KEY2CHEM

Distribution of Electrons in an Atom

Electrons are negatively charged particles that are attracted to the positively-charged nucleus in an atom. The closer the electrons are to the nucleus, the more strongly attracted they are (this is Coulomb’s law, where the force of the attraction is proportional to the magnitude of the charges, and inversely proportional to the square of the distance separating the charges). Photoelectron spectroscopy (PES) data can provide information regarding how tightly held an electron is, based on how much energy is required to remove it from an atom or ion. This is a measure of the ionization energy; electrons that are more tightly held (closer to the nucleus) require the input of more energy and have higher ionization energies. Electrons that are more loosely held (farther from the nucleus) require less energy to be removed and have lower ionization energies. The energy of the photon needed to move an electron between two energy levels is given by $$E = h\nu$$, where $$E = \text{energy of the photon}$$ , $$h = \text{Planck's constant} = 6.626 \times 10 ^{-34}\text{ J}\cdot\text{s}$$, and $$\nu = \text{frequency of the photon}$$. Example 1.

PES data shows ionization energies for $$3$$ of the electrons in $$\require{mhchem}\ce{Ne}$$ to have values $$839\text{ MJ/mol}$$, $$4.67\text{ MJ/mol}$$, and $$2.08\text{ MJ/mol}$$. Which electron is furthest from the nucleus?

A. $$839\text{ MJ/mol}$$

B. $$4.67\text{ MJ/mol}$$

C. $$2.08\text{ MJ/mol}$$

Solution

C. $$2.08\text{ MJ/mol}$$

The electron with the smallest ionization energy (energy needed to completely remove the electron from the nucleus) is furthest from the nucleus. Since it is less tightly held, it requires less energy to remove it.

Example 2.

For the photoelectron spectrum of $$\require{mhchem}\ce{Ar}$$, what is the ratio of intensities of the highest energy and lowest energy peaks?

A. $$1:1$$

B.  $$1:2$$

C.  $$1:3$$

Solution

C. $$1:3$$

The electron configuration of $$\require{mhchem}\ce{Ar}$$ is $$1s^22s^22p^63s^23p^6$$. The highest energy peak corresponds to the $$1s^2$$ electrons, while the lowest energy peak corresponds to the $$3p^6$$ electrons. As such, the intensity ratio will be 2:6 (or 1:3).

Example 3.

Which second period element is likely to give the following photoelectron spectrum? A. $$\require{mhchem}\ce{Li}$$

B. $$\require{mhchem}\ce{N}$$

C. $$\require{mhchem}\ce{Be}$$

Solution

A. $$\require{mhchem}\ce{Li}$$

There are two peaks in the spectrum. The peak at higher energy ($$6.26\text{ MJ/mol}$$) is twice as intense as the lower energy ($$0.52\text{ MJ/mol}$$) peak. This tells us that the ratio of electrons between the higher energy and lower energy peaks is $$2:1$$. This corresponds to Li, with its electron configuration of $$1s^22s^1$$, where the $$1s$$ electrons are ionized at higher energy compared to the $$2s$$ electrons, which are less tightly held.