What is the Second Highest Ionization Energy?

AI Thread Summary
The discussion centers on identifying the excited state of an oxygen atom and determining which element has the second highest ionization energy. The configurations for the excited oxygen atom are analyzed, with options a and c dismissed, while option d is confirmed as the ground state. The question regarding second ionization energy leads to confusion, with participants debating whether to consider first or second ionization energies. The consensus leans towards potassium (K) as the element with the second largest ionization energy due to its electron configuration. Overall, clarity on the question's intent is deemed necessary for accurate responses.
Miike012
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1. Which one of the following configurations depicts an excited oxygen atom?
a. 1s2 2s2 2p2
b.1s2 2s2 2p2 3s2
c.1s2 2s2 2p1
d.1s2 2s2 2p4

a. is out of the question
c. is out of the questioin

I know d. is the ground state of oxygen

but how did they get 3s2 for b.?
It must be b. But I don't understand it.

2.Which of the following has the second largest Ionization energy?
a. ca
b. K
c.Ga
d. Ge

I am guessing it is K because after its first V.E. is gone the second one would be in the core, making it much harder to remove...
??
 
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Miike012 said:
I know d. is the ground state of oxygen

OK

but how did they get 3s2 for b.?
It must be b. But I don't understand it.

Excited means there is still 8 electrons, but they not necessarily occupy lowest orbitals (that would be the ground state).

2.Which of the following has the second largest Ionization energy?
a. ca
b. K
c.Ga
d. Ge

I am guessing it is K because after its first V.E. is gone the second one would be in the core, making it much harder to remove...

Question is ambiguous to me. First possibility - take all elements, order them according to their FIRST ionization energies, tell which is the second on this list. Second possibility - take SECOND ionization energies, tell which one is the largest. Could be it is my English playing tricks on me.

In the latter case your answer seems to be right.
 
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