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Specialty Chemistry Forums => Nuclear Chemistry and Radiochemistry Forum => Topic started by: madisonwi on January 27, 2005, 12:48:03 PM

Title: Atomic Radii
Post by: madisonwi on January 27, 2005, 12:48:03 PM
Hello all -

In comparing atomic sizes, Rb - as 5s1 - should be larger than Kr - as 4p6.  

The logical problem is with Rb1+.  Rb1+ is isoelectric with Kr and still has one more proton.  Also, Rb1+ no longer has a fifth energy level.  The additional proton in Rb1+ - in other words the stronger positive pull of the nucleus as compared to Kr - should make Rb1+ smaller than Kr.  But the opposite is true.  In fact, in pm Rb1+ is almost twice as large as Kr (161 pm vs 88 pm).

What am I missing?

Title: Re:Atomic Radii of isoelectric atomic particles
Post by: Donaldson Tan on January 30, 2005, 09:01:41 AM
OK. I check out your values for the ionic radii at to see if they are right. I think your arguement is based purely on their composition.. I am thinking in the sense with a higher proton number, the d orbitals of Rb+ are more withdrawn into the nucleus than that for Kr. This results in much higher shielding effect for Rb+ than the valence electrons in Kr. Hence, the ionic radius of Rb+ > ionic radius of Kr
Title: Re:Atomic Radii
Post by: madisonwi on January 30, 2005, 08:21:52 PM
geodome - I'll be honest and say that I don't buy your answer.

Rb1+ and Kr are isoelectric, but naturally the ionic radius of each are different.  When the only thing that is different between them is the number of protons, it gives a clear starting point for the discrepency.

When comparing Rb to Kr, Rb is naturally larger for obvious reasons.  The 5s1 orbital should be the only difference between Rb and 4p6 Kr.  The atomic radius for Rb is 265pm and it is 88pm for Kr.

That's where the confusion starts.  The differnce of 187pm is being accounted for by only one electron difference.  Is the electron in 5s1 really flying that far from the 4th energy level.  And when it's ionized at only 400kJ/mol, why does the atomic radii stay relatively large at 161pm?

See, I think there is something going on in the structuring of Rb that:
1) lends to it's large size
2) lends to the large size of its ions

I understand the d orbital as the dumping ground, but with only 3d available - and quite buried by the time we are playing with the 5s orbital - I'm not sure how much impact it has.  Perhaps, then Rb is dumping things into 4d?  That might be believable, but why?  And how much energy is it really saving doing that?

Something about the 5 energy level coming into play just screams to me electron density reorganization.

Title: Re:Atomic Radii
Post by: madisonwi on January 30, 2005, 08:30:42 PM
Continuing my thinking from the previous post, would it be possible that when 5s1 is ionized, that the following state might be more stable:


would this explain the relatively large size of Rb1+?
Title: Re:Atomic Radii
Post by: Mitch on January 30, 2005, 10:45:59 PM
no way
Title: Re:Atomic Radii
Post by: Demotivator on January 31, 2005, 12:32:27 PM
1) I agree, "no way". Rb+ is simply [Kr] electronic config. The one extra electron in 5s1 does cause a large increased size over Kr because in going from left to right in the fourth row the atoms get progressively smaller with increased Zeff. Jumping to the fifth row causes a sharp decrease in Zeff.

2) If you believe webelements data, then this behavior of Rb+/Kr is present for all group 1 starting with Li+/He. Therefore, it might not be a matter of extra shielding due to the pull of the extra nuclear charge since even Li+ has the simplest 1s2 config.

I'm not totally certain but,
The reason may lie in how they calculate ionic radius. It looks like the calculations are based on bond lengths with F- or O2-. But even with highly electronegative anions, does that make the cation a pure ion? I don't think so.

