KEY2CHEM

Describing Electron Energies with Coulomb's Law

Electrons in an atom are distributed in energy based on Coulomb’s law: the force of interaction between charged particles is proportional to the product of their charges (\(q_1 \) and \(q_2\)) and inversely proportional to the square of the distance (\(r\)) between the charges. \(F\; \alpha \;\frac{q_1 \times q_2}{r^2}\)

Since work (a form of energy) is expressed as a force multiplied by a distance (\(w = F\times d\)), Coulomb's law can be expressed as the energy of interaction charged particles as: \(E_{electrostatic}\; \alpha \;\frac{q_1 \times q_2}{r}\).

Electrons closer to the protons in the nucleus are lower in energy (and therefore require greater ionization energies to remove them from the atom) and electrons further from the nucleus are higher in energy (and require smaller ionization energies). 


Example 1.

In an atom of \(\require{mhchem}\ce{B}\), which electron is easiest to ionize: \(1s\), \(2s\) or \(2p\)?

A. \(1s\)

B. \(2s\)

C. \(2p\)

 

 

Solution

C. \(2p\)

The electron in the \(2p\) subshell is easier (requires less energy) to ionize. The electron in this subshell is further from the nucleus, so it experiences a lower net nuclear charge and is easier to remove.


Example 2.

What is true about the energies of the electrons in the oxygen atom?

A. The energy of the \(2s\) and \(2p\) electrons is the same.

B. The \(2s\) electrons are higher in energy than the \(2p\) electrons.

C. The \(1s\) electrons are lower in energy than the \(2s\) electrons.

 

 

Solution

C. The \(1s\) electrons are lower in energy than the \(2s\) electrons.

Electrons are negatively-charged and repel one another. Electrons in different orbitals repel to varying extents; this difference in electron-electron repulsion is responsible for the differences in energy between electrons in different orbitals in the same shell. The \(1s\) require more energy to be removed from the atom (to be ionized). As such, they are lower in energy (more stabilized due to their closer position to the nucleus) compared to the \(2s\) electrons.


Example 3.

Which atom will have the lowest energy \(1s\) electron?

A.  \(\require{mhchem}\ce{Si}\)

B. \(\require{mhchem}\ce{P}\)

C. \(\require{mhchem}\ce{S}\)

 

Solution

C. \(\require{mhchem}\ce{S}\)

Since \(\require{mhchem}\ce{S}\) has the greatest number of protons, each electron will experience a larger net nuclear charge. This results in electrons being pulled closer to the nucleus, lowering their energy.