What is the #d# configuration of the metal in #["Mn"("H"_2"O")_6]^(2+)#? Will it be low spin or high spin?

1 Answer
Jan 18, 2016

Answer:

This complex is a #d^5# .

Explanation:

#[Mn(H_2O)_6]^(2+)# is a complex ion with #Mn^(2+)# as its central ion. The 6 #(H_2O)# that surrounds the central ion as known to be the ligand.

Crystal Field Theory concentrates on the splitting of the d orbitals of the metal center. It could either have low spin (strong field) or high spin (weak field) spin states.

Low spin (strong field) happens when crystal field splitting energy (CFSE) is greater than the pairing energy. This would make the electrons fill the lower energy orbitals (completely) first before filling the higher ones.

High spin (weak field) on the other hand happens when the pairing energy is greater than the CFSE. Electrons would fill all the orbitals first before pairing up. This is just like Hund's rule.

For this specific example,

First, we have to know the electron configuration of our metal ion which is #Mn^(2+)#. It can be seen in the picture below that it is a #d^5# metal which means it has 5 d electrons as you say it.

Now, use this information to "fill" the CF energy diagram depending on what is required. I did both the weak field and strong field to show you the difference.

enter image source here