Ionic and atomic radius question I'm confused on

[virtualmaster999]

Hey everyone

So I got confused on how to compare ionic and atomic radius. Do you first check to see if there is an isoelectronic series, and then compare those? Then once you know the order, you compare it to the others?

What's the best way to go about these? Like for example:

Cl, s2-, k, k+, o

Or s2-, f-, ne, k+, and ca2+

How do you guys go about these?

Thanks in advance!


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[virtualmaster999]

Could someone explain the order of possible btw?


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[FeralisExtremum]

I'm not sure if I'm right, but this is how I would approach it:

First set (Cl, S2-, K, K+, O). Check if this is an isoelectronic series: not all of them are. Shell will usually determine size to the greatest extent (e.g. 2nd shell will always be a lot bigger than the 1st shell) and anions/cations are always a lot bigger/smaller than the neutral species, respectively. In this set O is in the 2nd shell, so it will be the smallest. K is in the 4th shell, so it will be largest. The remaining (Cl, S 2-, K+) are all in the third shell in terms of electrons. K+ and S 2- have an equal number of electrons but we know that K+ has a larger/stronger nucleus, so it must be smaller than S 2- (pulls an equal amount of electrons in tighter). So far that makes the ranking in size from largest to smallest: K > S 2- > K+ > O. Cl doesn't have an equal number of electrons to the rest in its row, so I have no idea how to place it: it's likely smaller than S 2- because it has a stronger nucleus and less electrons, but I'm not sure if it can be easily distinguished from K+ in terms of size. This leaves me with: K > S 2- > ?Cl, K+? > O

Second set (S 2-, F-, Ne, K+, and Ca 2+). Check if this is an isoelectronic series: not all of them are. Based on # of electrons, F- and Ne will be in the 2nd shell (smaller) and the other three will be in the 3rd shell (larger). This one is a bit easier because the ones in the same shell/row are isoelectronic: F- and Ne have the same # of electrons but Ne has the larger/stronger nucleus, pulls its electrons in tighter, so F- > Ne in size. S 2-, K+, and Ca 2+ are also isoelectronic, so once again we use the strength of the nucleus to distinguish size: Ca 2+ has the largest/strongest nucleus (so it will be the smallest ion), then K+ will be larger, then S 2- largest. This makes our final ranking here from largest to smallest: S 2- > K+ > Ca 2+ > F- > Ne.

I distinguish them on the basis of shell (in terms of # of electrons), and then if they are within the same shell, I hope it's an isoelectronic series and figure it out that way.

 

[virtualmaster999]

I'm not sure if I'm right, but this is how I would approach it:

First set (Cl, S2-, K, K+, O). Check if this is an isoelectric series: not all of them are. Shell will usually determine size to the greatest extent (e.g. 2nd shell will always be a lot bigger than the 1st shell) and anions/cations are always a lot bigger/smaller than the neutral species, respectively. In this set O is in the 2nd shell, so it will be the smallest. K is in the 4th shell, so it will be largest. The remaining (Cl, S 2-, K+) are all in the third shell in terms of electrons. K+ and S 2- have an equal number of electrons but we know that K+ has a larger/stronger nucleus, so it must be smaller than S 2- (pulls an equal amount of electrons in tighter). So far that makes the ranking in size from largest to smallest: K > S 2- > K+ > O. Cl doesn't have an equal number of electrons to the rest in its row, so I have no idea how to place it: it's likely smaller than S 2- because it has a stronger nucleus and less electrons, but I'm not sure if it can be easily distinguished from K+ in terms of size. This leaves me with: K > S 2- > ?Cl, K+? > O

Second set (S 2-, F-, Ne, K+, and Ca 2+). Check if this is an isoelectric series: not all of them are. Based on # of electrons, F- and Ne will be in the 2nd shell (smaller) and the other three will be in the 3rd shell (larger). This one is a bit easier because the ones in the same shell/row are isoelectric: F- and Ne have the same # of electrons but Ne has the larger/stronger nucleus, pulls its electrons in tighter, so F- > Ne in size. S 2-, K+, and Ca 2+ are also isoelectric, so once again we use the strength of the nucleus to distinguish size: Ca 2+ has the largest/strongest nucleus (so it will be the smallest ion), then K+ will be larger, then S 2- largest. This makes our final ranking here from largest to smallest: S 2- > K+ > Ca 2+ > F- > Ne.

I distinguish them on the basis of shell (in terms of # of electrons), and then if they are within the same shell, I hope it's an isoelectric series and figure it out that way.

Thanks for that detailed explanation it helped a lot! In general, would you say I could first rank the elements of an isoelectronic series, and then compare it to the others?


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[FeralisExtremum]

If they're all isoelectronic, then you can jump into it straight away with that method. If it's not isoelectronic, I would separate them out on the basis of which shell they are in (going by their # of electrons) first. In the second set you gave for example, they're not all isoelectronic with each other, so I would first separate out F- and Ne into one isoelectronic group recognizing that they are in a smaller shell, then the other three into the larger isoelectronic group. Then rank the groups internally.