Gchem Lecture 1: Periodic table Trends

The worst way to learn periodic trends (or anything else in the PS portion of the MCAT) is to memorize them. This is because memorization without understanding will only allow you to solve problems that are asked in the same format that you have memorized. However, the MCAT is infamous for forcing you to take information that you've learned in your college classes and apply it to new scenarios with which you have no previous experience.

The periodic trends that you should know for the MCAT are atomic radius, ionization energy, electron affinity, and electronegativity. We will consider each one in turn in a second post, but first, we must consider a fifth trend upon which the other four all depend, which is the effective nuclear charge, Z.

Effective Nuclear Charge (Z or Zeff): This trend is probably the hardest for students to understand, but once you do understand it, all of the other trends will seem very logical and intuitive to you. Z is basically a book-keeping method, where we cancel out each core electron with one of the protons in the nucleus. (Core electrons are the ones in the inner shells; in other words, they are the non-valence electrons.) We do not cancel the valence electrons. (Remember that valence electrons are the ones in the outer shell.) When we have completed this exercise, we will find that, for a neutral atom, we are left with the same number of uncancelled protons in the nucleus as we have valence electrons. The number of uncancelled protons is equal to the value of Z, and it will be the same as the group number for a neutral atom.

The reason why we only cancel out core electrons is that these electrons are considered to be shielding the valence electrons. This means that they block the valence electrons from easily having access to the nucleus. Remember that the nucleus is positive, and all electrons are negative. Since unlike charges attract, the valence electrons are attracted to the nucleus. But the problem is that there are some core electrons in the way, and they too are negatively charged. Since like charges repel, core electrons will repel the valence electrons, somewhat countering their attracting to the nucleus.

Let's look at some actual atoms to make this concept clearer.

Calculating Z for Li: We'll start with lithium, which has three protons and three electrons as a neutral atom. If you begin to write the electron configuration of lithium, where do the electrons go? Well, the first two go into the 1s orbital. These are two core electrons. The third goes into the 2s orbital, and it's a valence electron. Doing our book-keeping, we find that we have cancelled two protons for our two core electrons, and this leaves us one proton left over. Thus, the value of Z for Li is approximately 1.

Calculating Z for Be: The next element after Li is beryllium, which has four protons and four electrons as a neutral atom. Its configuration is 1s2, 2s2, which means that again we have two core electrons as with Li. However, we now have two valence electrons, and when we have finished cancelling our core electrons with two of the protons, we now have two protons left over. Thus, the value of Z for Be is approximately 2. (Note that in doing this, we are assuming that electrons in the same shell do not shield one another. That is, we are assuming that the two valence electrons in Be do not shield one another. Technically that is not exactly true, but for our purpose, the effect of electron-electron repulsion by electrons in the same shell can be ignored.)

Calculating Z for B: After Be comes boron, which has five protons and five electrons. Its configuration is 1s2, 2s2, 2p1. As we have already seen, we have two core electrons from the 1s-orbital, which we can cancel with two of the five protons. This leaves us with three protons left over, and three valence electrons. (Again, we are assuming that the 2s electrons do not shield the 2p electron, but technically this is not completely true.) So now our Z value is 3.

Calculating Z for the rest of the elements in row 2: As we continue to fill the 2p orbitals, one electron at a time, we will find that the Z value continues to increase by one as we move from left to right. Here are the Z values for the rest of the elements in row 2:

Element Z-value
C.............4
N.............5
O.............6
F.............7
Ne............8

Calculating Z for the Group I Metals: Now, let us consider what happens to Z when we go down a group. We'll use Group I, the alkali metals. We've already seen that Li has a value of 1 for Z. What about Na, right below Li? Well, sodium has 11 protons and 11 electrons as a neutral atom. Its configuration is 1s2, 2s2, 2p6, 3s1. All of the electrons in the second shell are now core electrons; only the one electron in the 3s orbital is a valence electron. If we cancel out the ten core electrons with ten of the protons, we are left with one proton, just as with Li. So the value of Z for Na is also approximately 1, just as it is for Li. What about K, element 19? Its configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 as a neutral atom. Just as before, there is only one valence electron. The other 18 electrons are all part of the core. So they each cancel a proton, leaving us with 1 proton left over, and a Z value of about 1. Clearly, then, ALL of the alkali metals have a Z value of approximately 1. This same pattern will hold true for each of the other main groups in the periodic table.

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Here is a summary of the trends for Z.

As we travel from left to right across the periodic table, Z increases. This occurs because we are adding more protons as we move from left to right, but we are not adding more shielding core electrons.

As we move down a group from top to bottom, Z stays approximately equal. This occurs because we are not changing the net number of protons left over after we have cancelled out all of the shielding core electrons.

This post was compliments of QofQuimica on SDN.