Periodic Properties

Electron Shielding

A 1s electron spends more time near the nucleus than a 2s or a 2p electron, so the 2s electron is shielded from the nucleus by the 1s electron and is thus held less tightly by the nucleus than the 1s electron. Lithium has the electron configuration 1s²2s and the nucleus has a +3 charge. The 2s electron experiences a nucleus of charge +3 -[some portion of the -2 charges from the 1s electrons] whereas the 1s electrons experience roughly a full +3 charge. Therefore it is easier to remove the 2s electron rather than one of the 1s electrons.

For the same n value, the penetrating power of the electron decreases as l increases: s > p > d > f > .... Penetrating power refers to the ability of the electron to penetrate through the inner electrons to be closer to the nucleus. If the electron spends an appreciable portion of time near the nucleus then it is more difficult to remove it than for the electron which does not have as great a penetrating power. Therefore it is easier to remove a 2p electron than a 2s electron.

Removing an electron refers to forming an ion from the atom and this process is called ionization. The energy to remove the electron is called the ionization energy or the ionization potential. We can calculate the ionization potential for a one-electron species by using the energy result from the SE for a one-electron atom (to change the energy expression for the hydrogen atom into one for any one-electron atom we need only multiply by Z² where Z is the nuclear charge of the one-electron atom):

E = -13.6 eV*Z²/n² = -(1310 kJ/mol)*Z²/n²....eq 1

The above is then the energy of the electron in any one-electron atom. The ionization energy would then just be the negative of that electron's energy. For example,

I.E. for H = 1310 kJ/mol

I.E. for He⁺ = 2²/1² * 1310 kJ/mol = 5240 kJ/mol

Both of the above are for a n = 1 electron. Remember for a one-electron atom the energy values of all levels with the same n value are the same.

We can also use equation 1 in the reverse manner by using as input the experimental ionization energy for a particular electron and then solving for a Z_effective where

Z_eff = Z_actual - effect of inner electrons

The Z_eff will be the nuclear charge that the out electron experiences which is reduced by the inner electrons. Our equation 1 then becomes

E = -(1310 kJ/mol)*Z_eff²/n²

For example, the experimental ionization energy for the 1s electron in Na is 1.39 x 10⁵ kJ/mol and for the Na 3s electron it is 4.95 x 10² kJ/mol. Then for the 1s electron we have

1.39 x 10⁵kJ/mol = Z_eff²/1² * 1310 kJ/mol

or Z_eff = 10.3

Na has 11 electrons (1s²2s²2p⁶3s) so there are two 1s electrons. If one of the 1s electrons totally screened the nucleus for the other 1s electron then the 1s electron would experience a nuclear charge of 10. You see that since 10.3 is greater than 10, the other electron does not totally screen the nucleus for the other 1s electron. Another way to state this is that each 1s electron somehow penetrates the shielding of the other 1s electron to some extent.

Now let's look at the 3s electron. We have

495 kJ/mol = Z_eff²/3² * 1310 kJ/mol

or Z_eff = 1.84

From the electronic configuration we see that there are 10 electrons "below" the 3s electron and if they all totally screened the nucleus the 3s electron would experience a charge of 1 from the nucleus. But we see that the Z_eff is 1.84 rather than 1 so the 3s electron does penetrate the shielding of the other electrons and does spend some time nearer the nucleus than would be expected totally from it's n-value. The radial probability diagrams show this effect for you will see two nodes in the diagram for the 3s electron. The 2s electron has one node and the 1s electron has no nodes. The 3p electron has 1 node and the 3d electron has no nodes. See the pattern here? There are n-l-1 nodes in the radial probability diagrams.

Trends in size.

Now we are ready to understand some trends in size and ionization energy of atoms. The size of an atom is difficult to state because they do not behave like marbles. We can estimate the size of diatomic atoms because we can determine the bond distance between the two atoms and then estimate the size of each atom by dividing by two. Why would this estimate of the size of the hydrogen atom as determined from the bond length of H₂ be less than the estimate of the 99% radius of the electron distribution in the 1s orbital of the H atom? Sizes of other atoms can be estimated from compounds that they form.

When you then look at periodic table trends in sizes you see that size increases from right to left and increases from top to bottom. Why? Well going from the top of the periodic table to the bottom in a given column you see that the outer electron n-value is increasing. And what property of the atom does the n-value relate to? That's right, size. But what about increasing from right to left as we go across a row? Look at it as decreasing in going from the left to the right and see if that helps you understand. Remember that for the same n value, the penetrating power of the electron decreases as l increases: s > p > d > f. As we go from left to right in the main group elements, we are adding electrons with the same n-value but different l values. As the penetrating power of the electrons decrease, the outer electrons experience more of the nuclear charge and thus are drawn closer to the nucleus -- hence the size decreases as we move across the row. So now we understand size trends in the periodic table, right? Then as you look closer you see that the trend is not smooth and, in fact, there are some atoms in which the trend is reversed for them. We will not worry about these out-of-trend values at this point.

