Zeff

Part 1: GCSE physics

It seems obvious what the charge of a nucleus is - it's the total charge of all the protons, right? And yes, that's true. Except this isn't the charge the valence electrons orbiting the nucleus wil experience.

The most basic atom is the hydrogen atom - one electron orbiting one proton:

In this case, the charge of the nucleus is equal to the charge of one proton - 1.60x10^-19 C. And it's worth noting here that every atom will have a charge equal to a multiple of 1.60x10^-19 C, since that's the major* way elements differ - each successive element has one additional proton in its nucleus to the one before it. For simplicity, I'll refer to the charge of a nucleus from hereon out as Z, and the charge of a hydrogen nucleus is Z=1.

Let's now move onto helium, the second element. This time, we have two protons, and the charge is Z=2. Pretty basic stuff. But we also have two electrons - again, everyone knows this:

And now things start to get a bit more complicated. 

Part 2: Quantum Irritation 

You see, the way I've drawn the helium atom isn't quite right. The electrons don't orbit the nucleus at fixed distances, and they also don't orbit at fixed distances from each other. We've known this ever since the Bohr model was disproved nearly 100 years ago in favour of a quantum model, where electrons exist in orbitals. You can calculate the probability of finding an electron at a given radius from the nucleus using the radial distribution function (RDF), which is outside the scope of this blogpost. But the main idea is that the electrons aren't really fixed in place like this - the highest probability of finding the electron will be at about 53pm - the Bohr radius - when dealing with hydrogen, but it's not the only probability.

So...electrons move in the atom. They can also repel each other, because they are like charges. The electrons are also attracted to the protons - they are opposing charges. But if the electrons can be anywhere in the atom, then it stands to reason that some electrons will prevent the other electrons from experiencing the full attraction to the nucleus. This effect can be experienced in helium, but will be more pronounced when dealing with atoms where the valence electrons are in higher energy orbitals, from lithium onwards, and the innermost electrons closer to the nucleus weaken the nuclear attraction. This effect is known as shielding.

So not all electrons experience the attraction from the nucleus equally. Now I can finally discuss the main topic of this blogpost.

Part 3: Z_eff

The magnitude of this shielding can be measured using the shielding constant, σ - the larger the effect of shielding, the greater σ is. And if shielding is particularly strong, then it stands to reason that the nuclear charge experienced by the valence electrons will be far smaller than that experienced by electrons closest to the nucleus, where there is nearly no shielding. The difference between the actual nuclear charge, Z, and the shielding effect, σ, is known as the effective nuclear charge, or Z_eff. 

As an example, the Z_eff value for helium equals 1.69, whereas its Z value equals 2. This isn't that big a difference, because the electrons in helium aren't particularly good at shielding each other. They're in the same atomic orbital, after all. But when we go up to lithium, where it suddenly has one valence electron in its 2s orbital, with two in its 1s orbital, the shielding effects increase drastically. The Z_eff of lithium is 1.28, whereas Z(Li) is 3. We can thus see that the 1s electrons in lithium are very effective shielders compared to in helium. 

But in general, as you go across a period, Z_eff increases when you're merely adding electrons to the same orbital. Which makes sense, in a way; take sodium as an example. Sodium has a valence electron in a 3s orbital, at least that's where you'll most likely find it, and as such, if it wants to maximise its chances of experiencing a full nuclear attraction, it has to get past the 1s, 2s, and 2p electrons. They're all of lower energies, and are closer to the nucleus - they're going to be shielding the poor 3s electron. Whereas with neon, which comes before sodium, the valence electrons merely face the 1s and 2s electrons, as well as any 2p electrons which come in their way. Fewer electrons means less interference means less shielding means higher Z_eff.

We can even generalise this - the closer the electron is to the nucleus, the better it is at shielding. Which makes sense - it will experience a stronger nuclear attraction since it's, well, closer. This also means we can say that 1s electrons are better at shielding than 2s electrons, which are better than 2p, and so on. And in general, the orbital shielding order of precedence, for a given energy level, is: s > p > d > f. Again, f orbitals are furthest out from the nucleus, they'll therefore experience a far weaker attraction than s electrons.

Part 4: Z_eff trends 

There are various more trends related to Z_eff in the periodic table:

• As you go across a period, you'll see an increase in Z_eff  due to weaker shielding effects. This means that the atom will contract due to greater electrostatic attraction between the valence electrons and the nucleus, so the atomic radius will decrease. This greater electrostatic attraction also means the energy required for an atom to give up an electron will increase, and therefore the first ionisation energy of an element (defined as the energy needed for one mole of a gaseous atom to give up one electron to form one mole of a gaseous cation) increases as you go across a period.
• Except that latter point isn't exactly true. It generally holds, but there are some exceptions. Take beryllium and boron for example - boron comes after beryllium in the periodic table, but has a lower first ionisation energy. The reason for this is simple - boron has a valence electron in a higher energy level (2p) than beryllium (2s). This means the valence electron experiences a weaker attraction due to greater shielding, so less energy is needed to lose the electron.
• That rule about Z_eff increasing...doesn't really hold either.

Part 5: Lanthanide contraction 

Remember earlier, when I said that d and f orbitals aren't as good at shielding as s and p orbitals? Well, this becomes especially clear when you look at their atomic radii. In general, their atomic radii will decrease steadily, but occasionally you'll get sudden peaks, such as with manganese in d block and europium in f block. The reason behind this is that these elements have additional stability since they have half-filled d/f orbitals; this therefore means the electrons in the orbitals will experience lower repulsion since they're not being paired up with other electrons.

When discussing a contraction with atoms with valence electrons in their d orbitals, it's a transition metal contraction. When discussing contractions with the lanthanides, which have valence electrons in the 4f orbital, it's the lanthanide contraction: 

Source: Atkins' Physical Chemistry, OUP Oxford: twelfth edition

The Lanthanide contraction is highlighted in the graph above - notice how the atomic radii decrease slowly, but compared to the decrease from Li to F or from K to Br, the change isn't as sudden. Also notice the sudden peaks for europium (Eu) and ytterbium (Yb). A similar trend, known as the actinide contraction, impacts the actinide elements (with valence electrons in the 5f orbitals).

Part 6: Epilogue

I don't have much else to add, I've written this blogpost for revision but it would feel a bit odd if I just ended it like that.