Ionization Energy

[Qester]

---

Members do not see this ad.

Removing an electron from which of the following would require the most energy

a) K
b )K+
c) K2+

Following the basic Ionization Energy trend (up and right), K+ best fits this trend for most energy required.

From what I have read Second Ionization Energy is always greater than the first so we can Eliminate K.

Now the options are K+ and K2+. K+ fits the trend, as well K+ has the electron configuration of a noble gas which have extremely high ionization energies due to their filled octet.

However the book gave the answer as K2+ with the reasoning that each additional electron remove becomes harder IE1<IE2<IE3.... Regardless of periodic trend or octect filled.

Is this always true?

 

[PantherPride]

What this question is really asking you is which ionization energy for K is the greatest: the first, the second, or the third? The periodic table trend for ionization energy is only for the first, so you can't rely on that here. As a cation becomes more and more positive, the stronger its affinity toward electrons is, the harder it becomes to remove an electron. So yes, IE1 > IE2 > IE3 for any given element. If you're comparing IE1 or IE2 between different elements, then you'd need to refer to the periodic table

Here's a list of ionization energies: http://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements

 

[Qester]

What this question is really asking you is which ionization energy for K is the greatest: the first, the second, or the third? The periodic table trend for ionization energy is only for the first, so you can't rely on that here. As a cation becomes more and more positive, the stronger its affinity toward electrons is, the harder it becomes to remove an electron. So yes, IE1 > IE2 > IE3 for any given element. If you're comparing IE1 or IE2 between different elements, then you'd need to refer to the periodic table

Here's a list of ionization energies: http://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements

Thanks! It all makes sense, my biggest reason for choosing + over 2+ was thinking that since it had a full octet it would have the highest ionization energy

 

[krogers21]

Thanks! It all makes sense, my biggest reason for choosing + over 2+ was thinking that since it had a full octet it would have the highest ionization energy

For K, think of it like this... K has only 1 valence electron, which is gladly loses to make K+. So now, the outer shell (sorta like the new valence shell) has eight electrons. Taking one electron away to make K2+ is extremely hard in comparison as before, but taking another electron away from that would be even harder because it's "new valence shell" wants to be full.

 

[redlight]

For K, think of it like this... K has only 1 valence electron, which is gladly loses to make K+. So now, the outer shell (sorta like the new valence shell) has eight electrons. Taking one electron away to make K2+ is extremely hard in comparison as before, but taking another electron away from that would be even harder because it's "new valence shell" wants to be full.

True

In general the first ionization energy is always going to be lower than the 2nd, 3rd etc because for the 2nd IE and afterwards you are trying to pull an electron off of a positively charged atom. The more positively charged it is, the more energy you'll need to pull off an electron.