KEY2CHEM

Electronic Structure of the Atom

The electronic structure of an atom is determined by its electron energies. The energy levels of electrons can be determined by photoelectron spectroscopy (PES), which indicates the values for ionization energies. The more tightly held an electron (closer to the nucleus), the more energy is required to remove it. Within an energy level, factors that alter the net nuclear charge an electron experiences can vary the energy. For example, shielding (also called screening) of one electron by another can reduce the net nuclear charge an electron experiences. This decreases the amount of energy needed to remove it.


Example 1.

Which atom will have a greatest first ionization energy: \(\require{mhchem}\ce{As}\)\(\require{mhchem}\ce{Se}\), or \(\require{mhchem}\ce{Br}\)?

A. \(\require{mhchem}\ce{As}\)

B. \(\require{mhchem}\ce{Se}\)

C. \(\require{mhchem}\ce{Br}\)

 

Solution 

C. \(\require{mhchem}\ce{Br}\)

\(\require{mhchem}\ce{Br}\) has the largest number of protons of these species, so each electron is held more closely to the nucleus compared to the others. Although \(\require{mhchem}\ce{Br}\) has additional electrons compared to the other species as well, electrons in the same energy level do not shield one another very effectively. This means each electron in \(\require{mhchem}\ce{Br}\) experiences a greater effective nuclear charge compared to the other atoms. 


Example 2.

Write the ground state electron configuration for \(\require{mhchem}\ce{Cl}\).

A. \(1s^22s^22p^63s^23p^3\)

B. \(1s^22s^22p^63s^23p^5\)

C. \(1s^22s^22p^63s^23p^6\)

 

Solution 

B. \(1s^22s^22p^63s^23p^5\)

The \(17\) electrons in a \(\require{mhchem}\ce{Cl}\) atom fill in increasing energy order, from lowest levels (also called shells) to higher levels. Within a level, there are sublevels (also called subshells) that fill in increasing energy order. 


Example 3.

Which third period element is expected to give the following successive ionization energy (IE) data? Units are \(\text{MJ/mol}\).

 

\(\text{IE}_1\) \(\text{IE}_2\) \(\text{IE}_3\) \(\text{IE}_4\) \(\text{IE}_5\) \(\text{IE}_6\) \(\text{IE}_7\)
\(0.79\) \(1.58\) \(2.75\) \(4.36\) \(16.1\) \(19.8\) \(23.8\)

A. \(\require{mhchem}\ce{Si}\)

B. \(\require{mhchem}\ce{Cl}\)

C. \(\require{mhchem}\ce{P}\)


Solution

A. \(\require{mhchem}\ce{Si}\)

Core (inner) electrons require much more energy to remove than do valence (outer) electrons. The data shows that the first four electrons are relatively easy to remove, while the remaining electrons require much more energy. This indicates the first four electrons are valence electrons while the subsequent electrons are core electrons. The element with four valence electrons in the third period is \(\require{mhchem}\ce{Si}\).