Let me write down a couple of facts and this will be easier for you to know the essence of the question.

  • Radius of boron is smaller than beryllium.
  • Ionization energy of boron is smaller than beryllium.

I see that boron has 1 electron at 2p. We know that 2p orbital have lower ionization energies because they are on average a little further from the nucleus. Due to this, it's easy to see why ionization energy of boron is smaller than beryllium.

Now, the fun part is that it's a fact that boron is smaller in size, which I don't get why. If 2p on average is further away than 2s and 1s, it means boron's radius must be bigger. Note that with Slater's rule, I can see it holds to true as effective nuclear charge for boron is bigger, hence more attraction, hence smaller in size

What I'm trying to understand is why being 2p further away than 2s or 1s don't suggest that boron has to be bigger in size ? I think the hint here is with the word: "average", but I still don't get it to be happy with it.

  • $\begingroup$ Not related to downvote but there is general shrinking tendency along periods. Also orbital energy has indirect relation to size. The 2nd radial probability maximum of 2s orbitals is farther from nucleus than 2p maximum. $\endgroup$
    – Poutnik
    Commented May 15, 2023 at 5:25
  • $\begingroup$ @Poutnik I took your suggestion and made my answer considerably shorter :) $\endgroup$
    – ananta
    Commented May 15, 2023 at 14:39

4 Answers 4


The Answer

The quantum mechanical model (shown at the end) explains both the facts: radius of boron is smaller than beryllium; ionization energy of boron is smaller than beryllium.

The electrostatic model and your reasoning fall short because these don't account for inter-electron interactions, which is why we had to develop quantum mechanical model in the first place.

So, we don't say an orbital is further away, or closer to the nucleus. It doesn't make sense because an orbital is not a point, it is a wavefunction. Nonetheless, there are a few ways to go about investigating your question.

Your reasoning is incorrect because, according to you, $\ce{B}$ should have a higher ionization energy than $\ce{Be}$ because $\ce{B}$ has a lower size than $\ce{Be}$; the latter is correct, but doesn't imply the former. According to the electrostatic model too, the ionization energy of $\ce{B}$ should be higher than that of $\ce{Be}$, which it is not. Therefore, you need to think in quantum mechanical terms to understand this.

Actually, as I reasoned in The Fun Part section, ionizing an electron from the singly filled $\mathrm{2p}$ orbital of $\ce{B}$ is easier than ionizing an electron from the fully filled $\mathrm{2s}$ orbital of $\ce{Be}$, just like ionizing an electron from quadropoly filled $\mathrm{2p}$ orbital of $\ce{O}$ is easier than ionizing an electron from the triply filled $\mathrm{2p}$ orbital of $\ce{N}$; fully and halvely filled orbitals are more stable reducing the ionizing energy required to attain such configurations.

Electrostatic Model

Radii of $\ce{Be}$ and $\ce{B}$

We assume the valence electron is orbiting (with potential and kinetic energy) at a distance $r$ from a positive charge ($Z_\text{eff}$ of the nucleus) of charge $Z_\text{eff}$. We can obtain $r_\ce{Be}$ and $r_\ce{B}$ using the Bohr theory:

$$ r_n = \pu{5.29 \times 10^{-11}} \dfrac{n^2}{Z}\pu{m}\\ \text{may }\implies (r_n)_\text{eff} = \pu{5.29 \times 10^{-11}} \dfrac{n^2}{Z_\text{eff}}\pu{m}\\ \implies \dfrac{(r_2)_{\text{eff},\ce{B}}}{(r_2)_{\text{eff},\ce{Be}}} = \dfrac{Z_{\text{eff},\ce{Be}}}{Z_{\text{eff},\ce{B}}} = 0.787 $$

Therefore, the valence electron of $\ce{B}$ is situated closer to that of $\ce{Be}$ because $\ce{Be}$ has a lower effective nuclear charge than $\ce{B}$, as you correctly reasoned.

