# Why ionisation enthalpy of early lanthanides are expected to be greater than early actinides?

In my book Chemistry Part I (NCERT XII), a statement on p. 232 goes like this:

It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids.

My thought on this matter is that $$Z$$ will increase and f-orbitals will shield electrons poorly and so ionisation enthalpy should increase. But I'm not able to understand what my book is trying to say. What is meant by less penetration to inner core of electrons? To be honest I wasn't able to understand all the above mentioned 3-4 lines. So, can anyone explain this properly to me, please?

• @HahaHahaha How did I what? Yes, I added a literature source since OP failed to do that, what about it? – andselisk Jan 4 at 9:48
• If anyone could confirm, if I understood the meaning. 5f orbitals being more on vicinity as compared to 4f, will have more effectively shielded, and since, as mentioned, they penetrate less into nucleus, connotes more ionisation enthalpy. @andselisk what's more accurate to say actinides or actinoids and similarly lanthanides or lanthanoids, I have never seen latter in either case... – Zenix Jan 4 at 12:42
• – andselisk Jan 4 at 12:46
• @HahaHahaha I don't see how this is relevant. If you search Google Books for the exact phrase, for example, "The 5f electrons, will therefore, be more effectively shielded from the nuclear charge", you only find this very NCERT book. Google still works for everybody regardless the nationality. – andselisk Jan 5 at 10:17
• I'd suggest you follow the book blindly if you're preparing for JEE, even if the data might be wrong. There have been instances before when JEE Mains has considered this book to be correct despite it having the wrong information. – paracetamol Jan 6 at 19:30

This data is incorrct, and I am saying so because the Ionisation energy of the actinoid is a decent bit higher than the corresponding lanthanide

In each case IE 1 is significantly higher for lanthanoids, IE 2 is still higher for lanthanoids and IE 3 is comparable.

$$\ce{Ce}$$ $$534.4$$, $$1050$$, $$1949$$

$$\ce{Th}$$ $$587$$, $$1110$$, $$1930$$

These were two examples where i showed all 3 IEs

For, the sake of further clarification, here is a the IE1 of the first few

$$\ce{Pr}$$ $$527$$, $$\ce{Pa}$$ $$568$$

$$\ce{Nd}$$ $$533.1$$, $$\ce{U}$$ $$597.1$$

$$\ce{Pm}$$ $$540$$, $$\ce{Np}$$ $$604.5$$

$$\ce{Sm}$$ $$544.5$$, $$\ce{Pu}$$ $$584.7$$

(units used are $$kJ/mol$$)

The reason you stated in your doubt is absolutely correct, as even if we were to conclude at these aren't accurately determined, we see this phenomenon elsewhere in the periodic table also, where the values are accurately determined.

1. $$\ce{Sn}$$ is the lowest IE element of GP 14 and not $$\ce{Pb}$$.

2. $$\ce{Fr}$$ has a higher IE than $$\ce{Cs}$$.

3. $$\ce{Ra}$$ has a higher IE than $$\ce{Ba}$$.

All due to the same reason poor shieling, which increases the effective nuclear charge, which increases nuclear attraction, which in turn leads to a higher IE.

References

Wikipedia data page

• @Andselisk I tried to format this in the best way possible, however, are there some improvements that you can help me with? – Haha Hahaha Jan 4 at 6:54
• I find it hard to track what values are taken from where and what do the represent, especially in the first array. When you cite a textbook, it's almost always obligatory to include the page numbers. Also note that symbols for chemical elements must be upright (use \ce{…}). – andselisk Jan 4 at 12:50
• Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. See also: MathJax basic tutorial and quick reference. – andselisk Jan 4 at 12:51
• @andselisk my stupid laptop does not have a backward slash or curly brackets lol, so idk what to do, as said in the references all values are simply taken from the Wikipedia data page, which has taken its from the two given books which is given in their references, so basically I study NCERT too, i was almost sure this statement was wring, so i just googled data and found this, then i looked up these books and their rewives to see if the wiki editor has taken daya from good books, turns out they are both well reputed books – Haha Hahaha Jan 5 at 7:21
• Switch to US international layout which will give you access to all ASCII symbols including {} and \, which are essential if you do slightly advanced posting or any programming. As for the rest, make sure you read the sources you cite and don't throw them in because they look relevant. – andselisk Jan 5 at 7:39

Nuclear charge is shielded by various orbitals on the order s>p>d>f

More the shielding less the ionization energy(IE).

In lanthanoids 4f electrons are shielded by 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s

In actinoids 5f electrons are shielded by four more orbitals 4f 5d 6p 7s.

So effective nuclear charge on the outermost electron decreases and IE decreases.

Trying to explain a more technical in terms of a layman, i if you consider IE of an atom, it is the force×displacement required to pull out that electron.

The atom may be approximated as a sphere. So for an atomic no. Z, Z-1 electrons shield the outermost electron.

In physics (electrostats) you may probably heard about the shell theorem which states that spherical charge distribution can be assumed to be concentrated at the centre. So each electron neutralizes a proton of the nucleus. Finally we have 1 electronic positive charge at the centre.

Force is given by KQq/(r^2)

Here Q=-q=e

And radius of actinoids is greater than lanthanoids Also as radius is greater so you have to move a lesser distance away from the nucleus to ionize the atom so F×d decreases

So IE is less.