Let us know about Atomic Volume Change in Periodic Table. 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 in turn depends on the phase and the arrangement of the atoms within it.
So the atomic number for a Z element does not differ from the one in the second phase that it exhibits at room temperature (liquid, solid or gas), or when it is part of certain compounds. Thus, the atomic volume of Z in compound ZA is different from that of Z in ZZ.
Why? To understand this, it is necessary to compare atoms, for example, stone. Marbles, like the blue color of the better image, have well defined material boundaries thanks to its brilliant surface. In contrast, the boundaries of atoms are diffuse, although they can be considered remotely spherical.
Thus, which determines a point beyond the atomic boundary, there is a null probability of finding an electron, and this point may be further or closer to the nucleus depending on how many atoms the atom considers.
Atomic Volume and Radius
The interaction of two H atoms in an H molecule defines the position of their nuclei as well as the distance (internal distance) between them. If both the atoms are spherical, then the radius is the distance between the nucleus and the diffusion limit:
In the upper image it can be seen how the probability of finding an electron decreases as we move away from the nucleus. Dividing the internal distance between the two, the atomic radius is obtained. Next, assuming a spherical geometry for the atoms, we use the formula to calculate the volume of a sphere:
V = ( 4/3 ) (Pi) R3
In this expression R is the atomic radius determined for the H molecule . The value of V calculated by this imprecise method may change, for example, it was assumed to be H 2 in the liquid or metallic state. However, this method is highly inaccurate because the shapes of the atoms are far from the ideal region in their interactions.
To determine the atomic quantity in a solid, several variables concerning the arrangement are taken into account, and they are obtained by X-ray diffraction studies.
additional formula
The molar mass expresses the amount of matter that contains atoms of a chemical element.
Its units are g/mol. Density, on the other hand, is the amount that one gram of an element occupies: g/ml. Because the units of atomic volume are mL/mol, you’ll need to play around with the variables to get to the units you want:
(g / mol) (mL / g) = mL / mol
or what is similar:
(molar mass) (1 / D) = V
(molar mass / d) = V
Thus, the amount of one mole of atoms of an element can be easily calculated; Whereas the volume of an individual atom is calculated with the formula for spherical volume. To reach this value beforehand, a transformation through Avogadro’s number (6.02 10) is necessary. -23 ).
How does the amount of an atom vary in the periodic table?
If the atoms are assumed to be spherical, their diversity will be similar to that seen in the atomic rod. In the upper image, which shows the representative elements, it is illustrated that dwarf atoms from right to left; Instead, they become more shiny from top to bottom.
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 outer electrons, which feel an effective nuclear charge Z eff , less than the actual nuclear charge Z.
The electrons in the inner layers repel those in the outer layers, which are less affected by the nucleus; This is known as the screen effect. In the same period the screen effect does not manage to counteract the increase in the number of protons, so the electrons in the inner layer do not inhibit the contraction of the atoms.
However, by descending into a group, new energy levels are enabled, which allow electrons to orbit further away from the nucleus. In addition, the number of electrons in the inner layer increases, whose shielding effect begins to diminish if the nucleus re-adds protons.
For these reasons it can be seen that group 1a has the largest number of gas atoms, in contrast to the smaller atoms of group 8a (or 18), which belong to the noble gases.
Atomic Quantities of Transition Metals
Atoms of transitional metals incorporate electrons into the inner orbitals d. This enhancement of the screen effect, as well as the actual nuclear charge Z, is almost uniformly canceled out, so that their atoms maintain their same size over the same period.
In other words: over a period, transition metals exhibit similar atomic volumes. However, these small differences are very important when defining metal crystals (as if they were metal stones).
Example
There are two mathematical formulas available to calculate the atomic volume of an element, each with its corresponding examples.
Example 1
Given the atomic radii of hydrogen -37 pm (1 picometer = 10 -12 m) – and cesium -265 pm-, calculate its atomic amounts.
Using the formula for spherical volume, we then have:
V H = (4/3) (3.14) (37 hrs) 3 = 212.07 hrs 3
V C = (4/3) (3.14) (265 hrs) 3 = 77912297,67 hrs 3
However, volumes expressed in pyrometers are excessive, so they are converted to units of angstroms, multiplying them by the conversion factor (1 in/100pm). 3 :
(212.07 pm 3 ) (1 100 / 100pm) 3 = 2.1207 × 10 -4 3
( 77912297,67 pm 3 ) (1 100 / 100pm) 3 = 77,912 3
Thus, the difference in size between the smaller atom of H and the heavier atom of Cs remains numerically clear. It should be borne in mind that these calculations have been made only under the claim that an atom is perfectly spherical, which faces reality.
Example 2
The density of pure gold is 19.32 g/mL and its molar mass is 196.97 g/mol. Applying the formula M/D to calculate the amount of one mole of gold atoms follows:
V Au = (196.97 g/mol) / (19.32 g/mL) = 10.19 mL/mol
That is, 1 mole of gold atoms is 10.19 ml, but what is the ratio of gold atoms in particular? And how to express it in units of noon 3 ? For this simply apply the following conversion factor:
(10.19 ml / mol) (mol / 6.02 10) -23 atoms) (1 m / 100 cm) 3 (1 at / 10 pm -12 m) 3 = 16,92 10 6 pm 3
On the other hand, the atomic radius of gold is 166 pm. If you compare the two volumes – calculated with the formula for another rounded quantity obtained by the previous method – you will find that they do not have the same value:
V Au = (4/3) (3.14) (166 hrs) 3 = 19.15 10 6 hrs 3
Which of the two is closest to the acceptable value? The one that is closest to the experimental results obtained by X-ray diffraction of the crystalline structure of gold.
