# Periodic Trends

1. Feb 14, 2005

### JimmyRay

What are the periodic trends? I need help with them.. Regarding atomic mass, (first) ionization energy, electronegativity and anything else im missing.

Specially atomic mass, well as you go down a group and atomic number increases, atomic radius increases, ok this makes sense.. number of protons and electrons increase, the valence electron is screened by all the other electrons and thus ISNT pulled towards the center and you have a larger radius.... But why does it decrease as you go left to right in a period?

Electronegativity all I know is flourine is the most electronegative and all the elements near the upper right of the periodic table are more electronegative then the ones near the bottom left....

Anything else?

2. Feb 15, 2005

### Gokul43201

Staff Emeritus
The increase in radius down a group is due to the addition of a new shell each time. As you move across a period, you increase the number of protons and electrons. You increase the attraction to the nucleus but also the repulsion between the electrons. Calculations show that the increase in nuclear attraction slightly exceeds the increase in electron-electron repulsion, so increasing the atomic number (within a period) causes a reduction of size.

The trend in ionization potentials depends on the electronic configuration to a some extent but for the most part, follows from the atomic radius. The closer the valence electrons are to the nucleus, the greater is the nuclear attraction, which makes it harder to remove a valence electron (so the IE increases across the period and up a group). However, there are a few exceptions to this general trend. Fully filled and half-filled subshells are more stable, so it's harder to ionize them. Take the second period consisting of elements from $Li^3~(1s^2~2s^1)$ to $Ne^{10}~(1s^2~2s^2~2p^6)$. The first IEs (kJ/mol) are :

$$Li~(1s^2~2s^1) = 520$$
$$Be ~(1s^2~2s^2) = 899$$
$$B~(1s^2~2s^2~2p^1)=801$$
$$C~(1s^2~2s^2~2p^2)=1087$$
$$N~(1s^2~2s^2~2p^3)=1402$$
$$O~(1s^2~2s^2~2p^4)=1314$$
$$F~(1s^2~2s^2~2p^5)=1681$$
$$Ne~(1s^2~2s^2~2p^6)=2081$$

Clearly the trend is broken by Be (which has a fully filled 2s subshell) and N (which has a half-filled 2p subshell) which are harder to ionize because of this extra stability.

Unless someone else takes it from here, I'll get to electronegativity tomorrow.

3. Feb 15, 2005

### JimmyRay

Thanks a LOT that REALLLLY helped me out in terms of the atomic radius and ionization energy trends...

We are asked only to look at the general trends in this course so the subshell stuff about ionization energy was extra... But why does having a half filled p subshell make N harder to ionize?

If you could get to electronegativity it would also help.

Thanks again.

4. Feb 17, 2005

### Gokul43201

Staff Emeritus
Electronegativity is a measure of the tendency of an atom to attract an electron to its valence shell. The noble gases have octets in their valence shells and are pretty happy as a result, and have very low electronegativities (Kr and Xe are unexpected exceptions).

The tendency to attract an extra electron roughly :

(i) increases with the number of protons in the nucleus (even though there is a repulsion from as many electrons, the attraction is known to be slightly more dominant, as explained before)

(ii) decreases with the mean radius of the valence shell. The smaller the valence shell, the stronger the attraction from the nucleus.

As you go across a period, you are increasing the atomic number while decreasing the radius. Both effects help to increase the electronegativity (EN). So, in general, the EN increases across a period.

As you go down a group, you are increasing the number of protons, but also increasing the size of the atom. These two effects want to counter each other. However, since the electrostatic force is proportional to the square of the radius but only the first power of the nuclear charge (recall Coulomb's Law), the effect of the radius dominates. As a result, EN decreases down a group.

So, from the above trends higher ENs will be found at the top of a group and the far right of a period (do not include He, Ne, Ar). So, it's not hard to see why F is the most EN element and Fr is the least (barring the above 3 exceptions).

Look at the trends displayed at the top of this page. I like the "Ball Chart" display mode for that graphic.

A similar graphic can be found for the atomic radii .

5. Feb 17, 2005

### GCT

You need to consider effective nuclear charge in your text to explain the left to right trend, try searching for it through the index of your text and a discussion of this trend will probably follow with the introduction of Zeff ("nac").

Basically, as one moves across a row the outer electrons become experience a higher effective nuclear charge Zeff. Zeff is defined by the atomic number, Z (due to attraction to the nuclei and protons), however depending on the magnitude of electron shielding, Zeff is decreased correspondingly, that is the amount of pull the outer electron experiences(I don't recall the symbol for electron shielding at this moment). As you may have guessed, Zeff increases from left to right, due to the relative decreasing rate of increase of electron shielding compared to the steady linear increase of Z (atomic number).

One can understand all of this in observing the trend with respect to an outer shell. With the addition of the second electron to a s orbital in a specific shell, Zeff is increased; before, the electron experienced the full attraction of 1 proton (corresponding to its electron), now it applies to attraction to 2 protons, with some shielding effect (one electron "shields" another from experiencing the protons, some may argue that it is the nuclei that the electron is attracted to, however, let's just focus on protons in this case). The next electron, will be placed in a p orbital, this electron experiences less shielding from the s electrons due to less repulsion, the proportionate rate increase of electron shilelding is once again decreased, however Z is increasing at the same rate, in this case 3. I've simplified the situation to illustrat a concept.

Electrongativity does not really have specific paramaters and thus is not complete in the sense of a definition, it is related to some aspects such as those one's in which Gokul described, however there is no definitive relation between them. Linus Pauling had determined the concept and its utility (through statistical measures), it is basically the relative ability of an atom of a specific element to "hoard" the electrons in comparison to others.

6. Feb 17, 2005

### JimmyRay

I see, I dont think in this course we dive that deeply into electronegativity... basically my teacher just mentioned the general trend across the table... But its good to know, thank you.