The atomic volume is a relative value that indicates the relationship between the molar mass of an element and its density. So, this volume depends on the density of the element, and the density depends in turn on the phase and how the atoms are arranged within it..
So the atomic volume for an element Z is not the same in a phase other than the one it exhibits at room temperature (liquid, solid or gaseous), or when it is part of certain compounds. Thus, the atomic volume of Z in compound ZA is different from that of Z in compound ZB.
Why? To understand it, it is necessary to compare atoms with, for example, marbles. The marbles, like the bluish ones in the image above, have a very well defined material boundary, which can be seen thanks to their shiny surface. In contrast, the boundary of the atoms is diffuse, although they can be considered remotely spherical.
Thus, what determines a point beyond the atomic boundary is the null probability of finding an electron, and this point may be further or closer to the nucleus depending on how many neighboring atoms interact around the atom under consideration..
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By interacting two H atoms in the H moleculetwo, the positions of their nuclei are defined as well as the distances between them (internuclear distances). If both atoms are spherical, the radius is the distance between the nucleus and the fuzzy boundary:
In the image above you can see how the probability of finding an electron decreases as it moves away from the nucleus. Then dividing the internuclear distance by two, the atomic radius is obtained. Next, assuming a spherical geometry for the atoms, the formula is used to calculate the volume of a sphere:
V = (4/3) (Pi) r3
In this expression r is the atomic radius determined for the H moleculetwo. The value of V calculated by this imprecise method may change if, for example, H is consideredtwo in liquid or metallic state. However, this method is very inaccurate because the shapes of the atoms are far from the ideal sphere in their interactions.
To determine the atomic volumes in solids, many variables concerning the arrangement are taken into account, and which are obtained by X-ray diffraction studies..
Molar mass expresses the amount of matter that has a mole of atoms of a chemical element.
Its units are g / mol. On the other hand, the density is the volume that one gram of the element occupies: g / mL. Since the units of atomic volume are mL / mol, we have to play with the variables to arrive at the desired units:
(g / mol) (mL / g) = mL / mol
Or what is the same:
(Molar mass) (1 / D) = V
(Molar mass / D) = V
Thus, the volume of one mole of atoms of an element can easily be calculated; whereas with the formula of the spherical volume the volume of an individual atom is calculated. To arrive at this value from the first, a conversion is necessary through Avogadro's number (6.02 · 10-2. 3).
If the atoms are considered spherical, then their variation will be the same as that observed in atomic radii. In the image above, which shows representative elements, it is illustrated that from right to left the atoms get smaller; instead, from top to bottom they become more voluminous.
This is because in the same period the nucleus incorporates protons as it moves to the right. These protons exert an attractive force on the external electrons, which feel an effective nuclear charge Zef, less than the actual nuclear charge Z.
The electrons of the inner shells repel those of the outer shell, reducing the effect of the nucleus on them; this is known as the screen effect. In the same period, the screen effect cannot counteract the increase in the number of protons, so the electrons in the inner shell do not prevent the atoms from contracting..
However, descending into a group enables new energy levels, which allow the electrons to orbit further from the nucleus. Likewise, the number of electrons in the inner shell increases, the shielding effects of which begin to be diminished if the nucleus adds protons again..
For these reasons, it is appreciated that group 1A has the most voluminous atoms, unlike the small atoms of group 8A (or 18), that of the noble gases..
Transition metal atoms incorporate electrons into inner d orbitals. This increase in the screen effect and, as well as in the real nuclear charge Z, cancel out almost equally, so that their atoms retain similar size in the same period..
In other words: in one period, transition metals exhibit similar atomic volumes. However, these small differences are enormously significant when defining metallic crystals (as if they were metallic marbles).
Two mathematical formulas are available to calculate the atomic volume of an element, each with its corresponding examples.
Given the atomic radius of hydrogen -37 pm (1 picometer = 10-12m) - and cesium -265 pm-, calculate their atomic volumes.
Using the formula for the spherical volume, we then have:
VH= (4/3) (3.14) (37 pm)3= 212.07 pm3
VCs= (4/3) (3.14) (265 pm)3= 77912297.67 pm3
However, these volumes expressed in picometers are exorbitant, so they are transformed into units of angstroms, multiplying them by the conversion factor (1Å / 100pm)3:
(212.07 pm3) (1Å / 100pm)3= 2.1207 × 10-4 TO3
(77912297.67 pm3) (1Å / 100pm)3= 77.912 Å3
Thus, the differences in size between the small H atom and the bulky Cs atom are numerically evidenced. It must be borne in mind that these calculations are only approximations under the statement that an atom is totally spherical, which wanders in front of reality.
The density of pure gold is 19.32 g / mL and its molar mass is 196.97 g / mol. Applying the M / D formula to calculate the volume of one mole of gold atoms, the following is obtained:
VAu= (196.97 g / mol) / (19.32 g / mL) = 10.19 mL / mol
That is, 1 mole of gold atoms occupies 10.19 mL, but what volume does a gold atom occupy specifically? And how to express it in units of pm3? For this, simply apply the following conversion factors:
(10.19 mL / mol) · (mol / 6.02 · 10-2. 3 atoms) · (1 m / 100 cm)3(1 pm / 10-12m)3= 16.92 · 106 p.m3
On the other hand, the atomic radius of gold is 166 pm. If both volumes are compared - the one obtained by the previous method and the one calculated with the spherical volume formula - it will be found that they do not have the same value:
VAu= (4/3) (3.14) (166 pm)3= 19.15 · 106 p.m3
Which of the two is closest to the accepted value? The one that is closest to the experimental results obtained by X-ray diffraction of the crystal structure of gold.
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