Atomic volume change in the periodic table: Let us know about How does atomic volume change in the 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 how the atoms are arranged within it.
So the atomic volume for a Z element does not differ from the one in the other phase 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 with, for example, stones. Marbles like Better Image’s blue color have well defined their material range thanks to its brilliant surface. In contrast, the boundaries of atoms are diffuse, although they can be remotely considered spherical.
Thus, one who sets a point beyond the atomic boundary has zero probability of finding an electron, and this point may be further or closer to the nucleus depending on how many atoms are considered in the atom.
atomic volume and radius
The interaction of two H atoms in the H molecule 2 , defines the position of their nuclei as well as the distance between them (intrinsic distance). If both atoms are spherical, then the radius is the distance between the nucleus and the diffusion boundary:
In the upper image it can be seen how the probability of finding the electron decreases as we move away from the nucleus. Dividing the intrinsic 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)r 3
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 very imprecise because the sizes of the atoms are far from the ideal sphere in their interactions.
To determine the atomic volume in a solid, several variables concerning the arrangement are taken into account, and they are obtained by X-ray diffraction studies.
The molar mass expresses the amount of substance that consists of atoms of a chemical element.
Its units are g/mol. On the other hand, density 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 with the variables to reach the desired units:
(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; While the volume of an individual atom is calculated with the formula for the spherical volume. To reach this value from the first, a conversion via Avogadro’s number (6.02 10) is necessary. -23 ).
How do atomic masses vary in the periodic table?
If the atoms are assumed to be spherical, their diversity would be similar to that seen in the atomic radius. In the upper image, which shows representative elements, it is illustrated that dwarf atoms from right to left; Instead, they get brighter 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 experience an effective nuclear charge Z eff , less than the actual nuclear charge Z .
The electrons of the inner shells repel those of the outer shells, 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 prevent the contraction of the atoms.
However, by landing in 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 adds protons again.
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 Volumes of Transition Metals
The atoms of transition metals contain electrons in the innermost orbitals d. This enhancement of the screen effect, as well as the actual nuclear charge Z, cancel out approximately equally, so that their atoms retain their same size over the same period.
In other words: In a period, transition metals exhibit similar atomic volumes. However, these small differences are very important when defining a metal crystal (as if it were a metal rock).
Two mathematical formulas are available to calculate the atomic volume of an element, each with its corresponding examples.
Given the atomic radii of hydrogen -37 pm (1 picometer = 10 -12 m) – and cesium -265 pm-, calculate their atomic masses.
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, the 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 Å 3
(pm 77912297,67 Å 3 ) (1 100 / 100pm) 3 = 77,912 Å 3
Thus, the difference in size between the smaller atom of H and the heavier atom of Cs remains numerically clear. It should be kept in mind that these calculations are made only under the claim that an atom is perfectly spherical, which defies reality.
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 is:
V Au = (196.97 g/mol) / (19.32 g/mL) = 10.19 mL/mol
That is, that 1 mole of gold atoms is 10.19 mol, but what proportion of gold atoms specifically? And how to express it in units of pm 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 atm / 10 at -12 m) 3 = 16,92 10 6 at 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 volume obtained by the previous method – you will find that they do not have the same value:
V Au = (4/3)(3.14)(166 at) 3 = 19.15 10 6 at 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.