3/27/2023 0 Comments Atomic radius![]() ![]() Number of shells: it is common that the higher the number of shells in an atom, the farther out the outer electron will be from the nucleus and the greater the atomic radius will be. ![]() Shielding effect of inner shell electrons: the shielding effect of inner shell electrons decreases the effect of nuclear charge on outer shell electrons and hence an increase in shielding effect is associated with an increase in atomic radius. Higher the nuclear charge smaller will be the atomic radius. Nuclear charge: nuclear charge tends to decrease the atomic radius of the atom. There are 3 things which affect the atomic radius of an atom. It is further away because that is just the form of the radial wavefunction obtained by solving the Schrodinger equation - period. Notes and referencesģ I'm talking about $\langle r \rangle$, not the radius of a Bohr orbit, although that does also increase going from 1s to 2s.Ĥ I'm avoiding the circular logic that "higher energy means further away". It's likely that the exact dependency is very complicated and nowadays, such atomic properties are mostly calculated computationally anyway, which removes the need for such a mathematical formula. The description of $Z_\mathrm$ (going from 1.28 to 2.51), although it's sadly difficult to find an exact mathematical formulation in any textbook. So what am I missing here? Why does shielding matter? Is the force on a single valence electron a function of the net nuclear charge or do we have to look at it from a more piecewise level, considering how other electrons are in between it and the nucleus and how it interacts with those individual electrons? I find this contradictory, since as we go down a group, the effective nuclear charge remains constant (equal numbers of protons and non-valence electrons are added). Thus with more energy levels present and more shielding, valence electrons begin to reside further and further away from the nucleus.īut at the same time, textbooks make a big deal about the concept of effective nuclear charge, which states that the "effective" force a given valence electron feels is a function of the net charge of the nucleus and non-valence electrons combined-the effective nuclear charge. The general explanation seems to state that as we go down a group, there are more electron shells present to contribute to a "shielding" phenomenon, where inner electrons cancel out part of the inward force from the nucleus with a repulsive force. These trends explain the periodicity observed in the elemental properties of atomic radius, ionization energy, electron affinity, and electronegativity.I realize this should be a fairly basic question, but I'm still not quite satisfied with what I've been told from numerous sources. This happens because the number of filled principal energy levels (which shield the outermost electrons from attraction to the nucleus) increases downward within each group. Second, moving down a column in the periodic table, the outermost electrons become less tightly bound to the nucleus. As this happens, the electrons of the outermost shell experience increasingly strong nuclear attraction, so the electrons become closer to the nucleus and more tightly bound to it. ![]() First, electrons are added one at a time moving from left to right across a period. In addition to this activity, there are two other important trends. Stable octets are seen in the inert gases, or noble gases, of Group VIII of the periodic table. Elements tend to gain or lose valence electrons to achieve stable octet formation. These trends can be predicted merely by examing the periodic table and can be explained and understood by analyzing the electron configurations of the elements. The periodic table arranges the elements by periodic properties, which are recurring trends in physical and chemical characteristics. ![]()
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