What are Atoms?

Atoms consist of three types of particles: positively charged protons, neutral neutrons, and negatively charged electrons.

The protons and neutrons are bound together to form the atom’s nucleus, while the electrons surround the nucleus at different energy levels called orbitals.

Effective Nuclear Charge

The effective nuclear charge of an atom is the net positive charge experienced by its outermost electrons (also called valence electrons). Effective nuclear charge can be thought of as the strength of attraction between an electron and the nucleus. This effective charge is always lower than the actual nuclear charge of an atom because the effective charge accounts for electron shielding effects.

Electron shielding results from the inner or core electrons (i.e., the ones closest to the nucleus) shielding the outer electrons from the positive charge of the nucleus. Another way of saying this is that electron shielding causes a decrease in the amount of nuclear attraction felt by valence electrons.

So, the effective charge experienced by the valence electrons is lower than it would be if the inner electrons were not present. This is because the inner electrons’ negative charge locally counteracts the positive charge from the nucleus.

The actual nuclear charge of an atom is equal to the number of protons in the atomic nucleus. The number of protons is also referred to as the atomic number, which is the unique elemental quantity on which the periodic table is organized.

How to Calculate Effective Nuclear Charge

The effective nuclear charge (\(Z_{eff}\)) is calculated by subtracting the magnitude of electron shielding (\(S\)) from the actual nuclear charge (\(Z\)).

 
As stated, the value of \(Z\) (the actual nuclear charge) is the number of protons in the nucleus. The value of \(S\) is approximated based on the number of core electrons separating the outer (valence) electrons from the nucleus. However, more precise calculations of \(S\) require information about the shapes of the valence electron orbitals and the repulsive forces between valence electrons.

We can perform a simple example calculation of effective nuclear charge by considering one atom of elemental lithium. Lithium has an atomic number of 3, meaning it has three protons in its nucleus; therefore, its actual nuclear charge is 3. Lithium has two core electrons (that is, two electrons in the 1s orbital) and one valence electron (that is, one outer electron in the 2s orbital). For the valence electron in this case, we can estimate the electron shielding (\(S\)) as the number of core electrons. So, our equation here would be:

 
So, for the outermost electron of lithium, the effective nuclear charge (\(Z_{eff}\)) is approximately 1. In other words, the net attractive force felt by the lithium valence electron from the nucleus is equivalent to 1.

Now, if we consider the element beryllium, which is to the right of lithium on the periodic table, we see that it has an atomic number of four (i.e., \(Z = 4\)). As with lithium, beryllium has two core electrons (two electrons in the 1s orbital). However, unlike lithium, beryllium has two valence electrons (that is, two outer electrons in the 2s orbital). To approximate \(Z_{eff}\) for the valence electrons using the number of core electrons, we have:

 
So, beryllium has a higher effective nuclear charge than lithium.

Now skipping across the periodic table from left to right to fluorine, we have the following scenario: An atomic number of nine (i.e., nine protons), two core electrons in the 1s orbital, and seven valence electrons (two in the 2s orbital and five in the 2p orbital). Again, assuming that the electron shielding can be approximated as the number of core electrons (an assumption that becomes less accurate for elements with more electrons), we have the following for valence electrons:

 
So, fluorine’s effective nuclear charge for valence electrons is higher than lithium and beryllium. If we look up the tabulated value for the effective nuclear charge of fluorine, we find that it is actually around 5.1+. This discrepancy between our calculation and the tabulated value is due to our simplifying assumption that the electron shielding constant \(S\) can be approximated as the number of core electrons.

In fluorine, both the shape of the valence electron orbitals (the 2p orbitals) and the repulsive forces within the valence electron shells make the tabulated value substantially lower than our estimate. However, the basic principles of our simplified calculations are accurate and thus accurately represent why effective nuclear charge changes as we move from left to right across the periodic table.

Effective Nuclear Charge Periodic Trend

The observation that effective nuclear charge increases as we move to the right on the periodic table has important implications for the properties of elements. We find that a fluorine atom has a significantly smaller diameter than a lithium atom. This is because the valence electrons in fluorine experience a stronger attractive force toward the nucleus.

In a sense, the valence electrons of fluorine are held tighter than the valence electrons of lithium. This periodic trend in effective nuclear charge has important implications for many elemental properties such as atomic sizes, binding energies, and ionization energies.

Review

Okay, let’s review what we’ve learned! Understanding effective nuclear charge is important because it correlates well with many of the periodic trends in element properties. The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom and is always lower than the actual nuclear charge of the atom.

The actual nuclear charge is equal to the number of protons in the atom’s nucleus. The effective nuclear charge is lower than the actual nuclear charge because it accounts for the electron shielding effect. Electron shielding results from core electrons effectively blocking the outer valence electrons from the attractive force of the atomic nucleus.

Review Questions

Okay, before we go, let’s go over a few review questions.

1. Using the formula for effective nuclear charge, \(Z_{eff} = Z – S\) (where \(Z\) is the number of protons and \(S\) is the number of core electrons), calculate the effective nuclear charge for the valence electrons of silicon.

 
2. Moving from left to right across a row in the periodic table, why do atoms become progressively smaller?

 
3. Why is the effective nuclear charge always lower than the actual nuclear charge, which is the number of protons in the atomic nucleus?

 
All right, that’s all for this video. Thanks for watching, and happy studying!