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For writing the electronic configuration of the $\ce{Fe^2+}$ ion, you need to remove two electrons (as it is an ion) which are from the highest energy level.

Apparently, this would be $\mathrm{4s}$ from $n = 4,$ but I see that $\mathrm{4s}$ is actually lower than $\mathrm{3d}$ in the shell $n = 3.$ Is there a flaw in my understanding?

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Yes, generally, 4s orbital indeed has higher energy than 3d orbital. Based on what I read, that 4s orbital is at lower energy level than 3d orbital is a wrong concept, as that has been proven to be wrong spectroscopically and by calculation.1 But as Poutnik has pointed out, the order of orbital energy is not fixed.

If then, why we normally say that 4s orbital will be filled before 3d orbital?

Although 4s orbital has higher energy than 3d orbital, the most probable distance of 4s electron from the nucleus is longer than than of 3d electrons, so if you put two electrons in 3d orbitals, the two electrons will be closer together and the repulsion will be more significant. Else if you put two electrons in 4s orbitals, yes, you first need to pay higher energy because 4s higher in energy, but the cost is less than the one incurred by strong repulsion when placing two electrons in 3d.1

Why do we remove electron from 4s but not 3d?

Remember that the energy level is not fixed. The 3d repulsion effect mentioned above can be compensated when your effective nuclear charge is big enough (especially when you are discussing about cations).1 Just like you put electron in further 4s, but it got pulled down to 3d which is closer to nucleus (in terms of most probable distance).


1 Peter Atkins, Julio de Paula. Physical Chemistry (8th Edition). OUP. 2006. Page 342.

P.S. I don't know why, based on the reasoning above, K is 3d0 4s1 but not 3d1 4s0

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  • $\begingroup$ Because for K, 4s has lower energy than 3d. Their energy mutually switches later, for scandium upwards. So does Ca configuration is 3d0 4s2. By other words, for 3d+4s, there is a kind of filling/removing hysteresis. $\endgroup$ – Poutnik Aug 23 '20 at 13:38
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The order of orbital energies is not fixed, but depends on electron filling, because of different levels of kernel screening and mutual electron repulsion.

When both 3d and 4s are empty, 4s orbitals have lower energy than 3d and get filled first. But being filled , somewhere at scandium 4s orbitals get higher energy than 3d, so they get away first. This get few exceptions when 3d is half or fully filled.

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As per $(n+l)$ rule, $\mathrm{4s}$ as a value $n=4$ and $l=0$, therefore $n+l =4$.

In the case of $\mathrm{3d}$, $n=3$ and $l=2$, therefore $n+l=5$.

So, low energy orbital is $\mathrm{4s}$ and not $\mathrm{3d}$. Therefore electrons are easily lost from the $\mathrm{4s}$ level.

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