[Ni(NH3)6]Cl2 paramagnetic but [Co(NH3)6]Cl3 is diamagnetic

Why is [Ni(NH₃)₆]Cl₂ paramagnetic but [Co(NH₃)₆]Cl₃ is  diamagnetic ?

In case of [Ni(NH₃)₆]Cl₂ complex, the oxidation state of nickel atom is +2 .The atomic number of nickel atom, 28 and that of Ni(II)ion is 26 .

Valence shell electronic configuration of Ni(II) ion is [ Ar ] d⁸ . Coordination number of central metal Ni(II)ion : 6 . Arrangement of [Ni(NH₃)₆]Cl₂ complex is Octahedral.

Now, the paramagnetic character of above said octahedral Ni(II) complex does not explained on the basis of crystal field theory.

Because,  the ligand NH₃ is a strong field ligand. So, under the influence of the octahedral crystal field, the five degenerate d-orbitals of Ni(II) ion are splitted into two sets of  energetically different orbitals.

These two sets of orbitals are energetically lower t_2g orbital and energetically higher e_g orbital.

Now, the ligand NH₃ is a strong field ligand and hence it is a low spin complex. Therefore, the electronic arrangement should be t2g⁶ eg² .

Here, 10Dqo > P (pairingenergy) and hence all the electrons are paired.

Consequently, octahedral Ni(II) complex with strong field should be diamagnetic.

But,  actually the [Ni(NH₃)₆]Cl₂ complex  is paramagnetic in nature. Therefore, the paramagnetic character of [Ni(NH₃)₆]Cl₂ complex can be explained on the basis valence bond theory.

According to valence bond theory, the electronic arrangement of Ni(II)ion is as follows,

Under the strong field effect, the two unpaired electrons of 3d-orbital has to be shifted to higher 4d-orbitals in order to form low spin inner orbital complex.

This transfer of electrons from lower 3d to higher 4d-orbital is not energetically feasible.

Thus only outer orbital high spin complex is formed in Ni(II) six coordinated complex is formed through sp3d² hybridization.

Under this condition, nickel ion in [Ni(NH₃)₆]Cl₂ complex contain two unpaired electrons. Hence [Ni(NH3)6]Cl2complex is paramagnetic.

On the other hand,  in case of [Co(NH₃)₆]Cl₃complex, the oxidation state of cobalt is +3 .The atomic number of cobalt : 27 and that of Co(III) ion : 24

The ligand NH₃, which is a strong  field ligand. Co-ordination number of central metal Co(III) ion : 6 

Valence shell electronic configuration of Co(III) ion : [ Ar ] d⁶ .

Since, ligand NH₃ is a strong field ligand, hence the complex is low spin one. Therefore, under the influence of the octahedral crystal field, the possible electronic arrangement of Co(III) ion is t2g⁶, eg⁰.

The octahedral crystal field splitting of [Co(NH₃)₆]Cl₃complex is as follows,

From the above crystal field splitting diagram of Co(III) ion, it is evidently shown that, the Co(III) ion have no unpaired electrons in its outer 3d-orbital.

That is, all electrons are paired, hence [Co(NH₃)₆]Cl₃complex is diamagnetic in nature.

• Why is [Ni(NH₃)₆]Cl₂ complex paramagnetic ?
• Why is [Co(NH₃)₆]Cl₃ complex diamagnetic ?
• Why is [Ni(NH₃)₆]Cl₂  paramagnetic but [Co(NH₃)₆]Cl₃ is a diamagnetic?
• Calculate the magnetic moment of [Ni(NH₃)₆]Cl₂ complex .
• Calculate the number of electron of [Ni(NH₃)₆]Cl₂ complex .
• Calculate the number of electrons of [Co(NH₃)₆]Cl₃ complex .

Outer orbital complex definition with examples