Question

Which is the CFSE of a free $Co(II)$ ion on forming the tetrahedral chloro complex ${[CoC{l_4}]^{2 - }}$(in the units of \[\Delta o\])?
A.$ - 0.6$
B.$ - 1.2$
C.$ - 1.8$
D.$ - 2.4$

Answer 1

Hint: To solve or get the CFSE energy of any complex compounds we should be able to get the configuration of the compound. Also we should know the geometry of the compound whether it is octahedral or tetrahedral or any other type of geometry.

Complete step by step answer:
We are given with tetrahedral chloro complex which is ${[CoC{l_4}]^{2 - }}$. Firstly we will check its co electronic configuration.
Since we can see that cobalt has an oxidation state of $ + 2$ adding to it there are four chlorine molecules. Hence overall charge of ${[CoC{l_4}]^{2 - }}$ is $ - 2$. So with an oxidation state of$ + 2$, Cobalt is a ${d^7}$ metal and its electronic configuration is $[Ar]3{d^7}$.
In the below image we can see the relation of energy in tetrahedral geometry:

Hence it is tetrahedral geometry so the CFSE for tetrahedral geometry is
$4( - 0.6) + 3(0.4)$= $ - 1.2{\Delta _0}$
Hence option B is correct which is $ - 1.2{\Delta _0}$

Additional Information
CFSE stands for Crystal Field Stabilization Energy. It is defined as the energy of electronic configuration in the ligand field subtracted from the energy of electronic configuration in the isotropic field.
\[CFSE = \Delta E = {E_{ligand{{ }}field}} - {E_{isotropic{{ }}field}}\]
CFSE depends on multiple factors such as:
On geometry (which changes the d-orbital splitting patterns)
Number of d-electrons
The spin pairing energy
Ligand character
It is the additional stabilization which is gained by most of the splitting of orbitals. It is more energetic and favors forming six bonds rather than four.
The main point is that the CFSE of an octahedral is usually greater than the CFSE of tetrahedral.
Crystal Field Stabilization Energy is usually less than or equal to zero. If this is equal to zero then the complex is definitely unstable. The magnitude of Crystal Field Stabilization Energy basically depends on the number and nature of ligands. Mostly the s and p block are not used in CFSE theory on d orbital or block used in it.

Note:
Mostly the CFSE depends on geometry of compounds so first always check the geometry of the given compound. Also check electronic configuration of the ion or molecule for which we have to check the CFSE.