Ionization Energies of $\ce{Be}$ and $\ce{B}$

Assume that the electron, situated at distance $r$ from the nucleus is orbiting, the energy of this electron may be approximated using Bohr theory as:

$$ E_n = -13.6\dfrac{Z^2}{n^2}\pu{eV}\\ \text{may }\implies (E_n)_\text{eff} = -13.6\dfrac{Z_\text{eff}^2}{n^2}\pu{eV}\\ \implies \dfrac{(E_2)_{\text{eff},\ce{B}}}{(E_2)_{\text{eff},\ce{Be}}} = \dfrac{Z_{\text{eff},\ce{B}}^2}{Z_{\text{eff},\ce{Be}}^2} = 1.613 $$

Therefore, the ionization energy of $\ce{B}$ should be more than that of $\ce{Be}$ because $\ce{B}$ has a higher effective nuclear charge than $\ce{Be}$, but this isn't the case. The electrostatic model doesn't account for inter-electron interactions, which is why we had to develop quantum mechanics in the first place.

The Fun Part

First Things First

The Data

The data (obtained from sources 1, 2, and 3) for the atomic radii, ionization energies, and effective nuclear charges of Group 2 elements is:

atomic radius, ionization energy, effective nuclear charge data for group 2

The Graphs

The following is graph for elements of Group 2, that is $\ce{Li}$, $\ce{Be}$, $\ce{B}$, $\ce{C}$, $\ce{N}$, $\ce{O}$, $\ce{F}$, and $\ce{Ne}$.

Let us also see how the radius varies with ionization energies and effective nuclear charges.

ionization energy and effective nuclear charge varying with radius for second period element

Let us also compare the

ionization energy and effective nuclear charge of second period elements

So, even though, effective nuclear charges are increasing on going from $\ce{B}$ to $\ce{Be}$, we see that the ionization energies are decreasing. A similar anomaly is observed in going from $\ce{N}$ to $\ce{O}$. This is because extracting the valence electron makes the configuration halvely filled ($\mathrm{2p^3}$) for $\ce{O}$ or fully filled ($\mathrm{2s^2}$) for $\ce{B}$, which are considered stable.

Quantum Mechanical Model

I wanted to compare the orbital energies of $\ce{B}$ and $\ce{Be}$ to see which atom has the highest energy orbital. I created a simple model in ORCA 5.0.4 where $\ce{B}$-$\ce{Be}$ distance was large, $7 \overset{\circ}{\mathrm{A}}$, to avoid any significant interaction between the two elements. The highest-occupied molecular-orbital (HOMO) of the system will show us which atom is easier to ionize.


The HOMO of the system, as shown below, corresponds to $\ce{B}$ $\mathrm{2p}$ orbital from where it will be easiest to ionize. Thus, Ionization energy of $\ce{B}$ is less than that of $\ce{Be}$.

comparison of orbital energies for beryllium Be and boron B molecular orbitals

Notice that the size of $\ce{Be}$ 2s orbital is larger than that of $\ce{B}$ 2s orbital, verifying that $\ce{Be}$ is larger than $\ce{B}$.


The quantum mechanical model explains both the facts:

  • Radius of boron is smaller than beryllium.
  • Ionization energy of boron is smaller than beryllium.
  • $\begingroup$ Hello ananta. If you are using Excel (I think you are), and you have two magnitudes that share the same x-axis value (elements in this case), you can use a secondary axis. Selecting the series, you can go to format, and on the left you click on series options→secondary axis (I am translating from Spanish to English, but it should be something like that). That way, the orange curve in the first two graphs will be more clear, because the magnitude of $Z_\pu{eff}$ is much lower than $r$. A humble tip. $\endgroup$ Commented May 15, 2023 at 22:25
  • 1
    $\begingroup$ Good answer, but unfortunately it doesn't state the reason why it's actually like this. in your ionizing energy section of B and Be, you use Bohr model which affirms my viewpoint, but true, bohr model is wrong in its own, I get that, but explanation of what i'm. asking is not stated in your answer. Seems like Until I fully learn schrodinger equation, i won't be able to understand this at all. I knew about half filled and fully filled rule, but still, no good explanation why that is. Again, i mightn't be understanding it as i don't understand schrodinger $\endgroup$
    – Matt
    Commented May 16, 2023 at 11:09
  • $\begingroup$ @ananta Hello ananta. There are things in this answers that seem to me completely wrong or maybe I am crazy, and you can help me to clarify. (1) The value $E_n$ there is the ground state energy of the $\ce{H}$ atom, there is no such thing as the energy of the second, third, and so on electron when you get approximate solutions of the the Schrödinger's equation. $\endgroup$ Commented May 16, 2023 at 12:06
  • $\begingroup$ (2) The quantum number $n$ makes sense in the $\ce{H}$ atom, because it appears when we solve the radial part of the differential equation (for those who like solving it with series, we need to kill that series or it will diverge). For another atoms, it doesn't make any sense of speaking of any $n$ number (what is the $n$ number of $\ce{Be}$?), we rather speak of the lowest energy eigenvalue which are equal to the ground state. So, when you take the ratio of $E_n$ of $\ce{Be}$ or $\ce{B}$, what are you actually dividing there? $\endgroup$ Commented May 16, 2023 at 12:12
  • $\begingroup$ (3) Even if you apply the Bohr's model, you cannot use that formula to estimate the radius of any other element, there are two many electrons and you get nonsensical results. Even if you use $Z_\pu{eff}$, you can try that formula: for $\ce{B}$ the exp. value is $85$ $\pu{pm}$, with the Bohr's formula (using $n=1$, which still doesn't make any sense to talk about $n$ in another atom rather than $\ce{H}$) you get $11.30$ $\pu{pm}$. $\endgroup$ Commented May 16, 2023 at 12:23

