in the ground state electron configuration of Fe3+, how many unpaired electrons are present? I thought it was 1, but im getting a wrong answer.

Ground state of 26Fe.

1s2 2s2 2p6 3s2 3p6 3d6 4s2
So to make the 3+ ion, we remove the 2 outside electrons (the 4s2 electrons) + 1 more from the 3d orbital. Therefore, the ground state for the 3+ ion is
1s2 2s2 2p6 3s2 3p6 3d5
Hund's rule of maximum multiplicity says that these electrons will go into unfilled shells for the ground state. So how many d orbitals are available? There are 5 d orbitals and we place 5 electrons in them; therefore, there must be 5 unpaired electrons.

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To determine the number of unpaired electrons in the ground state electron configuration of Fe3+ (iron ion with a +3 charge), we first need to find the electron configuration of Fe3+.

The electron configuration of neutral iron (Fe) is [Ar] 3d6 4s2. Since Fe3+ has a +3 charge, three electrons have been removed from the neutral atom.

To obtain the electron configuration of Fe3+, we will remove three electrons from the neutral iron configuration. Since electrons are removed from the highest energy level first, we start by removing two electrons from the 4s orbital (now empty) and then one electron from the 3d orbital.

So, the electron configuration of Fe3+ is [Ar] 3d5.

To determine the number of unpaired electrons, we look at the filled orbitals and count the pairs of electrons. In the case of Fe3+, the 3d orbital has 5 electrons, which means there is only one unpaired electron, not three as you initially thought.

Therefore, there is one unpaired electron in the ground state electron configuration of Fe3+.

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