I am trying to find the ground state for, Pb2+
I went back two on the chart and came to Hg. And entered it's ground state.
Why is this wrong?
Are you trying to determine the term symbol? I think Pb^+2 has the same electron configuration as elemental Hg. And elemental Hg has the term symbol below.
1S0
The ground state electron configuration for Pb2+ can be determined by looking at the electron configuration of the neutral Pb atom and then accounting for the loss of two electrons to form the Pb2+ ion.
The electron configuration of the neutral Pb atom is: [Xe] 4f14 5d10 6s2 6p2.
When Pb loses two electrons to form the Pb2+ ion, the two outermost electrons (6s2 6p2) are removed. Therefore, the electron configuration of Pb2+ is: [Xe] 4f14 5d10.
So, the correct ground state electron configuration for Pb2+ is [Xe] 4f14 5d10.
Hg has a different electron configuration: [Xe] 4f14 5d10 6s2. It is not the correct electron configuration for Pb2+.
The ground state electron configuration for Pb2+ cannot be determined by simply going back two elements on the periodic table and using the ground state electron configuration of Hg.
To determine the ground state electron configuration of Pb2+, you need to consider its atomic number and the number of electrons it has lost.
Pb (lead) has an atomic number of 82. When it forms a cation with a charge of +2 (Pb2+), it loses two electrons from its neutral state.
To find the ground state electron configuration for a cation like Pb2+, you remove electrons from the highest energy level (outermost shell) first. The electron configuration of Pb in its neutral state is [Xe] 4f14 5d10 6s2 6p2.
By removing two electrons from the highest energy level (6p), the electron configuration of Pb2+ becomes [Xe] 4f14 5d10.
So, the correct ground state electron configuration for Pb2+ is [Xe] 4f14 5d10.