Atomic radius decreases as the group increases. Consider that a Group I element, such as Li or K, has a lone electron ready to "drop off". Halogens, though, are ready to "pull in" that last electron to approximate a noble gas shell.

  • $\begingroup$ Well, the actual problem is with slater rule, boron is smaller, but my question is that if Boron's valence electron is in 2p(which is farther away from nucleus than 2s, then this contradicts the slater rule and suggests boron should be bigger in radius). $\endgroup$
    – Matt
    Commented May 15, 2023 at 0:19

Having a partially(or fully) filled p orbital increases the size of a atom if the nuclear charge is the same which it isnt in your case.The size of a atom is determined by the wavefunction of the valence electron(s).As the nuclear charge is increase the rate of change of the complex wavefunction $\frac{d\Psi(r,\theta,\phi)}{dr}$ is increased as we slowly start moving away from the nucleus and reaches its absolute maximum faster then is descreased faster than in the case of a nucleus with less nuclear charge resulting in a wavefunction which probability density is located much closer to the nucleus which means by definition atom radius decrease.

  • $\begingroup$ My worry is that with slater rule, boron is smaller, but my question is that if Boron's valence electron is in 2p(which is said to be farther away from nucleus than 2s, then this contradicts the slater rule and suggests boron should be bigger in radius). Isn't there any way to understand this without schrodinger ? $\endgroup$
    – Matt
    Commented May 15, 2023 at 0:20
  • $\begingroup$ hmm atomic structure is described by quantum mechanics and quantum mechanics is described by Schrodinger equation , Slater's rules are rules of thumb and cannot explain the Why? $\endgroup$
    – Volpina
    Commented May 15, 2023 at 0:30
  • 1
    $\begingroup$ @Volpina Hello Volpina. Why choose $(\partial \Psi/\partial r) (r,\theta,\phi)$ instead of $\partial (\Phi^* \Psi)/\partial r (r,\theta,\phi)$? Would you extend your argument to a molecule also? $\endgroup$ Commented May 15, 2023 at 10:21
  • $\begingroup$ @MetalStorm because we find $\Psi(x)$ from the Schrodinger equation not the probability density. $\endgroup$
    – Volpina
    Commented May 15, 2023 at 14:48
  • $\begingroup$ @Volpina Yes but analyzing the wavefunction like that is extremely misleading. I will give a simpler case than the N-electron wavefunction. For the 1D particle-in-the-box, $\psi_2(x) = \sqrt{2/L}\sin(2\pi x/L)$. The highest rates of change of $\psi_2$ are in $x \in (0,L/2,L)$. Guess the probability of finding the particle here? They are all zero. Do you agree with me? $\endgroup$ Commented May 16, 2023 at 0:59

The trend of the atomic radius across the period is decreasing of course but if we consider the Hund's rule which states that the "completely filled or completely half filled orbitals" are more stable than the incompletely filled, incompletely half filled orbitals and the more stable orbitals usually tend to have lower energies.

If we look into the configurations of the II and XV th groups which are ns2 and ns2,np3 where IInd group is completely filled s-orbital and the XV th group completely half filled p-orbital as they are more stable compared to the groups before and after they have less energies comparatively more stability and hence comparatively a bit larger size than the surrounding groups.

Using the same thing Boron falls under the III rd group which has a configuration of 1s2,2s2,2p1 which is a completely filled 1s and 2s -orbitals but an incompletely filled 2p orbital making it have comparatively more energy, less stability and more radius.


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