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Chapter 2 - Electronic Structure and the Periodic Table

Chapter 2 - Electronic Structure and the Periodic Table

Postby ssnk1234 » Wed Jul 06, 2016 5:54 pm

Hi,
This query relates to Ionisation Energy, found in Chapter 2 of General Chemistry.
It says on page CHM-22: "The ionisation energy potential increases from left to right within a period and decreases from the top to the bottom of a group or column of the periodic chart."

It then says on page CHM-23, "...the 1st IP of Na+ is therefore expected to be greater than the 1st IP of Al+"

However, doesn't this contradict the sentence above, whereby the ionisation potential is increasing from left to right across a period? Therefore, by this logic, shouldn't Na+ have a lower ionisation potential compared to Al+, since Al+ is located in IIIA and Na+ is located in IA?

Appreciate any help/clarification.
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Re: Chapter 2 - Electronic Structure and the Periodic Table

Postby goldstanda3269 » Wed Jul 06, 2016 9:12 pm

Just to be clear: You don't need to memorise 2nd IP but this kind of reasoning is exactly what shows up on the real GAMSAT.

ssnk1234 wrote:It says on page CHM-22: "The ionisation energy potential increases from left to right within a period and decreases from the top to the bottom of a group or column of the periodic chart."


Correct. This point is underlined with the arrows shown around the periodic table on CHM-27.


ssnk1234 wrote:It then says on page CHM-23, "...the 1st IP of Na+ is therefore expected to be greater than the 1st IP of Al+"

However, doesn't this contradict the sentence above, whereby the ionisation potential is increasing from left to right across a period? Therefore, by this logic, shouldn't Na+ have a lower ionisation potential compared to Al+, since Al+ is located in IIIA and Na+ is located in IA?


It is not true that Al+ is located in IIIA and Na+ is located in IA. It is true that Al is located in IIIA and Na is located in IA. And therefore Na has a lower ionisation potential compared to Al. But this is the whole point about 2nd IP compared to 1st IP. Now (2nd IP) we are looking at removing an electron from an atom that is already missing an electron. The electronic configuration of Na+ is similar to Ne (neon, in the last group in the periodic table), and Al+ is below that, similar to Mg, and so Na+ has the higher IP.
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