The atomic radii of zirconium (160 pm) and hafnium (159 pm) are nearly identical. Explain.

Video transcript

All right. Hi, everyone. So this question says, considering measurements rounded to the nearest PICO meter, the atomic radius or radii, excuse me of Niobium at 146 Peters and tantalum at 146 Peters can be practically identical. Explain this observation. Now, here we have four different choices. And what I want to point out here is that all of them discuss either shielding or lack thereof of electrons in either the six S or the four F shells which therefore affect the nuclear charge. So please keep that in mind. As we proceed through this question, I want to start off by actually providing the electronic configurations of both of our elements because recall that Niobium has an electronic configuration of Krypton 44 five S one where is talum has an electronic configuration of Xenon 4 F-14 five D three six S two. And additionally, tin talum has an atomic number of Sony three, where is Neo beam has an atomic number, atomic number of 41. Now recall first and foremost that as you increase the number of electrons in the same shell, the attractive force is therefore going to increase. Additionally, as you increase the shell number, then the distance between the electron and the nucleus increases, which therefore actually decreases the attractive force. Now, the reason why I want to bring this up first is because I first want to point out here that tantalum has a higher atomic number and therefore more protons than niobium, which means at tantalum is going to have more of a nuclear charge because there are simply more protons inside of the nucleus. Additionally, tantalum has more electrons, more electrons in shells and four and five compared to neo. So that is going to increase the attractive force as well. Now recall that shielding is a phenomenon observed which is supposed to decrease the effective nuclear charge because the electrons themselves are supposed to surround the nucleus and therefore shield the protons inside of the nucleus. Right. Now, the shielding effect exerted by the inner electrons actually decreases as you go from S to F. Right. So the s show is going to have the highest shielding effect followed by PD and then the F shell is going to have the least shielding effect. So the 14 electrons inside of the F shell in tantalum aren't going to do much shielding at all, which means that the shielding effect in tantalum is going to be considered less effective, right, because it is less able to counter the decrease in radius caused by the increase in nuclear charge. So essentially, right, the atomic radius of tantalum is going to decrease due to the increase in nuclear charge given the fact that there are simply more protons inside of the nucleus. And in addition to this, the poor shielding of four F electrons means that the shielding effect is not going to be sufficient enough to counter this drastic decrease in the atomic radius. So our answer here, if I scroll back up is going to be option C in the multiple choice, right? Due to the poor shielding of four F electrons, the shielding effect is insufficient to counter the decrease in radius caused by increasing nuclear charge allowing tantalum to have an identical radius to neo and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.