# Why does being “bonded” decrease atomic size?

I have a furiously overwhelming question-

Why is the atomic size of noble gases larger than the atomic size of the previous halogen when atomic sizes decrease across a group?

I did search it on the net and asked my teacher but everyone says that because noble gases are usually not bonded.

But:

Why does being "bonded" decrease atomic size?

Also we are talking about the atomic sizes of isolated elements?

• What data are you using? According to this Wikipedia article the calculated atomic radius is smaller for a noble gas than it is for the preceding halogen, just as you would expect. – ron Jan 18 '17 at 18:42
• sorry I mean the "previoius halogen", I edited the question. :) – MartianCactus Jan 19 '17 at 7:43

The first thing you should realize is that there are various definitions of the size of the atom. One distinguishes for example the covalent radius, the Van der Waals radius and the ionic radius. The covalent radius is based on the binding of atoms into molecules and on the resulting bond length. The van der Waals radius is half the minimum distance between two atoms that not form a covalent bond and the ionic radius is derived from crystal structures of ions in salts. The covalent radius cannot be determined for noble gasses as they do not bind under normal conditions. The Van der Waals radius however can be determined and since the noble gasses are not very polarizable, the Van der Waals radius does increase with respect to the preceding halogen. You'll find more in the wikipedia article as mentioned by ron.

• I dont really understand what the van der walls force is...why would you take the half of the distance between 2 non bonded elements? – MartianCactus Jan 19 '17 at 7:58
• van der Waals forces are actually attractive forces due to dispersion of electron clouds, but they do not play a role in this discussion. The point here is one of the assumptions of the van der Waals gas equation: atoms occupy space and two of them can't occupy the same volume at the same time; so if you model them as perfect spheres, the closest they can get is actually just twice the radius of that sphere. If both atoms are of the same element, this radius is the van der Waals radius for this element. – Felipe S. S. Schneider Jan 19 '17 at 22:56

As Paul has pointed out, there are many different ways of defining atomic radii. ron has also pointed out that the trend goes in the opposite direction from what you said. For instance, according to Ptable,

|Calculated radii |
|  |r(pm)|  |r(pm)|
|H | 53  |He| 31  |
|F | 42  |Ne| 38  |
|Cl| 79  |Ar| 71  |
|Br| 94  |Kr| 88  |


I have added hydrogen here as the natural "halogen" for helium. Empirical and covalent also follow the same trend as above. But the van der Waals radii, as shown below, do not,

|    vdW radii    |
|  |r(pm)|  |r(pm)|
|H | 120 |He| 140 |
|F | 147 |Ne| 154 |
|Cl| 175 |Ar| 188 |
|Br| 185 |Kr| 202 |


I am assuming this is the data set you've mentioned. The reason for the disagreement is that electron clouds may relax or contract depending on their chemical environments and so do atomic radii. Thus, what this shows is that:

1. H, F, Cl and Br are larger than He, Ne, Ar and Kr when bonded.
2. H, F, Cl and Br are smaller than He, Ne, Ar and Kr when not bonded;

The reason for the above is the following:

Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell. However, at the same time, protons are being added to the nucleus, making it more positively charged. The effect of increasing proton number is greater than that of the increasing electron number; therefore, there is a greater nuclear attraction. This means that the nucleus attracts the electrons more strongly, pulling the atom's shell closer to the nucleus. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases.

What is being described is also called shielding effect.

2. For van der Waals radii (source),

[While] van der Waals radius is used to define half of the distance between the closest approach of two non-bonded atoms of a given element.

[...]

[Based on the ideal gas law above], van der Waals equation takes the molecular size and the molecular force into account. As a result, the attractive force (a/V2) is added into the pressure part. Likewise, the volume is subtracted by the molecular volume (b), which is determined by the van der Waals radius.

Here the shielding effect is much less important. van der Waals radii are actually used to express how much volume each atom occupies on its own. It is a minimum interatomic contact distance, less than that the atoms are somehow bonded. And here the most important effect is plain electrostatic repulsion between atoms.

Thus it seems reasonable to me that it is harder to put two Kr atoms close together than two Br atoms, simply because Kr has more electrons and the electrostatic repulsion will be stronger between them.

• and what do you mean by "those atoms"? If you mean He, Ne, Ar and K then thats my question. Why are they smaller when those atoms are not bonded? – MartianCactus Jan 19 '17 at 7:46
• I edited the original post to clarify it. – Felipe S. S. Schneider Jan 19 '17 at 11:45