7

First, a point that is more like a comment: I'm not sure that just because electrons get put in to d-orbitals (and thinking of crystal band structures in terms of atomic orbitals is of questionable value anyway) that it would not be a metal. Many elements (one could even say most elements) with d-electrons are quite happy as metals. You might want to start ...


6

TL; DR: coordination number of carbon atom in methyllithium tetramer is indeed 7, arising from 6 intramolecular interactions, as you suggested, and an extra bond with the lithium atom of the next tetramer. The key point is that you need to look beyond the tetramer. I'm sorry to inform you that the image from Wikipedia of the standalone cubane $\ce{Li4(CH3)4}...


6

There is no need (or possibility, really, in terms of standard lab capabilities) to oxidize sodium(I). In fact, one method relies on sodium(I) reduction to metal as a method of eliminating unwanted chloride. Method 1 Electrolysis of molten sodium chloride: $$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)}$$ Oxidation of sodium metal to oxide by burning: $$\ce{4 Na + ...


5

Another interesting paper (arXiv:1310.4718 [cond-mat.mtrl-sci]) on this topic. I think you should be able to access this one, but I found three passages worth noting in the article. Peculiarities in structural behaviour of K under pressure are based on specific features of its electron configuration: K opens in the Periodic table the first long period ...


5

The group 1 elements are the so-called alkali-metals. The bonding between the atoms is caused by the interaction of the nuclei with the delocalized electrons. With increasing number of electrons and protons, the atomic radii get bigger and hence this interaction becomes weaker as the average distance between nuclei and electrons increases as you go down in ...


5

One candidate for reaction with the rubidium is silica, $\ce{SiO2}$. Reference [1] reports that black powders are formed when alkali metals are contacted with silica. Such powders are more stable in dry air than ordinary alkali metals. The reference suggests that electrons are imparted from the alkali metals into the silica forming an electride complex. ...


4

I can only give you a qualitative answer based on my experience. But I'm doing a lot of Europium based chemistry where I use Europium metal as reagent, often together with sodium as a fluxing agent. The sodium is then removed using liquid ammonia by constantly dissolving it, decanting it off and warming it up. Just to give you an idea, I'm using about 90 - ...


4

The reactions are ongoing this way: Relatively free electrons of potassium reduce water: $$\ce{2 e- + 2 H2O -> H2 + 2 OH-}\tag{1}$$ That leaves metal positively charged. Liquid ammonia, if exposed to alkali metal, reacts with electrons much slower than water, forming a dark blue solution of solvated electrons. As electrons progressively kick out ...


4

The first step you can take to make the spectral bands narrow is to reduce the slit width. It is slightly wide at this moment. We are not sure where you were looking for wavelengths? In the compiled tables in handbooks or schematic diagrams on the web, the emission wavelengths are listed for the excited atoms using very high temperature flames using air ...


4

As suggested in the comments, rubidium aluminates do exist. Aluminates such as $\ce{RbAlO2}$ and $\ce{Rb6Al2O6}$ are prepared at higher temperatures (above 550 °C [1]) or even from the melt. However, alkali metal vapors at elevated temperatures act as reducing agent, so I wouldn't expect aluminates anyway. On the other hand, aluminides such as $\ce{RbAl}$ (...


4

You are talking about alkalides, salts where the anion is an alkali metal. There is a very brief overview on Wikipedia which also provides a couple of examples, including $\ce{[Na(\text{cryptand[2.2.2]})]+Na-}$. A good university level inorganic text book such as Greenwood and Earnshaw or Housecroft and Sharpe will provide more detail. If you are interested ...


4

Reactivity of Group 1 and 2 elements increases as you go down the periodic table. So sodium is more reactive than lithium. Sodium will react with oxygen forming $\ce{Na2O}$ (sodium oxide). Lithium forms lithium oxide $\ce{(Li2O)}$. Basically, there are general trends (reactivity with acids, air, etc.) down a group or along a period.


4

As noted in the answer to the other question, alkali metal can exist in higher oxidation state when bonded with polycylic multidendate ligands like cryptands etc. This is an excerpt from an eBook: The chemistry of group 1 elements have been dominated by +1 oxidation states. However, there have been indications that caesium might form higher oxidation ...


4

As Mithoron says in the comments, tert-butyllithium is one choice. Like pure organolithium compounds generally, this is pyrophoric and reactive to many common atmospheric gases or vapors. An article by Schwindeman et al. [1] recommends using a nitrogen or argon atmosphere. In the case of alkaline earth metals, magnesocene provides a relatively volatile ...


4

I always suggest students to try Google Scholar (scholar.google.com) when a simple Google search fails. I just searched three keywords : alkali metals ammonia solutions and the third result is highly relevant. When your book talks about "in concentrated solution", it means more alkali metal in liquid ammonia. This paper, which you should search in Google ...


3

I agree. This is a dangerous reaction and should not be tried without proper equipment, including a lab hood and fire extinguishing equipment. The fumes could be "just" droplets of sodium hydroxide solution. I've seen similar fumes involving hydrochloric acid, just by adding water to rinse out traces of concentrated acid (in a hood, of course). They may ...


3

Here is a harder version of the answer above. Sodium has 3 electron shells while lithium has only 2 electron shells Sodium can donate its electrons more easily because the valence electron is further from the nucleus and the force of attraction between both is weaker. Whereas lithium donates its electron less easily because the valence electron is closer to ...


2

Crown ethers are typically not recycled because they remain in the organic phase where normally also all the byproducts are (so the purification is very complicated) and they are still complexing the alkali metal after the reaction. You want to recycle them to again complex some alkali metal, but then you have first to break down the previous complex, which ...


2

It's hard to tell what is the answer you were supposed to provide, but yes, generally speaking, you are correct. According to CRC handbook of chemistry and physics [1, p. 4-14] (emphasis mine): Francium, the heaviest known member of the alkali metal series, occurs as a result of an alpha disintegration of actinium. […] Because all known isotopes of ...


1

You can explain the trend in terms of increased charge density, and ensuing charge repulsion between nuclei and between electrons as you go down the group. An alternative explanation is provided in a libretext, which draws information from a popular educational chemistry site: When any of the Group 1 metals is melted, the metallic bond is weakened ...


1

These at two different scenarios and two different phenomena. Flame color is due to atomic emission and complexes are colored because of light absorption in the solid state and in solution. In a high temperature flame, an atom is considered to be a free and independent entity. With heat, the electrons in the atoms are excited to higher energy levels. ...


1

I agree completely with M. Farroq. The quality of your data might improve greatly if you reduce the slit width of your spectrometer, you will see less background light and obtain higher resolution. Lookig at your image, it seems to me that your design may have a problem with the angle of incidence of the light at the difraction grating. It should be at ...


1

Because, in the condensed phases, you don't have molecules of metallic elements. You have, as a first approximation, a "rice pudding" structure where cations are embedded (like raisins) in a "pudding" of delocalized valence electrons (metallic bonding). The main cohesive force in the condensed metals is then electrostatic attraction between the opposing ...


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