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The question is as follows:

What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers: $n=4, m_{l}=+3$

Since $m_{l}=+3$ I know that $l=3$ and then I wrote them all: $$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}$$ which means that the maximum number of electrons is $70$, but the actual configuration in the answer was: $$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 5d^1, 4f^{14}$$ which means that the maximum number of electrons is 71. What didn't I count for?

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Here are the electron configurations of the relevant elements:

enter image description here

Source: https://chemistrytheperiodictable.files.wordpress.com/2017/11/electronic-config-lanthanides.png?w=474&h=399

The first $f$ electrons get filled in the 6th period of the table of elements. First, the $6s$ get filled, then (for lanthanum) the $5d$ gets filled for the first time, and then the $4f$ subshell. As you go across the lanthanides, the $5d$ is sometimes filled with one electron and sometimes empty.

What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers: n=4,ml=+3

The answer depend on how you interpret "last electron". Going from Thulium to Ytterbium, you could argue the last electron is going into $4f$, whereas from Ytterbium to Lutetium, the last electron is going to $5d$. On the other hand, the $5d$ is filled before $5f$ going from Barium to Lanthanum. Comparing Lanthanum to Lutetium, all 14 electrons seem to be going into the $4f$ subshell, supporting the notion that the answer to the question is 71.

In the end, it is not the best question - it does not have a clear answer, and you don't really learn anything from the answer other than that the energies of $4f$ and $5d$ are close to each other.

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