3
$\begingroup$

While going through my book, I came across the following observation:

$\Delta H_{eg1}$ is positive for Be and Mg

Where $\Delta H_{eg1}$ is the first electron gain enthalpy.

It seemed quite reasonable as Be and Mg have fully filled $\ce{ns^2}$ configuration and also that the incoming electron would enter the higher energy p orbital which is well shielded by the s sub shell (we know that screening of s is maximum).

But why don't the other elements follow this? I thought maybe because they have d orbitals and thus get stabilised but Mg also has a d orbital right? So that doesn't seem a reasonable explanation.

$\endgroup$
2
  • $\begingroup$ All atoms have $d$ orbitals, but in general occupying them is energetically favorable only if all lower energy orbitals are filled. So you have to fill orbitals up to $3p$ first, therefore at least an argon core. $\endgroup$ Jun 7 at 12:22
  • 1
    $\begingroup$ Oh. So it is indeed because of the presence of... sort of 'accessible' d orbitals which help in stabilising the formed anion? $\endgroup$ Jun 7 at 13:04
5
$\begingroup$

The presence of $d$ orbitals can indeed provide a means for otherwise closed-shell atoms of alkaline-earth metals to accept electrons. Wu et al. [1] describe carbonyl complexes of Ca, Sr, and Ba in which the valence $d$ orbitals of these metal atoms are indeed engaged in (covalent) bonding.

In principle, all atoms actually have $d$ orbitals. Getting them to accept electrons in the ground state, however, is a different matter. With two units of orbital angular momentum and thus at least two nodes (corresponding to a principal quantum number of at least 3), all such orbitals are relatively high in energy and require a relatively deep potential well to make a bound state with the additional electrons. With 20 positive charges in the nucleus in a calcium atom the well is deep enough as the $3s$ and $3p$ orbitals privide only partial shielding from the nuclear charge. Magnesium with only 12 positive nuclear charges does not get there. It does have $3s$ electrons and perhaps bonding with $3p$ orbitals, these orbitals having two total nodes in a magnesium atom; but these have less angular momentum than the $3d$ orbitals we seek to occupy.

References

1. X. Wu, L. Zhao, J. Jin, G. Wang, M. Zhou, G. Frenking, "Observation of alkaline earth complexes M(CO)8 (M = Ca, Sr, or Ba) that mimic transition metals", Science 31 Aug 2018: Vol. 361, Issue 6405, pp. 912-916. DOI: 10.1126/science.aau0839

$\endgroup$
3
$\begingroup$

A simple, intuitive answer based solely on high-school chemistry and available emperical data; 1s < 2s < 2p < 3s < 3p < 4s < 3d

For Be, all the orbitals up to 2s are filled, so the "new" elctron has to go into the 2p orbital. Since there is a large gap between the 2s/2p energy levels of Be(it is one of the only elements in the 2nd period that cannot easily form pi bonds; BeF2 is polymeric while BF3 is monomeric with meaningful B-F pi bonds), the first EG enthalpy is positive. Same goes for Mg using its 3s/3p orbitals.

However, the situation changes when one goes to Ca. All the orbitals up to 4s are filled, but the next orbital is 3d instead of 4p. According to several papers, including this one, the 3d orbitals on Ca nearly act as frontier orbitals in well-defined compounds- exactly like the transition metals. Since the 4s/3d gap in the early 3d transition series is small, one can assume the same for Ca, and hence the first EG enthalpy of Ca is negative, just like the 3d transition metals. Again, the same goes for Sr using its 5s/4d orbitals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.