If there was some small covalent character in the ionic bond (it's about 10% in li+), then the ion really has some extra partial negative charge which has to be accomodated in an outer, partially mixed molecular orbital, resulting in a larger apparent size than one might expect for a pure ion.
Title: Re:Atomic Radii
Post by: jdurg on January 31, 2005, 01:51:21 PM
Actually, I don't think that the 'oddity' is the Rubidium +1 ion.  I think it's the netural Krypton atom that is out of place.  If you look at the radius of Br- compared to Rb+, it's 182pm to 161 pm.  That is pretty good since the Br- has the 'extra' electron which would account for the greater radius, and Rb+ has the greater proton count which would account for the smaller radius.  So it's the Kr with it radius of only about 88 pm that is off.  So I think for this question one should look further into the Krypton atom to figure out the solution.
Title: Re:Atomic Radii
Post by: jdurg on January 31, 2005, 02:07:29 PM
Also, Rb+, Br-, and Kr are isoelectronic.  They have the same shells filled up and the same number of electrons.  Rb+ and Br- are VERY close in size with the differences easily explained.  It's the Kr atom that just doesn't seem right.  The way they tend to determine what an atomic radius is, is to measure the distance between the nucleus of two touching atoms and dividing that by two.  I think this might explain the differences here.  With a charged ion, that charge is felt over a greater area than just the atom.  So it's like trying to put two similar poles of two different magnets together.  They just will not want to get close to each other.  As a result, their radius will appear to be bigger than it actually is.  For a neutral atom, you don't have this repulsion so the radius will be a bit smaller.  Hence why the Kr atom has the smaller radius than the Rb+ and Br- atoms.  When calculating/measuring the radius of Kr, they don't have to deal with the repulsion between two like charges.  When calculating/measuring the radius of an ionic species, you DO have to deal with that repulsion.
Title: Re:Atomic Radii
Post by: Demotivator on January 31, 2005, 02:38:39 PM
Br- has to be larger than Kr because it has one less proton, hence electron repulsion is more pronounced.

What repulsion? Ionic bonds are between opposite charged ions.
Title: Re:Atomic Radii
Post by: jdurg on January 31, 2005, 02:47:19 PM
But when calculating the radius of an atom, they take two of the same atom and divide the distance between the nucleus by two.  So for a neutral atom it poses no problem.  For a charged atom, I have no idea how they get them together in order to determine the radius, but taking the distance between the nucleus of a Cl- ion and a Rb+ ion and dividing that by 2 can only result in horribly inaccurate calculations.
Title: Re:Atomic Radii
Post by: Demotivator on January 31, 2005, 02:56:45 PM
They don't do it that way for ions.
They start with a base ion like O2- and assume a fixed size for it. Then that radius is subtracted from the total bond length of whatever it's attached to.
Title: Re:Atomic Radii
Post by: jdurg on January 31, 2005, 03:18:37 PM
Ahhhh.  Well, even that is kind of crappy.   :P  It looks like they need a better way to determine ionic radii!   ;D
Title: Re:Atomic Radii
Post by: madisonwi on January 31, 2005, 09:42:53 PM
Well, this is beginning to make some more sense.  I agree with jdurg that the point of comparison has more to do with Kr than Rb1+ due to Rb1+ and Br1- having similar ionic radii.

When checking for atomic radii on webelements, there are usually two figures - measured and calculated.  The measured atomic radii would make sense to set two side by side and divide by two.  I also see how repulsion (or for that matter attraction in the case of Br2 from 2 Br1-) wouldn't work for ions and that a lattice would be needed.  I also agree with jdurg that the measured figure seems rather crappy.

It's the calculated radii that interests me.

For the 5s orbital, the wave function is as follows:

Radial wave function R5s = (1/300?5) × (120 - 240? + 120?^2 - 20?^3 + ?^4) × Z^3/2 × e^-?/2

Angular wave function Y5s = 1 × (1/4?)^1/2

Wave function ?5s  = R5s × Y5s

Since ?=2Zr/n, for Rb1+ as compared to Rb, Z is remaining constant but n is decreasing from 5 to 4.  Ro would then be larger for Rb1+ and the second part of the expression - by 20p^3 - shouid make Rb1+ smaller than Rb.  Then my guess is that with Kr's Z being smaller, the radial wave fucntion comes out even smaller.  

Should I chugg the number to verify, or am I way off (again).