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Effective Nuclear Charge
Michael DeDominicis
#1 Posted : Wednesday, August 01, 2012 12:22:09 PM
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I know that effective nuclear charge increases down a group, but when you use the formula, Zeff=#protons-#core electrons, the Zeff is the same for all the atoms in the group. Is this formula just an estimation and we actually get a subtle increase?
As well, does the effective nuclear charge not take into account the radius and that's why the ionization energy actually decreases going down a group?
Thanks a bill.
vanessalittle
#2 Posted : Wednesday, August 01, 2012 3:09:51 PM
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Hey there,

Zeff is defined as the following in the class companion:

The nuclear charge that the valence electrons actually experience due to the repulsive nature
of core electrons is called the effective nuclear charge (Zeff).

A SIMPLE way to calculate Z (aka, an estimation)is using Zeff = Z-S, as you said. Likely a more intricate/detailed version of this equation is used to more accurately determine Z, one that would need to account for the increase in radius, etc. You would not be expected to know that for the MCATs, you'd only need to know Zeff = Z-S.

Now, they do mention the following:

The number of protons increases as we go from top to bottom in a group, causing a greater force of electrostatic pull inward on the valence electrons.

BUT

The number of core electrons increases as we go from top to bottom in a group, causing greater shielding and decreasing the effective nuclear charge.

So, unless it's discussed in an MCAT passage, you're expected to remember that although there are these competing components, effective nuclear charge still increases slightly moving top to bottom in a group.

Why?

You don't need to know why exactly for the MCATs, you'd just be expected to know the trend. But if you look at page 41 in your classroom companion (which is where I'm getting this information), you'll see a figure demonstrating how the electrons repel each other in an atom, but that when an outer electron has a 'clear view' of the nucleus, that attractive force is what keeps the valence electrons from being repelled away.

Now imagine you have a very large, diffuse atom like iodine. With the electron cloud being so diffuse, I can imagine it'd be easier for the electrons to avoid each other (reducing repulsion) and also easier for the outer electrons to get a 'clear view' of the nucleus, thereby increasing Zeff.

Of course, this is a very simplified thought experiment, and there are likely MANY more factors at play. But that will hopefully help you to more intuitively understand why Zeff increases (very slightly) as you go down a group.

Let me know if you have any more questions on this!

Vanessa
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