Biggest ionic radius trend

[virtualmaster999]

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Hey everyone!

So in Destroyer, it says that usually, the most negative is the biggest; does that mean you can apply this even for non-isoelectronic series?

I knew the trend applied for the series, but what I do otherwise is group each "bunch" into their respective series, and then compare, which so far has come out the same.

Any feedback helps. Thanks!

 

[goldcrown17]

I'm not sure if I understand exactly what you're asking, but let me try to help. The size of the atom can be understood in two parts. The first part is the standard trend for your day-to-day atom. When looking at the periodic table, as you move down size increases. Additionally, as you move from right to left, atomic radius increases.

The second part (which I think your question is about) deals with ions. In a cation, electrons have been removed. Lets say our atom has 10 electrons and 10 protons. When we remove an electron to create a +1 charge, there are now 9 electrons and 10 protons. Those 10 protons are just as positive as they were before and have more power to pull in the electrons. I like to think of it in terms of tug of war. If the same 10 electron, 10 proton atom were to gain an electron, the anion would be larger than the neutral atom. Again, the tug of war favors the electrons and they are held less tightly to the protons in the nucleus.

For a given atom: Largest X^-2> X^-1 > X > X^+1 > X^+2 Smallest

Not sure if that answers your question. Hope that helps.

 

[tyjacobs]

rules:

1) atomic radius trend increases from right to left and from top to bottom
2) cations are SMALLER than their atomic counterparts (Na+ < Na)
3) anions are LARGER than their atomic counterparts (F- > F)
4) if asked to rank radii given an isoelectronic set (isoelectronic meaning all atoms have the same number of electrons): calculate the proton/electron ratio for each atom. Whichever atom has the smallest ratio has the LARGEST radius, the 2nd smallest ratio having the SECOND LARGEST radius, and so on...

 

[tommyngu]

rules:

1) atomic radius trend increases from right to left and from top to bottom
2) cations are SMALLER than their atomic counterparts (Na+ < Na)
3) anions are LARGER than their atomic counterparts (F- > F)
4) if asked to rank radii given an isoelectronic set (isoelectronic meaning all atoms have the same number of electrons): calculate the proton/electron ratio for each atom. Whichever atom has the smallest ratio has the LARGEST radius, the 2nd smallest ratio having the SECOND LARGEST radius, and so on...

I agree with everything here except I think it's easier to just look at the atomic number for isoelectronic series. The greater the Z, the smaller the ionic radius.

I believe the OP is having trouble with comparing ionic radii of ions that are not an isoelectronic series, though.

ex. Rank Br-, Ca++, Rb+, and Sr++ from smallest to largest.

To rank this set, you would identify Br-, Rb+, and Sr++ as part of an isoelectronic set so rank them accordingly.
Sr++ < Rb+ < Br-

Now what do you do with Ca++? Here's another rule you can add to your list: When comparing ions of the same charge (ions within the same column), the ionic radii trend follows that of atomic radii. In other words, ionic radii will increase going down.

Since Ca++ is above Sr++, Ca++ would be smaller than Sr++.

So the final order would be: Ca++ < Sr++ < Rb+ < Br-

These rules work fine for a lot of problems, but we still haven't exhausted the possible scenarios.

ex. Rank Br-, Al+++, Sr++, F-, O--, I- from smallest to largest.

The answer would be Al+++ < Sr++ < F- < O-- < Br- < I- according to an ionic radius chart, but I would not know how to get this without looking at a table.
All I could get is:

Sr++ < Br-
Al+++ < F- < O--
and I- > F- > Br-
which are all true, but how would you get the exact order?