Hello Dr.Bob. Could you please take a look at the following question? I know the correct answer for it, but I'm not sure how to go about solving it.
Which ground state electronic configuration will most readily ionize to a 2 + cation?
A) 1s22s22p63s23p64s23d10
B) 1s22s22p63s23p64s23d104p4
C) 1s22s22p63s23p64s1
D) 1s22s22p63s23p64s23d104p2
E) 1s22s22p63s23p2
The correct answer for this question is A). Could you please let me know why that configuration is the right one? I thought that since A) has a complete outer electron shell, then it would be more stable and thus less likely to ionize to a 2+ cation.
Your notations are hard to read because you didn't space it properly. I have redone it below. Also, I think you should change the order, at least mentally, to what I show below.
A) 1s2 2s2 2p6 3s2 3p6 4s2 3d10
B) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
C) 1s2 2s2 2p6 3s2 3p6 4s1
D) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2
E) 1s2 2s2 2p6 3s2 3p2
A) 1s2 2s2 2p6 3s2 3p6 3d10 4s2
B) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4
C) 1s2 2s2 2p6 3s2 3p6 4s1
D) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2
E) 1s2 2s2 2p6 3s2 3p2
Why did I reconfigure this? Because the OUTSIDE shell, which is the one used for most ionizations, is the outside shell and that is 4 instead of 3. I ALWAYS write the shells in numerical order (and not filling energy wise order) because students try to ionize the 3d first over the 4s. Any, here is my best guess how to anwer this question and basically it is by elimination.
Eliminate E since the last electrons are in the 3rd shell and we know the 4th shell is easier to pick off an electron.
Eliminate D because we PROBABLY will get +4 and not +2.
Eliminate C since +1 is the obvious choice and not +2.
Eliminate B since you probably will get +4(could get +6) and not +2.
This leaves A as the only choice and +2 electrons are handy as the two outside electrons.
To determine which ground state electronic configuration will most readily ionize to a 2+ cation, we need to consider the concept of ionization energy and electron configuration.
Ionization energy refers to the amount of energy required to remove an electron from an atom or ion in the gas phase. Generally, it is easier to remove an electron from an atom or ion if the electron is farther from the nucleus, as it experiences less attractive force.
In this case, we want to ionize to a 2+ cation, which means removing two electrons from the atom. To make it easier, we want to start with an electronic configuration that already has a tendency to readily lose electrons.
Let's analyze each option:
A) 1s22s22p63s23p64s23d10
B) 1s22s22p63s23p64s23d104p4
C) 1s22s22p63s23p64s1
D) 1s22s22p63s23p64s23d104p2
E) 1s22s22p63s23p2
Option A) has a complete outer electron shell (4s23d10) with a total of 2 electrons. However, these electrons are in higher energy levels (higher principal quantum numbers) compared to the electrons in the earlier shells, which means they are less attracted to the nucleus. As a result, it requires less energy to remove them, making it easier to ionize the atom.
On the other hand, options B), C), D), and E) have incomplete outer electron shells. They will have a stronger attraction to the nucleus due to higher effective nuclear charge, making it more difficult to ionize these atoms.
Therefore, option A) is the correct answer as it has a higher tendency to lose the required two electrons to form a 2+ cation.
Remember, when assessing ionization tendency, it is crucial to consider not only the number of electrons in the outer shell but also the energy levels and effective nuclear charge experienced by those electrons.