II) has a $3d^6$ electron configuration. Following Hund's rule for a high spin state, we place three electrons in the $t_{2g}$ level, two in the $e_g$ level, and the sixth electron pairs up in the $t_{2g}$ level. This gives us the configuration $t_{2g}^4{e}_g^2$.

Now, let's execute the calculation:

Next, we turn our attention to $[NiC{l}_4{]}^{2-}$. With four ligands, it is tetrahedral. Chloride ($C{l}^{-}$) is also a weak field ligand. Nickel(II) has a $3d^8$ electron configuration.

In a tetrahedral field, the $e$ orbitals are lower. We fill them up: two electrons go into the $e$ level, three into the $t_2$ level, and then the remaining three electrons pair up. We end up with four electrons in the $e$ level and four in the $t_2$ level, giving the configuration $e^4{t}_2^4$.

Let's calculate the CFSE for this tetrahedral complex:

We have successfully navigated the crystal field theory for both complexes. The CFSE for the iron complex is $-0.4{\Delta }_o$, and for the nickel complex, it is $-0.8{\Delta }_t$. Matching these results with our options, we can confidently select option (b) as the correct answer.

Always remember, the geometry of the complex and the strength of the ligand dictate the electron configuration, which in turn determines the stabilization energy. Keep practicing, and these patterns will become second nature!