Dependence of the energy of an electron on distance from the nucleus

Question

I was going through an article given to me by my teacher which gave info about the influence of the positive nucleus on the electrons in the 's orbital', 'p orbital' and 'd orbital' respectively. Here I am quoting the part of the article which I didn't understand.

Despite the shielding of the outer electrons from the nucleus by the inner shell electrons, the attractive force experienced by the outer shell electrons increases with increase of nuclear charge. In other words, the energy of interaction between, the nucleus and electron (that is orbital energy) decreases (that is more negative) with the increase of the atomic number (Z).

Now after reading the article further I came across this paragraph..

The energy of electrons in s orbital will be lower (more negative) than that of p orbital electron which will have less energy than that of d orbital electron and so on.

Aren't both the above paras contradicting?

First, it says that as the electron is away from nucleus, the nucleus's influence decreases. (like it should) Then why is electron more negative? (see brackets in first para above). Does it mean that as the electron moves away then it has less positive charge and the negative charge dominates?

But in the second para it says that the electron in s orbital (which is near nucleus) is more negative? I mean what's going on in both of these paras?

All in all I want to know,

• Is electron near nucleus more negative/less negative and has lower energy/higher energy?

OR

• Is electron away from nucleus more negative/less negative and has lower energy/higher energy?

And why?

Please help me. I did genuine efforts to understand all this and surfed internet for many hours but couldn't get it.

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Reference for the text quoted above: NCERT Class XI Chemistry Book, chapter 2, §2.6.3 (Energies of orbitals), also available as ark:/13960/t0tr17s5t.

Answer 1

The two statements are independent and not at all contradictory. Specifically:

• For all subshells (but keeping the subshell fixed), as you increase the nuclear charge $Z$, the energy becomes more negative.
• For a fixed nuclear charge $Z$, as you look at the different subshells of the same atom, the $s$-shell electrons will be more tightly bound (have a more negative energy) than the $p$-shell electrons.

The only way to glean a contradiction from this is to compare apples with oranges.

Answer 2

Let me break it down for you .The statements quoted in that article talk about two types of energies: The one which is due to the nucleus of an atom, and the other which is the energy lost by the electron when it starts coming towards a nucleus.

When an electron is away from a nucleus it has zero influence over it(by the nucleus of an atom) , so when it gets attracted by a nucleus, in starts loosing energy and therefore it's energy becomes negative .So we can say that the electrons which are more close to the nucleus have more negative energy relative to other electrons away from the nucleus. Similar thing happens in case of s, p and d subshell. The electrons present in s subshell are more close to the nucleus then the ones present in p or d subshell. The electrons present in the s subshell decrease the nuclear factor on the electrons present in the outer shells in such a way that electrons present in the S subshell have relatively more energy then the electrons present in the outer shell.