2019年4月10日 · As we progress from atomic number 57 to 71, an electron is added in 4f energy level with increase in atomic number with the exception of some metals where the electron occupies a 5d energy level. In case of Cerium, energy of 4f and 5d subshells are quite similar and thus explains the electronic configuration stated above.

2020年4月26日 · I need to calculate the term symbol of the ground state configuration of Cerium. The electron configuration is: $[Xe] 4f^1 5d^1 6s^2$ Since the d and f electrons are in different subshelles, they are non-equivalent electrons and Pauli exclusion principle doesn't apply, so:

2019年3月9日 · Cerium in $0$ oxidation state has electronic configuration $$[\ce{Xe}]\mathrm{(4f)^1(5d)^1(6s)^2}$$ But when it gets oxidised to $+2$ state, it becomes $$[\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$ This is a change as in other elements in the lanthanides series get their $\mathrm{6s}$ electrons abstracted and the other shells remain at liberty ...

2017年12月10日 · Electron configuration is not the be-all and end-all of stability. It's just one term in the energy balance. Other factors like the electron affinity of the anion former (think of oxygen or fluorine versus sulfur or iodine), lattice or solvation energies, etc may override the electron configuration term; if so, then you find your europium in the +3 oxidation state after a

Now back to the element which the question was about, the problem with the question is that it did not state what environment the cerium was in. if the cerium is in the solid state as an oxide then cerium dioxide (ceria) CeO2 will be very nice and stable. The other lanthanides do not form dioxides with the same ease.

2018年4月10日 · The outer electron configuration for $\ce{Mo}$ is $\ce{5s^1 4d^5}$. The number of valence electrons would be 6 if I am correct. Wouldn't the Lewis dot structure look like the one for group 6A elements with six dots? EDIT: I spoke to my lecturer, he said I have to do research and draw the Lewis dot diagram of Mo and Ce. My textbook says that ...

So, I am studying electronic configuration but the elements of the series of the lanthanides confused me. The electronic configuration of $\ce{La}$ is $\mathrm{6s^2\,5d^1}$ and that of cerium is $\mathrm{6s^2\,5d^1\,4f^1}$. So, why does the extra electron that was added make the $\mathrm{5d}$ orbital have a higher energy?

2013年3月8日 · First of all the configuration of $\ce{Cu}$ you see has anomalous electron configuration because completely filled or half filled sub shells are more stable than any other configuration and you can see the configuration at last it is $\ce{4s^13d^{10}}$ and the basic configuration is $\ce{4s^23d^{10}}$ so you can see that $\ce{4s}$ is half filled and $\ce{3d}$ is completely filled.

2020年7月26日 · Lanthanum has electronic configuration [$\ce{Xe}$] $4\mathrm{f}^0$ $5\mathrm{d}^1$ $6\mathrm{s}^2$. Why is the $4\mathrm f$ sub-shell not filled even though we know that $4\mathrm f$ is filled before $5\mathrm d$ sub-shell? I know that for some elements energy level of $4\mathrm f$ > $5\mathrm d$. Why is this so?

$\begingroup$ Electron spin is not necessarily arbitrary in the absence of another electron. Interestingly, the nucleus is a point of reference for spin to be defined and there is a very real, physical consequence with respect to which spin the electron has. This comment is included just for fun. $\endgroup$ –