[Alexander850]

I tried using the search tool, but could not find a direct answer to my specific questions.  If I have missed it, I apologize in advance for repeating the questions (and for the length of this post).

Question 1:  What is the trend of effective nuclear charge as you move down a group, and why?

Question 2:  If the effective nuclear charge increases as you move down a group, why does the ionization energy decrease moving down the group?  

I understand very well why Z* increases moving across a period, but I am having trouble understanding the trend of effective nuclear charge as you move down a group.  I have found conflicting sources (credible sources) on this trend, some of which even conflict with themselves (several sources describe why Z* decreases down a group, but then show graphs or tables that indicate an increase in Z* moving down a group).  

Sources that state Z* decreases down a group cite two points as evidence: (1)electrons are adding to shells of larger n-values, so there are more electrons to shield the nuclear charge and (2) the increasing atomic radius causes the valence electron(s) to feel a decreased nuclear charge.  

Sources that argue that Z* increases down a group cite penetration effects as evidence.  They argue that, due to penetration differences between subshells, nuclear charge increases more rapidly than the shielding does--resulting in a slow increase in Z* going down a group.

[Darren]

1) effective nuclear charge decreases down a group due to shielding effect of inner electrons from the nucleus for the valence electrons to be removed more easily.

2) thus, ionisation energy decreases down the group.

[cheese (MSW)]

1) effective nuclear charge decreases down a group
This statement is incorrect despite what you might read in some
Introductory Texts.
Would anyone else like to try?

[cheese (MSW)]

Superficially it makes sense that the more e⁻s there are the greater the shielding of the nuclear charge from the valence e⁻s, but we are forgetting that the nuclear charge increases as well.  You may know that screening of the nuclear charge follows the order s>p>d>f and this gives rise to phenomena such as the Lanthanide contraction (Wikipedia).  Zeff = Z - S As I mentioned before chemists still use the Zeff from the Clementi-Raimondi calculations of the 60’s.  Here are some Zeff values:
Group 1: Li = 1.28, Na = 2.51, K = 3.50, Rb = 4.98, Cs = 6.36
Group 14: C = 3.14, Si, 4.29, Ge = 6.78, Sn = 9.10, Pb = 11.61
Group 17: F = 5.10, Cl = 6.12, Br = 9.03, I = 11.61
http://www.knowledgedoor.com/2/elements_handbook/clementi-raimondi_effective_nuclear_charge.html
Quite clearly Zeff increases down a chemical group because of the presence of d and, in the case of Pb, f e⁻s.
Rmax = valence shell orbital radii (e.g., http://www.webelements.com/lithium/orbital_properties.html)
Element   Zeff    Rmax          IP kJ mol^-1
Li           1.28    164.1     520.2
Na           2.51    179.4   495.8
K           3.50    230.0   418.8
Rb           4.98    249.0   403.0
Cs           6.36    282.0   375.7

The reason for the decrease in IP going down a chemical group I believe is because the valence e⁻ is further from the +ve nucleus as rationalized by the SWE.  (I am not aware of any treatment like this in a textbook, but I haven’t looked carefully recently.)

[Bryan Sanctuary]

First we need to define “effective nuclear charge”.  Indeed as some of the answers say as electrons are added to an atom, they shield the nucleus.  The first electrons are held more tightly and as more are added they are less tightly held.  The outer electrons are most weakly held and these are the valence electrons.  At most energies we deal with, these are the ones that jump around and lead to most of the chemistry.  The more tightly held electrons form the core.

If you remove electrons from an atom, each one has an ionization energy.  Remove the first, easy.  Then the other electrons re-arrange and the second is harder to remove.  When the first core electron is removed, there is a big jump in ionization energy.  See image of Ionization energy of Silicon.  Si has four valence electrons.  Note the big jump in removing the fifth electron.  That is where the Noble gas core starts.  Same for other atoms.

The core of electrons around an atomic number Z is the Noble Gas core.  The noble gas core is NOT the same as a noble gas.  For example Argon has 18 electron and not very reactive.  However the Noble gas core of Calcium is 18 electrons with Z = 20.  Hence two more protons than Argon.  Hence the Noble gas core of Calcium pulls the electrons in more tightly than Argon.

To see this look at the image attached of Argon vs the Argon core of electron for Calcium.  The latter is much more tightly bound.

So you can figure out the trends in ionization energies by thinking about shielding.  As you move down a group in the Periodic table, there are more electrons to shield the outer electrons.  Also the outer electrons are further from the nucleus.  So it is easier to ionize moving down a group.
 
Same argument moving across a period—as Z increases, so does the number of electrons, the shielding is greater and the electrons are further away.  Hence across a period the Ionization Energy increases.  There are exceptions due to stability of half-filled orbitals, but it all comes down basically to shielding.

In doing quantum calculation, it is a huge approximation to consider only the valence electrons.  For Calcium, for example, it is a good approximation to use Z_eff for the argon core which is a bit bigger that Z = 18, and consider two electrons moving in this effective positive charge.  This is a problem that can be solved accurately.  Imagine if you had to consider all 20 electrons.

Most of the trends in the periodic table can be argued through shielding.  For example as more electrons are added and less weakly bound than previously added electrons, means atoms get bigger across the table.  If you look at cations, then there are fewer electrons than Z and hence those electrons are pulled in more tightly and cations are smaller that their atoms.  Vice versa for anions.

Penetration:  finally if you look at the shapes of orbitals, then spherical s orbitals keep electrons at bay and they cannot penetrate much.  A noble gas core is also spherically symmetric and not much penetration.  However if you look at the p, d, f orbitals, see attached for p orbitls, then these all have nodes at the nucleus.  This means that electrons can penetrate deeply between these nodes.