Write the ground state electron configurations of the following ions. (Type your answer using the format [Ar] 4s2 3d10 4p2 for [Ar]4s23d104p2. If the configuration is a noble gas, enter the noble gas in brackets, for example [Ne] for F -.)
Pb2+
Au+
Mn2+
Here is Pb. The others are available at the same site. Go to the home page of www.webelements.com, click on the element, then scroll down the right side to find electron configuration and click on it.
http://www.webelements.com/lead/atoms.html
Pb2+ = [Xe] 6s2 4f14 5d10
Au+ = [Xe] 6s2 4f14 5d10
Mn2+ = [Ar] 4s2 3d5
Why did the electron take a taxi to the party? Because it didn't want to get ion-ized!
To write the ground state electron configurations of the given ions, let's first start with the neutral atomic configurations of the elements and then remove the necessary number of electrons to match the charge of the ions.
1. Pb2+:
The neutral atom of Pb has an electron configuration of [Xe] 6s2 4f14 5d10 6p2.
Since Pb2+ has a 2+ charge, we need to remove 2 electrons from the configuration.
Therefore, the ground state electron configuration of Pb2+ is [Xe] 4f14 5d10.
2. Au+:
The neutral atom of Au has an electron configuration of [Xe] 6s2 4f14 5d9.
Since Au+ has a +1 charge, we need to remove 1 electron from the configuration.
Therefore, the ground state electron configuration of Au+ is [Xe] 4f14 5d9.
3. Mn2+:
The neutral atom of Mn has an electron configuration of [Ar] 4s2 3d5.
Since Mn2+ has a 2+ charge, we need to remove 2 electrons from the configuration.
Therefore, the ground state electron configuration of Mn2+ is [Ar] 3d5.
To determine the ground state electron configurations of the given ions, we need to identify the atomic number and then remove the corresponding number of electrons from the neutral atom.
1. Pb2+ (Pb is lead)
The atomic number of Pb is 82. The neutral atom's electron configuration is [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p². To determine the ground state electron configuration of Pb2+, we remove two electrons from the neutral atom. The configuration becomes [Xe] 6s² 4f¹⁴ 5d¹⁰.
2. Au+ (Au is gold)
The atomic number of Au is 79. The neutral atom's electron configuration is [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p⁶. To determine the ground state electron configuration of Au+, we remove one electron from the neutral atom. The configuration becomes [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p⁶.
3. Mn2+ (Mn is manganese)
The atomic number of Mn is 25. The neutral atom's electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵. To determine the ground state electron configuration of Mn2+, we remove two electrons from the neutral atom. The configuration becomes 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³.