Selenium (Se) is a non-metal element that sits in Group 16 of the periodic table, located right below sulfur, and possesses an atomic number of 34. This means a neutral selenium atom contains 34 electrons. The arrangement of these electrons dictates nearly all of the element’s chemical properties, particularly its ability to form bonds. To understand selenium’s behavior, it is necessary to determine the exact number of “unpaired” electrons, which are ready to participate in chemical reactions.
Understanding Atomic Shells and Orbitals
Electrons exist within distinct energy levels referred to as shells, numbered starting from one. Higher numbers represent greater distance from the nucleus and higher energy. Within each shell, electrons occupy specific regions of space known as atomic orbitals, which represent the probability of finding an electron in that area.
These orbitals are designated by letters: \(s\), \(p\), \(d\), and \(f\). A single orbital of any type can hold a maximum of two electrons. Consequently, the one \(s\) orbital holds up to two electrons, the set of three \(p\) orbitals holds up to six electrons, and the set of five \(d\) orbitals holds up to ten electrons.
Determining Selenium’s Electron Configuration
To find the location of the 34 electrons in selenium, electrons must fill the lowest energy orbitals first. Starting from the innermost shell, the configuration begins with the \(1s\), \(2s\), \(2p\), \(3s\), and \(3p\) orbitals. These inner shells are completely filled, accounting for 18 electrons and matching the stable configuration of the noble gas argon, \([Ar]\).
The remaining electrons fill the orbitals of the fourth shell, which represents the outermost energy level, or valence shell. The filling sequence continues with the \(4s\) orbital, which accommodates two electrons, followed by the ten electrons that fill the \(3d\) orbitals. This brings the total count to 30 electrons, leaving four more to place.
The final four electrons enter the \(4p\) orbitals, resulting in the full electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^4\). The valence shell contains the \(4s^2\) and \(4p^4\) electrons, totaling six valence electrons. The next step is to determine the arrangement of these six outermost electrons within the partially filled \(4p\) subshell to find the unpaired count.
Applying Hund’s Rule to Count Unpaired Electrons
The number of unpaired electrons is determined by applying Hund’s Rule to the valence shell. The focus is on the four electrons in the \(4p\) subshell, as the \(4s\) orbital is full with two paired electrons. Hund’s Rule states that when filling a set of equal-energy orbitals, electrons will first occupy the orbitals singly before any orbital receives a second electron.
The \(4p\) subshell consists of three separate \(p\) orbitals. The first three of the four available \(p\) electrons will each enter a separate orbital with parallel spin. The fourth electron must then pair up with one of the electrons already present. This leaves the other two \(p\) orbitals with one electron each. Therefore, a neutral selenium atom possesses a total of two unpaired electrons in its ground state configuration.
How Unpaired Electrons Influence Chemical Reactivity
The presence of two unpaired electrons in the \(4p\) orbitals dictates selenium’s chemical behavior. Atoms react to achieve a stable electron configuration, typically by obtaining a full octet of eight valence electrons. Since selenium has six valence electrons, it readily seeks two additional electrons to complete its outermost shell.
This drive to gain two electrons explains why selenium commonly forms an anion with a \(-2\) oxidation state in selenide compounds, similar to oxygen and sulfur. The unpaired electrons are also used to form covalent bonds, typically creating two single bonds, as seen in hydrogen selenide (\(H_2Se\)). Selenium can also utilize its empty \(4d\) orbitals to expand its valence shell, leading to higher oxidation states of \(+4\) and \(+6\) in compounds like selenium dioxide (\(SeO_2\)).