Question

How many unpaired electrons would you expect on iron in [fe(h2o6] 3+?

Answer 1

The iron (III) has 5 valence electrons and the complex is in a octahedral geometry. Since water as a ligand has a small splitting energy, the complex will be a high spin and all of the five electrons will be unpaired. 

Answer 2

Answer:

Five unpaired electrons

Explanation:

This should be the real equation: [Fe(H20)6]3+

Firstly, Ligands are ions or neutral molecules, that bonds to a central metal atom or ions. In this case the ligand here is the aqua molecule(H2O)6.

According to Hund's rule,electrons always enter an empty orbital before they pair up. This rule holds because the orbital have the same energy. In a situation like this where we have a ligand coordinating to a metal ion, the energy of the orbital are no longer same.

What really affects Hund's rule is the crystal field splitting . In this case the ligand molecule(H2O) has a weak crystal splitting and is denoted as weak field ligands. The weak field ligand has a less field splitting energy than the pairing energy. So the electron filling pattern in the d shell follows the Hund's rule.

Therefore, iron(Fe) = 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Giving out the 2 electron from 4s shell and one electron from  the  only paired d electron( Fe3+) . The remaining five electron from the d shell becomes unpaired.