Trends in Ionization Energy

The first IE or IP is the energy required to remove the electron from an atom which is the easiest to remove. After removing that electron, the atom is now an ion with a positive charge so the remaining electrons are more difficult to remove. The energy required to remove the next easiest to remove electron is called the second IP. For magnesium we have the following so you can see considerable increase in the IP after the first electron is removed.

Mg(g) ---> Mg⁺(g) + e^- IE₁ = 738 kJ/mol

Mg⁺(g) ---> Mg⁺⁺(g) + e^- IE₂ = 1451 kJ/mol

Now look at Li and Na first IP:

Li(g) ---> Li⁺ + e^- IE₁ = 520 kJ/mol

Na(g) ---> Na⁺ + e^- IE₁ = 495 kJ/mol

Note that the first IP decreased in going from Li to Na which is going down a column in the periodic table. Also note that the first IP increased in going from Na to Mg which is going across a row in the periodic table.

The IP trends in the periodic table are easy to understand based upon all of the above discussion. As we go down a row in the periodic table the size is increasing so the outer electrons are getting further away from the positive nucleus. The further away from the nucleus they are, the easier they are to remove from the atom. So the IE required to remove the easiest electron to remove (the first IE) from an atom is decreasing as we move down a row in the periodic table.

Going across a row is a bit more complicated, but in general since we are decreasing in size as we go from left to right so the IP should increase -- and that is, in general, what we see. However look at:

Al(g) ---> Al⁺(g) + e^- IE₁ = 580 kJ/mol

Al⁺(g) --> Al⁺⁺(g) + e^- IE₂ = 1815 kJ/mol

In the first IP we see that the value decreased in going from Mg to Al rather than increasing as we just said. So what is going on? What type electron did we add in going from Mg to Al? That is right we added a 3p electron to the configuration [Ne]3s². So? Well the n value stayed the same but we increased the l value and that means that the penetration of the outer electron decreased. So? Well if the penetration decreased then the electron is mostly outside of the inner electrons and thus it is a bit easier to remove. But then as we continue across the row we see that the first IP for Si is greater than that for Al so our general trend now continues.

Trends in Electron Affinity

Affinity means an attraction for doing something. I have an affinity for KrispyKreme doughnuts. Some atoms have more of an affinity for attracting an electron to themselves than other atoms. Fluorine will form the F^- ion and S will form the S^-- ion. What is happening to the electron configurations of these ions? That is right, they are forming the Noble Gas configuration which has all of the subshells full. Some books will define the electron affinity (EA) of an atom to be

X(g) + e^_ ---> X^-(g)....EA is negative

and some will define it to be

X^-(g) ---> A(g) + e^-....EA is positive

If you are also using a textbook, be careful to determine which nomenclature is used. I will talk about trends in the absolute value of the EA to get away from this controversy since what we are interested in here is trends. When we look at trends in EA we see the same general results as for IP: the absolute value of the EA increases as we go across a row and it increases as we go down a column. The similarity in trend to that of IP is easier to understand with the second convention for with that convention both process are removing an electron and the same reasoning as above applies here.

Ionic Sizes

What happens to the size of a species when we add an electron? That is right, it increases. Why? We have not added another proton so the nuclear charge has not changed, but we have added an extra electron which sees a very shielded nucleus. Therefore the added electron is not very strongly attracted to the nucleus and it's average distance from the nucleus is greater than the original radius.

What happens to the size of a species when we take away an electron? Again we have not changed the nuclear charge so now we have more protons than electrons so that the electrons are strongly attracted to the nucleus and the size decreases from the original radius.

Paramagnetism and Diamagnetism

A substance which is repelled by an applied magnetic field is said to be diamagnetic. A substance which is attracted to an applied magnetic field is said to be paramagnetic. These are usually rather weak forces. Substances which have a very strong attraction to a magnetic field are called ferromagnetic. The cause of the paramagnetic and ferromagnetic phenomena is the presence of unpaired electron spins in the atom. Atoms which are diamagnetic have no unpaired electrons. You can now predict which atoms will be paramagnetic and which will be diamagnetic.

Which of the following atoms are diamagnetic? Na, N, Ca

Which of the following atoms are paramagnetic? Be, Ne, O

Now take a practice quiz to help you understand if you understand the basic concepts.

You must use your real name when it asks for a name.

The test will only submit when you have answers all of the questions correctly.

If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.