# Why does ["Co"("NN"_3)_6]^(3+) form an inner orbital complex but ["CoF"_6]^(3-) form an outer orbital complex?

##### 1 Answer
Oct 10, 2016

$\textsf{{\left[C o {\left(N {H}_{3}\right)}_{6}\right]}^{3 +}}$ is an inner orbital complex because it adopts a low spin electronic configuration.

I am assuming you mean $\textsf{N {H}_{3}}$ as the ligand in question in the 1st example.

#### Explanation:

One characteristic of the transition elements is their ability to form complex ions. These consist of a central metal ion surrounded by electron - donating species called ligands.

Cobalt(III) ions have the electron configuration:

Fig 1 (a)

The empty 4s, 4p and 4d orbitals are quite close in energy to the 3d orbitals.

This means that ligands with available lone pairs such as ammonia molecules are able to donate electrons into these empty orbitals and form co - ordinate bonds.

Quite a common stable arrangement is with 6 ligands.

An example is hydrated sf(Fe^(3+):

The d orbitals look like this:

Fig 1 (b)

Because of the octahedral symmetry of the complex the $\textsf{{d}_{{z}^{2}}}$ and $\textsf{{d}_{{x}^{2} - {y}^{2}}}$ orbitals point directly at negative lone pairs on the ligand and are destabilised.

The other 3d orbitals have lobes which project between the ligands and are relatively more stable. This is shown here:

These orbitals are given the symmetry terms $\textsf{{e}_{g}}$ and $\textsf{{t}_{2 g}}$ and the energy gap is denoted by the symbol $\textsf{\Delta}$.

The important point to note here is that the value of $\textsf{\Delta}$ varies with the nature of the ligand.

The spectrochemical series lists the ligands in order of $\textsf{\Delta}$:

Some are listed here:

I- < Br- < S2- < SCN- < Cl- < NO3- < F- < OH- < C2O42- < H2O < NCS- < CH3CN < NH3 < en < bipy < phen < NO2- < PPh3 < CN- < CO

This tells us that $\textsf{N {H}_{3}}$ has a larger $\textsf{\Delta}$ value than $\textsf{{F}^{-}}$ which has implications for the way the electrons occupy the 3d orbitals.

For the ammonia complex the value of $\textsf{\Delta}$ is relatively large and it is energetically more favourable for the 6 electrons to occupy the $\textsf{{t}_{2 g}}$ sub - levels with spins paired:

This is termed a "low - spin" complex.

You can see from Fig 3 that the orbitals shown in red are able to each accept a pair of electrons from each $\textsf{N {H}_{3}}$ ligand and form $\textsf{{\left[C o {\left(N {H}_{3}\right)}_{6}\right]}^{3 +}}$.

The 3d, 4s and 4p orbitals effectively reorganise themselves to form 6 equivalent orbitals which are described, therefore, as $\textsf{{d}^{2} s {p}^{3}}$ hybrid orbitals

Because inner 3d orbitals are used this can be referred to as an inner orbital complex .

In the case of the $\textsf{{\left[C o {F}_{6}\right]}^{3 -}}$ complex the relatively small value of $\textsf{\Delta}$ means that coulombic repulsions between the electrons means it is more energetically favourable for them to occupy the orbitals singly in a "high spin" arrangement:

Because the $\textsf{{e}_{g}}$ orbitals are now partially occupied the ligands can donate their electrons into the empty 4s, 4p and 4d orbitals which can reorganise themselves into $\textsf{s {p}^{3} {d}^{2}}$ hybrid orbitals.

Because the outer 4d orbitals are now being used this can be described as an outer orbital complex.