MCAT Gen Chem: Ionization Energy

[Winterbourne]

I am posting a question from an MCAT General Chemistry Review Book. I hope it is ok to post the question considering copyright and such.

The question is:

What is the expected trend in the first ionization energies of cadmium, indium, and tin?

My answer (Increasing ionization energy): Cd, In, Sn

My reasoning: Ionization energy trend: Increasing from left to right and up the table.
Sn, has smallest atomic radius out of three three elements, therefore it is more difficult to remove electron (strong effective nuclear charge) and so HIGHEST ionization energy.



Please help!

The book has (increasing ionization energy) Sn, In, Cd


and their explanation is "Ionization energy increases with decreasing atomic radii, which occurs in a period as the effective nuclear charge increases." The only thing is….. I am pretty sure atomic radii DECREASE from left to right. So within this period (5), Cd should be larger than In and Sn.

Is this an error in the book? Do I have this concept all wrong?


** This is from a McGraw Hill Book

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[Teleologist]

Try writing out the electronic configurations for each element. You're missing the fact that one is in the d-block and the other two are in the p-block.

 

[Sammy1024]

I think what they're trying to say is that Cd has and orbital which makes it larger than you would expect?

I would have guessed the same thing as you.

 

[Winterbourne]

I think what they're trying to say is that Cd has and orbital which makes it larger than you would expect?

I would have guessed the same thing as you.

The reason is as Teleologist said "Cd is in the d-block. It was a full d orbital (10 electrons)." So this is very stable, will be difficult to ionize.

 

[Winterbourne]

Try writing out the electronic configurations for each element. You're missing the fact that one is in the d-block and the other two are in the p-block.

Right, but with In and Sn….

In: [Cd]5p1
Sn: [Cd]5p2

 

[Teleologist]

The reason is as Teleologist said "Cd is in the d-block. It was a full d orbital (10 electrons)." So this is very stable, will be difficult to ionize.

Meh not really. I meant that on a time-average basis, 4d electrons are closer to the nucleus than 5s electrons.

Also spin coupling of two 5s electrons in the same orbital produces electron-electron repulsion. This explains why the book is correct.

 

[Winterbourne]

Meh not really. I meant that on a time-average basis, 4d electrons are closer to the nucleus than 5s electrons.

Also spin coupling of two 5s electrons in the same orbital produces electron-electron repulsion. This explains why the book is correct.

Ok, all things considered…. knowing the periodic trend doesn't really help in this question. When seeing a question that involves this topic, is it then better to think about electron configuration rather than the periodic trend? (Last post, I promise!)

 

[Teleologist]

Ok, all things considered…. knowing the periodic trend doesn't really help in this question. When seeing a question that involves this topic, is it then better to think about electron configuration rather than the periodic trend? (Last post, I promise!)

We have to consider the issue from all angles. We must consider as you considered 1) size and 2) orbitals.

Size does decrease from left to right on the periodic table but once you get to the d-block things get screwy so don't rely on that size trend. In the d-block we have n level electrons and both immediately before and after the d-block we have n+1 level electrons. So just saying that size decreases from left to right across the periodic table top to bottom doesn't get you anywhere because that's wrong. n+1 level electrons will be further from the nucleus than n level electrons.

 

[Winterbourne]

We have to consider the issue from all angles. We must consider as you considered 1) size and 2) orbitals.

Size does decrease from left to right on the periodic table but once you get to the d-block things get screwy so don't rely on that size trend. In the d-block we have n level electrons and both immediately before and after the d-block we have n+1 level electrons. So just saying that size decreases from left to right across the periodic table top to bottom doesn't get you anywhere because that's wrong. n+1 level electrons will be further from the nucleus than n level electrons.

Ouch! Well you're right. I appreciate your help.

 

[Winterbourne]

Meh not really. I meant that on a time-average basis, 4d electrons are closer to the nucleus than 5s electrons.

Also spin coupling of two 5s electrons in the same orbital produces electron-electron repulsion. This explains why the book is correct.

Sorry to bring this question up again. I just want to clear it up if someone looks at it

1. your point about the d-orbital is correct (closer to nucleus)
2. Ionization Energy Trend: The single np electron of the G3A element is more easily removed than one of the two ns electrons in the preceding G2A
3. Does not apply to this question, but in general: G6A (ns2np4) have smaller ionization energies than G5A in 2nd, 3rd, and 4th periods because of the repulsion b/w electrons in p-orbital.

So the answer should actually be:

Ionization energies of:

Cd: 868
Sn: 709
In: 558

Ok…time to move on.

 

[DrknoSDN]

http://ptable.com/#Property/Ionization
If a practice question does not agree with a trend, check to see that the book isn't using bad logic.
Bad logic would be trying to rationalize things that are not true in reality.. A lot of scenarios are idealized for the MCAT but an example problem should still use trends that are based on factual data.

 

[Winterbourne]

http://ptable.com/#Property/Ionization
If a practice question does not agree with a trend, check to see that the book isn't using bad logic.
Bad logic would be trying to rationalize things that are not true in reality.. A lot of scenarios are idealized for the MCAT but an example problem should still use trends that are based on factual data.

Yes, the book's answer is using "bad logic." Very annoying! Thanks for the link.