Periodic Trends - AP Chemistry

Atomic radius expands down the columns and to the LEFT.

Atomic radius INCREASES DOWN and TO THE LEFT in the periodic table

Electronegativity increases across a period (going right) and decreases down a group \[not including noble gases\]. The element closest to the top right is Phosphorous

If we look at elements of the same period the prinicipal quantum number for each one of these elements is the same. Thus the outermost energy level for the electrons of each atom is the same. If we consider moving through the periodic table in numerical order, the left side of the periodic table features atoms that have just begun to add electrons to new energy levels. On the right side of the periodic table the elements are moving closer to filling the energy level. The differences in atomic radii are the result of differing amounts of protons between atoms whose electrons are in the same energy level. K has one electron in the fouth energy level and Kr has eight electrons in the fouth energy level. K has 19 protons with which to generate pull on the electrons. Kr has 36 protons with which to generate pull on the electrons. Thus Kr has a smaller atomic radius becasue of its ability to have a tighter grasp on its electrons becasue of the stronger charge generated by its nucleus.

Electronegativity measures the ability an atom to attract shared electrons
in bond. It follows a trend that it increases moving from left to right and down to up across
the periodic table.

The trend for radius is that it increases down and to the left

Atomic radius decreases as one moves across the periodic table from left to right, since effective nuclear charge increases and the electrons are held more tightly to the nucleus.

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.

The transition metals are capable of losing various numbers of electrons from the s and d orbitals of the valence shell. Metals such as Cu, Fe, and Mn have various oxidation states and can form many different ionic compounds.

Electronegativity increases as one moves across a period (row) from left to right, or up a group (column) from bottom to top. Following these trends, fluorine is the most electronegative element.

Effective nuclear charge increases as one moves across a period (row) from left to right, so Cl will have a higher effective nuclear charge. Anions of an element have a lower effective nuclear charge than their parent atom, since they have more electrons than protons and feel less of a pull from the protons in the nucleus. Thus, Cl will have the highest effective nuclear charge.

Ionization energy is the energy required to completely remove an electron from an atom. The electron that would be removed is the atoms most loosely held electron. Some atoms are more likely to give up an electron to get to a more stable electron configuration. This is the reason that Group I elements have the lowest first ionization energy because after loosing one electron, these elements have now acheived an octet configuration. The second ionization energy is the energy to remove a second electron from an atom. This will be lowest for the Group II elements because these elements acheive an octet configuration after loosing two electrons.

Electronegativity increases as one moves left to right across the periodic table, it also increases as one moves up the periodic table. This is because of the changes in the atomic radii of elements that occurs as one move in these directions. Flourine is the most electronegative element because the positive charge of its nucleus exerts the strongest attractive force on it's electrons when compared to the other elements. Conversely, the nuclei of the elements in the lower left hand side of the periodic table exert the weakest attractive force on their electrons when compared to the other elements. As one moves from left to right within the same period, one electron and one proton are added to each element. The electrons will not be any further away from the than the previous element but there will be one more proton with which to generate attractive force on the electrons. So the upper left hand elements have the most protons within their respective period with which to generate pull on the electrons. The electrons of these species are also the closest in proximity to the nucleii.

Ionization energy goes up across the period and down as you go down the group. This is the energy required to remove an electron. Of the listed elements it would be hardest to remove an electron from Flourine as the positive nucleus close to the electron orbitals has greater attractive force.

The atomic radius of elements decreases as one moves from left to right across the periodic table. The degree to which lithium has a larger atomic radius than fluorine is most similar to the difference between another pair of elements within the same groups, that are also found on the left and right sides of the table.

Fluorine and chlorine are both halogens, and lithium and sodium are both alkali metals. Fluorine and lithium are both in the second period, and sodium and chlorine are both in the third. As you move from lithium to fluorine across the table, you will see similar changes and patterns as if you were moving from sodium to chlorine.

Helium has the smallest atomic radius. This is due to trends in the periodic table, and the effective nuclear charge that holds the valence electrons close to the nucleus. Atomic radius decreases as you move across a period from left to right and decreases as you move up a group from bottom to top. Since He is at the upper right-hand corner of the table, it must have the smallest atomic radius.

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:

Na, hspace{1 mm}Mg,hspace{1 mm}S,hspace{1 mm}Cl

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:

O,hspace{1 mm}Ne,hspace{1 mm}F

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

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.

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.