Looking at Figure 1, we see the first four ionization energies for three distinct elements on the third peroid. We can now disregard every element except those on the third peroid. Looking at the ionization energies of element Z, we see a massive jump in energies between IE₂ and IE₃.

The jump in energies derives from the fact that the second ionization energy removes the final valence electron from the element, and the third ionization energy begins to remove core electrons. Removing core electrons takes more energy than removing valence electrons. The jump in energies shows the transition of removing valence electrons to removing core electrons. We then look at the periodic table and look for an element on the third period which has two valence electrons (Group 2), giving us the answer of magnesium.

We see that element Z is magnesium due to its ionization energy discrepancy between IE₂ and IE₃. It is in the second group, therefore making it an alkali earth metal. Remember that elements in the same group exhibit similar chemical properties.

Looking at Figure 1, we see the first four ionization energies for three distinct elements on the third peroid. We can now disregard every element except those on the third peroid. Looking at the ionization energies of element Y, we see a massive jump in energies between IE₃ and IE₄.

The jump in energies derives from the fact that the third ionization energy removes the final valence electron of the element, and the fourth ionization energy begins to remove core electrons. Removing core electrons takes more energy than removing valence electrons. The jump in energies shows the transition of removing valence electrons to removing core electrons. We then look at the periodic table and look for an element on the third period which has three valence electrons (Group 3), giving us the answer of aluminum.

Looking at Figure 1, we see the first four ionization energies for three distinct elements on the third peroid. We can now disregard every element except those on the third peroid. Looking at the ionization energies of element X, we see a massive jump in energies between IE₁ and IE₂.

The jump in energies derives from the fact that the first ionization energy removes a single valence electron, and the second ionization energy begins to remove core electrons. Removing core electrons takes more energy than removing valence electrons. The jump in energies shows the transition of removing valence electrons to removing core electrons. We then look at the periodic table and look for an element on the third period which has one valence electron (Group 1), giving us the answer of sodium.

All these elements have shells more than half-full, so they will tend to gain electrons.  and  only require 2 electrons, versus 3 electrons for  and , to attain a stable full shell of electrons. Due to the energy requirement of pulling an electron, it is easier to pull 2 electrons than it is to pull 3. Between  and ,  has a smaller atomic radius; its nucleus will therefore more easily pull electrons towards itself. The effect of smaller atomic radius overrides the fact that  has a larger nucleus.

Electronegativity is the tendency of an atom to pull an electron. Halogens are most likely to pull an electron because they need only 1 to complete a stable, full valence shell, or "noble gas configuration". Noble gases already have full valence shells and do not tend to gain or lose electrons, while alkali metals and transition metals tend to lose electrons to attain full or half-full shells.

Each of these species has the same electron configuration (that of ). The chloride nucleus has the fewest number of protons, meaning a weaker attractive force between it and the electron cloud. As a result, the atom is more disperse than the others which have larger atomic numbers (and more protons) with the same number of electrons.

From left to right on the periodic table, atomic radius decreases because the number of protons increase while electrons are being added to the same shell; the attraction between the nucleus and the electron cloud override the shielding effect of adding electrons to the same shell. Therefore, those on the left are larger. From top to bottom on the periodic table, atomic radius/size increases because electrons are added to new valence shells significantly further from the nucleus than the previous shell. Therefore, those on the bottom left are largest.

From left to right on the periodic table, atomic radius decreases because the number of protons increase while electrons are being added to the same shell; the attraction between the nucleus and the electron cloud override the shielding effect of adding electrons to the same shell. Therefore, those on the left are larger. From top to bottom on the periodic table, atomic radius/size increases because electrons are added to new valence shells significantly further from the nucleus than the previous shell. Therefore, those on the bottom left are largest. Sulfur is the largest since it is in period 3. The remaining atoms are in period 2, with lithium being the largest, followed by nitrogen, oxygen, and fluorine.

The tendency to lose an electron is related to ionization energy, which is principally the opposite of electronegativity. The higher the ionization energy, the harder it is to pull an electron from the atom (takes more energy to ionize). Elements on the right side of the periodic table have higher ionization energies because they tend to gain electrons to achieve a full valence electron shell rather than lose them. Elements on the left side of the periodic table have lower ionization energies because they tend to lose electrons in their valence shell to achieve a full shell. Alkali metals need to lose 1 electron to achieve noble gas electron configuration, whereas the first electron lost from an alkaline earth metal does not confer this stability.