New answers tagged

2

Due to two contradictory factors, one is the effective nuclear charge which is greater in $\mathrm{5d}$ making it of lesser energy and the second one is a nodal factor. Nodal factor depicts localization of electrons, which indicates of higher energy. Hence to balance these two effects out, for Ce the second electron goes to $\mathrm{4f}$ and its electronic ...


2

Sodium carbide isn't hypothetical. It does exist and can be synthesized. In @Oscar's answer, there is a patent which shows the synthesis reaction of sodium carbonate and carbon at temperature 1050 °C-1200 °C and contacting the resulting gaseous mixture of carbon monoxide and sodium vapor on steel surface. There is another patent which shows a different ...


4

As the comments imply, sodium carbide certainly does exist. However, it is difficult to get because it is only metastable. Therefore it cannot be made from direct combination of the elements in thermodynamically stable form, a procedure often used (in some cases indirectly) with stable binary compounds. Sodium carbide can be made by using the sodium/CO ...


0

Yes, there are a large number of negatively charged metal ions, many of them containing several covalently bound atoms. The positively charged counterion is typically a group 1 metal. Wikipedia defines a zintl phase as the product of a reactio between a post transition metal or metalloid, but most of the examples are based on elements that are not ...


0

To make a capacitor out of metal, you have to have negatively charged metal atoms on one of the plates (although I guess one could argue that the negative charge is not tied to a particular atom, but shared among all the atoms).


6

The electron affinity, defined as the energy released in the reaction $$\ce{X(g) + e^- -> X^- (g)}$$ is positive for all but the noble elements and a few elements with half-filled shells (see a data table here), which is consistent with the anion being more stable than the neutral atom. This makes it difficult to understand the motivation for the first ...


10

Absolutely! You will find these mostly in electride systems and off these, mostly in alkali metals. Here is an example research paper: "Superakali-Alkalide Interactions and Ion Pairing in Low-Polarity Solvents, J. Am. Chem. Soc., 2021, 143(10), 3934–3943 (https://pubs.acs.org/doi/10.1021/jacs.1c00115) Remember, metals have a positive charged when ...


0

After the electron configuration, the last shell of the beryllium atom has two electrons. In this case, both the valence and valence electrons of beryllium are 2. We know the details about this. The elements that have 1, 2, or 3 electrons in the last shell donate the electrons in the last shell during bond formation. The elements that form bonds by donating ...


1

One way of looking at this is to consider that if $L = 1$ (for example), then there must be a state with $M_L = -1$ (because $M_L$ ranges from $-L$ to $L$ in integer steps). However, $M_L$ is not obtained by a Clebsch–Gordan series: instead, it's just a simple addition of the individual $m_l$'s of each electron. That is: $$M_L = m_{l,1} + m_{l,2} + m_{l,3} + ...


Top 50 recent answers are included