It's a relatively unknown Bangladeshi book. It's called "Chemistry-First Paper (Class XI-XII)" by Professor Haradhan Nag

According to my book, "Electrons of s and p orbitals produce more shielding than electrons of d and f orbitals. As a result, atoms of s and p-block elements experience lower effective nuclear charge. Consequently, their atomic sizes are greater than those of atoms of d and f-block elements".

My book's claim makes no sense to me. It's true that the electrons of s and p orbitals produce more shielding than electrons of d and f orbitals due to penetration, but why would that make atomic sizes of s & p block elements greater than d and f-block elements?

Let me take s-block as an example. In period 4, $\ce{K}(220 \pu{pm})$ is an s-block element and $\ce{Sc}(160\pu {pm})$ is a d-block element. It's true that the atomic size of K is greater than Sc; however, I don't think it's because s-orbitals have a stronger shielding effect than d-orbitals. If anything, we could ironically argue that Sc should have a greater atomic radius than K as s-orbitals have a stronger shielding effect than d-orbitals because the 4s-orbital is inner in Sc. In contrast, the 4s-orbital is situated outermost in K. So, why is my book making the claim that an outermost orbital's shielding power will determine the size of the atom? Outermost orbitals have no contribution to shielding!



It turns out that the 4s orbital is not actually inner in Sc. However, my point still stands. Why is my book making claims about shielding based on the outermost orbital (which has no contribution to shielding!)


1 Answer 1


Let's have a look at some illustrations:

  1. Size of atoms by atomic number:

size by atomic number

  1. The size of Li, C and Ne with their orbitals (blue is completed and yellow is empty):

3 examples

  1. The shielding effect, illustrated with the density of probability of the electrons' presence (green):

Shielding effect

  1. Hydrogen probability densities of 2p and 2s orbitals:

graph: 2s and 2p

We will need to use quantum physics to answer this question, so it gets further than the classical chemistry problems. I therefore consider electrons as matter waves, their position being non-deterministic, the probability of their presence at such and such a place determined by the Schrödinger equation.

On the illustration n°1, you notice that the atomic radius decreases with the number of electrons until reaching a new orbital, which necessarily increases the range of the electronic cloud (each orbital being more extended than the previous one).

The illustration n°2 shows you that the atomic radius decreases with the stability of the electron (number of completed orbitals).

The third and fourth ones just illustrate the consequences of the shielding effect.

Now, here is the answer to your question: within a set of neutral atoms that have the same number of orbitals, the charge of the nucleus and the electronic cloud increases with the atomic number Z. As the shielding effect of the first sublayers is limited to the maximum number that these orbitals can hold, the charge felt by the farthest electrons will be stronger and stronger, which will attract them more towards the center and decrease the atomic radius. Concerning the shielding effect of the higher orbitals, I have a little trouble understanding what your book means since the s and p orbitals always have a shielding effect (never replaced by the shielding of the other orbitals, but added to it). Generally speaking, a smaller shielding effect gives a higher charge of the nucleus and therefore a smaller radius. Atoms with d- and f-layers have a larger radius because the range of these orbitals is simply larger according to the Schrödinger equation (you can compare the range of 2s and 2p on picture n°4). I hope this helped you!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.