Question

In a multielectron atom, the lowest$-l$state for each$n\left(2s,3s,4s,etc.\right)$is significantly lower in energy than the hydrogen state having the same $n$. But the highest$-l$ state for each $n(2p,3d,4fetc)$ is very nearly equal in energy to the hydrogen state with the same $n$. Explain.

Short answer

The given statement is proved below.

Step-by-step solution

Given Information  

We have to given the lowest$-l$state for each $n\left(2s,3s,4s,etc.\right)$, and the highest$-l$state for each $n(2p,3d,4fetc)$.

Explanation

High- $l$states correspond to classical orbits that are nearly circular, so a high- orbit stays mostly outside the orbits of the inner electrons. The inner electrons "shield" the nucleus, then the outer electron "sees" a net charge of only ${}^{+}e$. The$z$protons are shielded by $z-1$electrons. The force on the outer high- electron from this net charge of${}^{+}e$is, to a good approximation, the same as the force on an electron in a hydrogen atom. Therefore the energy of a high-state is very close to that of the hydrogen state with the same $n$.

A low-l state corresponds to a classical orbit that is highly elliptical. These penetrating orbits dive to the nucleus. During the close part of its orbit, the electron is no longer shielded by the inner electrons and it "sees" the full nuclear charge ${}^{+}Ze$. Because the potential energy of opposite charges is negative, this increased interaction with the nucleus lowers the electron's energy; it is harder for a low-$l$electron to escape the pull of the nucleus. It is true that an electron in a penetrating orbit also gets closer to the other electrons and experiences an increased repulsive force. But the electrons are spread out and try to keep their distance, so the increased repulsion from the inner electrons is small in comparison to the increased attraction to the nucleus. The net effect is that the energy of a low-$l$state is less than that of a hydrogen state with the same $n$.