# Why are higher-energy orbitals larger in size/radius?

I know the different atomic orbitals of atoms are derived from the wave function in the Schroedinger equation. Because I'm not a mathematician, though, I can't decipher the Schroedinger equation and therefore can't simply explain why higher-energy orbitals are larger than lower-energy orbitals (and therefore contain electrons that are, on average, further from the nucleus). For instance, the 2s orbital is larger than the 1s orbital for any atom, but why?

Is there a relatively simple answer to my question (that doesn't require complex mathematical understanding)? Or is the answer simply that the higher-energy atomic orbitals are larger because that's the solution given by the Schroedinger equation? I'm assuming this is the case, but I'm not entirely sure.

• Addressing the title, applying the classical analogy of the central quadratic force (F = k/r^2).: It is like when a planet has higher mechanical ( potential + kinetic ) energy, if its orbit is farther from the Sun. Commented Jan 19, 2022 at 9:52
• Roughly speaking, the potential energy is $-\frac1r$. Larger orbital means greater $r$ means smaller $\frac1r$ means greater $-\frac1r$ means higher energy. Commented Jan 19, 2022 at 9:53
• $2p$ orbitals are higher in energy than $2s$ orbitals. Yet, $r_{mp}$ (most probable radius) is smaller for the $2p$ orbital than for the $2s$ orbital. ($r_{mp}$ can be checked on the RPD diagram). Commented Jun 17 at 8:31

The highest energy state ($$n=\infty$$) is a free electron, i.e. an electron at "infinite" distance. The excited states orbitals are merely stops on the way from the ground state ($$n=1$$) to infinity, both in terms of energy and average distance from the nucleus.

You might compare it to lifting a stone from the ground. It costs you energy to lift the stone, and therefore at any distance from the earth the stone has a higher (potential) energy. If you stop carrying the stone at any height, it will spontaneously fall back to the ground state; that's why we say the ground state has the lowest energy.

In an atom, the picture is more complex (for starters, it's quantized) but directionally the same.

In order to determine whether an orbital corresponds to a higher energy state you need to look at the combined potential+kinetic energy.

Although dot-density diagrams are very informative about the potential energy of an electron in an orbital, they tell us nothing at all about its kinetic energy. It is impossible, for example, to decide from Figure 5.6 whether the electron in a 1s orbital is moving faster on the whole than an electron in a 2s orbital, or even whether it is moving at all! Fortunately it turns out that this difficulty is unimportant. The total energy (kinetic + potential) of an electron in an atom or a molecule is always one-half its potential energy.

(A dot-density diagrams is a graphical representation of the electron density described by an orbital.)

As noted in the excerpt there is a simple relationship between the energies that allows us to relate the total energy to the potential energy (if you know the potential energy of an electron in a particular orbital, you know it's total energy). The relationship between kinetic, potential and total energy follows from the virial theorem. So how do you determine the potential energy?

The average potential energy of an electron can be computed as an average over all space of the Coulombic interaction energy between the electron in its orbital and the nucleus:

$$\left< PE \right> = \int |\psi|^2 \frac{-e^2Q}{r} dv$$

Here $$\psi$$ is the wavefunction of the electron derived by solving Schroedinger's equation, and $$|\psi|^2$$ describes the electron density about the nucleus. These are functions of the distance from the nucleus.

For an orbital with larger principal quantum number, the electron density $$|\psi|^2$$ at greater distances from the nucleus is larger, so that regions of space where 1/r is smaller are weighted more, and the overall integral becomes smaller (meaning less negative).

I know the different atomic orbitals of atoms are derived from the wave function in the Schroedinger equation. Because I'm not a mathematician, though, I can't decipher the Schroedinger equation and therefore can't simply explain why higher-energy orbitals are larger than lower-energy orbitals (and therefore contain electrons that are, on average, further from the nucleus). For instance, the 2s orbital is larger than the 1s orbital for any atom, but why?

More commonly an orbital refers to an approximate representation of the probability density of an electron, which describes the likelihood of finding the electron somewhere around the nucleus. A larger orbital contains more of its density spread further from the nucleus. (The word "orbital" is also used as a synonym for a 1-electron wavefunction, which is an approximate representation of the probability amplitude of an electron. The wavefunctions are approximate solutions of the Schroedinger equation.)

Is there a relatively simple answer to my question (that doesn't require complex mathematical understanding)? Or is the answer simply that the higher-energy atomic orbitals are larger because that's the solution given by the Schroedinger equation?

The simple answer is that electrons are attracted to the nucleus and placing them further from the nucleus requires that the energy of the electrons be increased in order to overcome the nuclear attraction. Generally this means that the average kinetic energy also increases (which also follows from the virial theorem).

Reference

1. Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn (2020). Electron Density and Potential Energy, in: General Chemistry, ChemPRIME at Chemical Education Digital Library (ChemEd DL). Accessed online 2/2022.