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# What is the electron configuration of ${{F}^{-}}$ ?

Last updated date: 15th Jul 2024
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The following must be kept in mind before writing any electronic configuration. Principal Quantum number: Electrons in an atom are characterized by a set of four quantum numbers and the maximum number of electrons that can be accommodated in a shell (energy level) is based on principal quantum number (n). In this case, we must find electron configuration of fluoride anion, ${{F}^{-}}$ so start by writing electron configuration of a neutral fluorine atom; $F.$ Fluorine is located in period $2,$ group $17$ of periodic table and has an atomic number of $9.$ This tells we that neutral fluorine atom has a total of $9$ electrons surrounding its nucleus. Its electron configuration will be $\,F:1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$ Now, ${{F}^{-}}$ anion is formed when $1$ electron is added to a neutral fluorine atom. Notice that the $2p$-subshell of the neutral atom contains $5$ electrons. Its maximum capacity is actually $6$ electrons, two electrons for each p-orbital.
This means that extra electrons will be added to one of three $~2p$ orbitals, let's say to $2py.$ $2p-$ subshell will now be completely filled, i.e. it will hold $6$ electrons. Electron configuration of fluoride anion will thus be $F:1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}.$
Remember that fluoride anion has a total of eight electrons in its second shell, outermost shell. This tell we that anion has a complete octet because fluoride anion is isoelectronic with neon, $Ne$, we can write its electron configuration using noble gas shorthand notation as ${{F}^{-}}:[Ne]$