What is an explanation as to why the ionization energy changes as it does for other periods and groups?

#"(i) nuclear charge........"#it should be HARDER to remove an electron from an atom of higher atomic number, #Z#, because of greater electrostatic attraction.......

#"(ii) shielding by other electrons........"#unfilled electronic shells shield the nuclear charge VERY ineffectively. The result is that as we move across a row, a Period, of the Periodic Table, ionization energy INCREASES substantially. This is also reflected by the Periodic decrease in ATOMIC size in the same direction. As we fill an electronic shell, shielding of the nuclear charge becomes more effective, and this is responsible for the electronic structure of a given atom, as the valence electrons move out to large radii.

So what you gots to remember. Ionization energy INCREASES across a Period, from left to right as we face the Table. Ionization energy DECREASES down a Group, a column of the Periodic Table.

Remember we interrogate the reaction.......

#M(g) + Delta rarrM^(+)(g) + e^(-)#.

And as chemists, as physical scientists, we should consider actual data......

Does this graph support what I have argued? Why or why not? Why should the Noble Gases have the largest ionization energy of their Period? The #"y-axis"# has units of electron volts.

Note that it might seem that I suggest that these ionization energies are a result of the position of the element in the Periodic Table. It is more correct to say that the modern Periodic Table reflects the physical and chemical properties of the individual elements. Metals tend to be electron-rich materials, that tend to be OXIDIZED, i.e. lose electrons, and their position on the Periodic Table (to the left as we face it!), reflects this. On the other hand, non-metals, say fluorine, and oxygen, tend to be reducing, i.e. they tend to accept electrons. Fluorine, in particular is the most oxidizing element on the Periodic Table:

#1/2O_2(g)+2e^(-) rarr O^(2-)(g)#

#1/2F_2(g)+e^(-) rarr F^(-)(g)#