As you move across a single period (row) on the periodic table, the atomic radius of each successive atom decreases. As you move down a single group (column), the atomic radius of each successive atom increases. As you move down a group, the maximum energy level of the valence shell increases, thus increasing the size of the electron cloud and atomic radius. As you move across a period to the right, the number of protons in the nucleus increases. This also increases the attraction between the positively-charged nucleus and negatively-charged electrons, pulling the electrons in tighter and reducing the atomic radius.

The smallest atoms are going to be located toward the upper right of the periodic table. Of our given answer choices, fluorine is the closest to the upper right, and thus has the smallest radius.

Atomic radius will decrease as you move to the right, because the atomic number of the element will be increasing. This results in a more positively charged nucleus that pulls the electrons closer to the center. As a result, atomic radius will notably decrease from left to right.

Electronegativity, ionization energy, and electron affinity all increase to the right of the periodic table.

Within the same period of the periodic table, atomic radii decrease as there are more charged particles to attract one another, and within the same group, atomic radii increases. The increase from the ascending group generally speaking is larger than the decrease down a period.

Four of the elements listed are within the same period, so we will place those four elements in order of decreasing atomic radii:

Now we simply have to place Neon, Fluorine, and Oxygen, which are in the same period. The trend of decreasing radii with increasing atomic number is not true for noble gases, as they have a complete octet and are slightly larger to offset electron-electron repulsion from the octet. In order of decreasing atomic radius:

The increase from the octet is less than the increase from electron-electron repulsion.

The alkali metals are found in the first group (column) of the periodic table, on the leftmost side. They have only 1 loosely bound electron in their outermost shells, and their effective nuclear charge values are low, giving them the largest atomic radii of all the elements in their periods.

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.