It seems to me that the addition of electrons and protons as you move across a period would cause an atom to become larger. However, I'm told it gets smaller. Why is this?
-
1$\begingroup$ A nice periodic chart showing the effect you mention can be found here. It's worth noting that compared to the halogens, the noble gases are quite large. If you think in terms of satisfying shell requirements, that's actually a bit counter-intuitive. Also, while in chemistry one usually speaks in terms of filling shells, the deeper reason why electrons sometimes seem to repel each other more than they are attracted to protons is because of Pauli exclusion of identical-state fermions. $\endgroup$– Terry BollingerApr 26, 2012 at 0:23
-
1$\begingroup$ Where did You read about that fact, without finding the explanation for it? $\endgroup$– GeorgApr 26, 2012 at 20:26
-
$\begingroup$ Chemistry and physics have developed different ways of describing phenomena that are really the same thing. Pauli exclusion, played against charge attraction by a point-like positive nucleus and the low mass (high quantum location uncertainty) of electrons, is what produces the complex patterns and geometries that chemistry summarizes as "shells." Shells are very convenient, since they exhibit simple rules (e.g. 8 electrons "fill" a shell) that are much easier to deal with than the lower-level exclusion analysis. Nonetheless, for a question like this, you need to drop to that deeper level. $\endgroup$– Terry BollingerApr 27, 2012 at 2:11
-
$\begingroup$ Also: I'm serious about the sudden jump from the very small size of a fluorine atom to the far larger size of neon -- sort of like a baseball expanding to beach ball size -- and yet becoming hugely more stable is quite surprising in may ways. If electrons are extremely happy to reach 8 (vs 7) in a group, shouldn't they show it by doing something like getting into a tight little sphere? Instead, just the opposite happens! I've never seen that explained. I may even ask that one over in Physics just to see if someone knows why "stability = bloat" and "incomplete = tiny". $\endgroup$– Terry BollingerApr 27, 2012 at 2:21
-
1$\begingroup$ Georg, you are correct. The table I used apparently was based on binding radii, which are bogus in this context. The discussion of this over in Physics S.E. can be found on this link. The less visual but correct neutral atomic radii table can be found here. Note that in this table, element 10 (neon) is correctly listed as smaller in radius than element 9 (fluorine). $\endgroup$– Terry BollingerApr 28, 2012 at 6:42
3 Answers
As you move from left to right across a period, the number of protons in the nucleus increases. The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons).
As you move down a column, there are more protons, but there are also more complete energy levels below the valence electrons. These lower energy levels shield the valence electrons from the attractive effects of the atom's nucleus, so the atomic radius gets larger.
-
6$\begingroup$ Can you explain why "this attraction is much stronger than the relatively weak repulsion between electrons"? Why are the repulsions weak? $\endgroup$– YannApr 25, 2012 at 18:30
-
$\begingroup$ Just as a side note, there can also be more complete energy levels as one moves left across a row (the $d$ and or $f$ shells can get filled) but these will NOT be the outer shell. $\endgroup$– soandosApr 25, 2012 at 18:31
-
$\begingroup$ @Yann, I believe that it has to do with the basic principles of magnetism, where the distance between "individual" electrons in the cloud is large, and their individual charges are small, where the protons can "combine" their charges, allowing it to have far more strength over the distance. $\endgroup$– soandosApr 25, 2012 at 18:34
-
$\begingroup$ The forces are the same. the fierce repulsion of electrons is why you cannot put your hand easily thru a door. The arrangement of electrons in an atom minimizes distance to the nucleus while maximizing distance between the electrons the two electrons in a 1s2 configuration, hydride ion, each think that they have the nucleus to themselves, but we know better. That is why hydride ion is larger than the H atom and a halogen ion is larger than a halogen atom $\endgroup$– jimchmstOct 3, 2022 at 1:43
Remember, the 'size' of an atom has nothing to do with the size of the nucleus. It has to do with the size of the valence shell (which itself is not well-defined*).
So, if we neglect change in electrical attraction, the size should stay the same—a shell is a shell and it need not 'expand' to accomodate electrons.
Now, as we add more protons and electrons, the attraction between the nucleus and shell increases and the shell contracts. Thus the atom gets smaller.
*Shells reach 'til infinity, so it's better to define the size on the basis of 'the electrons of the outermost shell will be within this region x% of the time.' This only changes this answer a tiny bit. Now, we say 'the probability of finding the electrons closer to the center increases due to increased nuclear charge.'
-
$\begingroup$ Your ultimate and penultimate paragraphs are trying to emulate electrons one cannot enlarge and should stay the same the other is contracting. which is it? perhaps a bit of both. the additon of d and f electrons result in reasonably constant sizes so rather than constant E level sizes the two effects almost balance. $\endgroup$– jimchmstOct 3, 2022 at 1:57
There is an important tool developed to answer this question, which is effective nuclear charge, which is the effective nuclear charge experienced by an electron, as an electron in an outer shell will always be shielded by inner electrons.
So, as we move from left to right the number of electrons in outer shells increases whereas the number of electrons in inner shells remains the same.
Hence, effective nuclear charge increases, forcing electrons to come closer.