Ionization and exchange energy

Question

As we all know that 'Nitrogen' has higher 'Ionization Energy' than its neighboring elements both carbon and oxygen, opposite to the trend of increasing ionization energy across a period. The reason mentioned everywhere I have read is - the extra stability of half and full-filled orbitals due to exchange energy. Now let's discuss exchange energy. Many of you say that exchange energy is - The amount of energy released when electrons with the same spin swap positions in degenerate orbitals and since exchange is possible maximum in half and full-filled orbitals they are more stable. This statement raises a lot of questions in my mind -

1.What kind of energy is this? - electrical, magnetic, nuclear, or anything else that I have not come across yet

2.When we go across a period let's say from boron to carbon, a electron enters in carbon p orbital with the same spin as of another and that leads to release 'exchange' energy so from where did that exchange energy come to electron?

3.When we ionize hydrogen (no exchange energy in hydrogen) by supplying energy to it, electron use it to break the attraction between the nucleus and itself and comes out of the reach of the nucleus. But In nitrogen valence electron releases two kinds of energies when it enters in 2p orbital -

(I). Electric potential energy

(II). Exchange energy

So when we ionize nitrogen, the electron must gain all the energy that it has released before coming out and it gains electric potential energy by using the supplied energy through breaking the bond between the nucleus and itself but in what form and how it gains exchange energy. This means when an electron gets exchange energy does it stop exchanging its position or does anything else happens to the electron after getting Back exchange energy?

4.I also found someone https://chemistry.stackexchange.com/a/58628/129242 who is explaining exchange energy with coulomb repulsion between electrons but I think that's wrong because we already have used that repulsion factor when calculating effective nuclear charge and Nitrogen has less effective nuclear charge than oxygen https://en.m.wikipedia.org/w/index.php?title=Effective_nuclear_charge and according to that nitrogen will have lower ionization energy and we know that is not true so I think exchange energy is not related to electronic repulsion.Correct me if I am wrong

At the end please consider the fact that this question is asked by a 12 class student so please avoid higher mathmatical stuff.

Please comment if any point is not clear to you.

Answer 1

I admit that this exchange energy stabilizing a half-full shell is very questionable especially when the spins are aligned and this interaction bring a positive energy contribution. If I am wrong, feel free to downvote or add a comment.

Exchange interaction

Without trying to do deeper, there are fundamental properties related to an associated interaction. Most of the interactions can be described by a general potential of the form $\propto PP'\frac{1}{r}$, where $P$, $P'$ can be an electric charge, a magnetic charge, or a mass. A magnetic interaction is a relative dynamic effect of the electric interaction. An approximation will be from a classical rotational motion of a charge particle to deduce an intrinsic magnetic dipole generating a field rather than a static magnetic charge, but none of the classical representations will describe the symmetry of the intrinsic angular momentum generating a magnetic moment for a charge particle like an electron. Another approach will be to create a potential derived from the interactions of this intrinsic angular momentum (spin) : the exchange potential, which has the form of the Coulomb potential, but there is a permutation of the positions of the electric charges. The use of a permutation operator and the antisymmetry of the total WF after exchange gives the normal derivation of this contribution. Unlike the mass with always has a positive value, the charge and the spin can be either positive or negative, therefore the exchange interaction energy can be positive or negative, the positive coupling leads to an interaction that reduces the Coulomb repulsive interaction and promotes an overlap with antiparallel spins and a negative coupling avoids an overlap and creates parallel spins.

1.What kind of energy is this? - electrical, magnetic, nuclear, or anything else that I have not come across yet

The exchange energy in this case is an electrostatic or Coloumb energy but does not release any energy nor create a force. Instead, it couples spins by allowing two electrons with different spins to share a single orbital state (singlet state). To consider the exchange interaction as a true exchange of position is correct, an important case is an anyion, where a theoretical path of this exchange is important. That is all I can say about this interaction without using math.

Half-filled stability

The $s$ subshell is normally always an innermost subshell because of the orbital angular momentum $ \ell =0$. The orbitals with $\ell >0$ have a "centrifugal force" and are farther to the nucleus compared to $s$ orbitals. The solution of an electron SE in an effective potential gives orthogonal solutions, the orthogonality or two states ($|1\rangle$ and $|2\rangle$) implies that an electron measured in one state cannot be in another state, this means that the probability that an electron is in a state $|1\rangle$ position and after move to another state $|2\rangle$ is zero, there is no possible exchange between $|1\rangle$ and $|1\rangle$. Let's take the case of a half-filled $p$ orbitals, they are several configurations, the most stable configuration is the first Hund's rule, where the spins are aligned and each has one of the three $p$ orbitals as a state.

\begin{array}{|l|c|c|c|c|}\hline m_l & +1 & 0 & -1\\\hline & \uparrow & \uparrow & \uparrow \\\hline \end{array}

This configuration avoids Coulomb repulsion when two electrons have the same orbital. The question is whether an electron can exchange a position from one $m_l$ orbital to another $m_{l'}$ in this configuration : the answer is no because the orbitals are orthogonal and the probability that an electron goes from one orbital to another is zero, even the weak dipole interaction will avoid this transition. As a consequence, I do not understand the stability due to an exchange interaction. If will consider an exchange, the coupling will be positive because the spins are aligned and the ionization energy will not change much because the electrons will always feel approximately the same potential before and after the exchange, this argument is not correct.

The stability of a half-filled subshell is mainly due to the symmetry of this configuration and quantitatively the orbital angular momentum. This process can be fully understood with a semi-classical approximation. Let's consider a half-filled $p$ subshell, the "centrifugal force" acts on the projections $m_l \neq 0$, as an orbital has $m_l = -1$ and another $m_l=1$, the centrifugal force is cancelled, therefore $L=0$, the geometry is spherical and similar to $s$ orbitals, the electrons are closer to the nucleus, feel a stronger potential of the nucleus, the ionization energy is higher.