# Solutions of Group 1 and Group 2 metals in Ammonia

When we add Group-1 and Group-2 metals to liquid ammonia, they dissolve to form metal cations and solvated electrons.

$$\ce{M + NH3(liq) -> M+ + e-}$$

Now, when the G-1 solutions evaporate, we get the metal back, but in the case of G-2 solutions (except Beryllium), they give a metal complex, $\ce{[M(NH3)6]}$.

"Evaporation of the ammonia from solutions of Group 1 metals yields the metal, but with Group 2 metals evaporation of ammonia gives hexammoniates $\ce{[M(NH3)6]}$ of the metals. These slowly decompose to give amides."

Page No. 335, Chapter 11-'Group 2 - the alkaline earth elements', Concise Inorganic Chemistry

and,

"The alkaline earth metals except beryllium form similar solutions [to alkali metals] readily, but upon evaporation, a solid 'ammoniate', $\ce{[M(NH3)x]}$ is formed."

What is the reason for this contrasting behaviour, and do all Group 2 metals(except beryllium) exhibit such a property?

Source:

1. Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998

2. Huheey, James E., Keiter, Ellen A., Keiter, Richard L.; Inorganic Chemistry: Principles of Structure and Reactivity, 4th Edition; Pearson Education Inc.; 1993

• – Mithoron Sep 19 '17 at 15:03

As per P. Bahadur's Objective Chemistry 1st Year Program for IIT-JEE and All Other Engineering Entrance Examinations, under the chapter "s-Block Elements - The Alkaline Earth Metals", under Solubility in Liquid Ammonia, it reads:

Like Alkali Metals, alkaline earth metals also dissolve in liquid ammonia and form coloured solutions. When such a solution is evaporated, hexa-ammoniate, $\ce{M(NH3)6}$ is formed, which slowly decomposes to amides. $$\ce{M(NH3)6 -> M(NH2)2 + 4NH3 +H2}$$

(I noticed that it is almost exactly the same as JD Lee's Concise Inorganic Chemistry text.)

So far, it doesn't specify whether this metal $\ce{M}$ is only Beryllium or is the representation of the other alkaline earth metals as well... Since this is the case, I can safely assume that this applies to all the alkaline earth metals, and not just beryllium.

Another point mentioned in the book by Bahadur, under the chapter "s-Block Elements - The Alkali Metals", under Complex Ion Formation is that the Group 1 metals, i.e., the alkali earth metals have a larger size, low nuclear charge and hence don't tend to form complexes too easily. On the other hand, the Alkaline Earth metals are smaller, have higher nuclear charge, and hence, have a greater tendency to accept electrons, forming complexes. In Bahadur's Book, under "s-Block Elements - The Alkaline Earth Metals", under Formation of Complexes, he writes that:

However, $\ce{Be^{2+}}$ on account of smaller size forms many complexes such as $\ce{BeF3-}$, $\ce{BeF4^{2-}}$ and $\ce{Be(H2O)4^{2+}}$. Of the others, only $\ce{Mg}$ and $\ce{Ca}$ show much tendency to form complexes in solution and these are usually with oxygen donor ligands.

$\ce{NH3}$ is a ligand, and hence acts as a powerful complexing agent. This is why the $\ce{NH3}$ coordinates with these metals, forming hexa-ammoniates.

Source:

1. Bahadur, P.; Objective Chemistry 1st Year Programme for IIT-JEE & All Other Engineering Entrance Examinations; GRB Books; 2016

2. Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998

• To add to this Eu and Yb, which form reasonably stable 2+ ions, form blue solutions in ammonia which according to doi.org/10.1016/0022-5088(69)90184-2 decompose to the amide and imide. So while I can't quickly find a reference to the hexa-ammoniates it is consistent with the above as Eu and Yb are smaller than Ba, and Eu2+ and Yb2+ are smaller than Ba2+. Other lanthanides dissolve in NH3 to form blu solutions, but after a full 5 minutes searching I can't find decomposition products for Sm in particular. – Ian Bush Oct 6 '17 at 9:17
• @abhigyan According to your answer, the group 1 metals won't form complexes due to their large size and lower charge density. But, the second book(Principles of…) explicitly states that beryllium won't form the complex with ammonia, which seems a bit contradictory to me. – Ayushmaan Oct 6 '17 at 17:45
• @Ayushmaan I agree with your point, but sadly, I don't have any way of providing you an answer for this. I've looked through just about any textbook I could find, but most of them are copied off either from Concise Inorganic Chemistry or off Inorganic Chemistry: Principles of Structure and Reactivity. P. Bahadur's book is a little different, but as per that book, beryllium forms various complexes. Therefore, I'm truly unaware... And I'm really sorry – AbhigyanC Oct 7 '17 at 0:54
• @Ayushmaan I found the answer to your question regarding whether beryllium forms complexes or not. It turns out, beryllium is able to form complexes of various types, and has a strong tendency to do so. However, it is unable to form the "hexa" ammoniate because it's maximum coordination number is 4, and can't be extended to 6. What I wrote about the alkali metals (group 1) still stands the same. – AbhigyanC Feb 4 '18 at 17:10

I can't answer your two questions. As to your first question, which if I'm understanding it, is "Why are Group I and Group II metals different?" - I don't understand it. Why are red rubber balls and blue rubber balls different? Why are giraffes and crocodiles different? We actually can get into the details of what their differences are, but that's so obvious as to suggest that that is not what you have attempted to articulate. As a broad rule, Science isn't very good at answering "Why" questions, it is much better at answering "How" questions. I suspect you know why I and II elements are different. While I am not a big fan of the Periodic Table, we can see that I elements have an s electron, and II elements have two (full sub-shell) s electrons (in their valence shells). Would you predict, knowing just that, that they'd behave the same? I sure wouldn't.
As to your second question, (which I also can't really answer), well either Huheey, et al are wrong or not. I couldn't find much on-line about it, especially for Ra and Fr. What little I did find was decades out of date and itself admitted that the chemistry wasn't (then) well understood. (Cotton & Wilkinson, Adv. Inorg. Chem. 1972(!!)) From that is claimed that A. The II ammines are only "moderately stable". B. Dilute solutions of I & II metals in NH3 are blue, more concentrated solutions (e.g. 3M) are yellow/gold. This and other evidence suggests that in more concentrated solutions metal clusters rather than lone M+ ions exist. C. M+n + ne- is in equilibrium with M(NH2)n + (n/2)H2 and D. the M° coordination compounds are non-stoichiometric and only approach M(NH3)6.
I re-emphasize that this info is 45 years old and I know that a lot of progress has been made since then, especially in the area of M° coordination cmpds.
As an aside, I was tempted to "answer" your question by pointing out that everything has an opportunity cost. That is, the "reason" why the group I metals didn't form M° hexammines would be either because either the metal and (liquid? gaseous?) ammonia has a more negative free energy OR the amide is somehow kinetically retarded, contradicting the claimed equilibrium OR because the amide has a more negative free energy - compared to the M° complex's crystal structure. Opportunity cost. But since that is obvious (I hope!) I didn't think it would be helpful. (It's a tautology)

G-2 metals are smaller in size, so more charge density and thus have more (deltaO) value, so they form more stable complexes compared to G-1. G-1 metals complexes are uncommon because they are so elctropositive and can rarely act as lewis acids. Although there are certain cage like complexes formed by G-1 metals. (Refer JD Lee Chapter-Alkaline metals). Berillium is very small so it cannot accomodate 6 ammonia molecules around it. Also on a CFT point of view larger metals have larger (deltaO) value so they are more stable. (You can refer to Huheey Keter , chapter- Coordination Compounds on what are the factors that effect (deltaO) values.)