Question

What is the electron configuration of \[{{F}^{-}}\] ?

Answer 1

Hint: We know that representation of the chemical reaction in the symbolic form is called the chemical equation. And the chemical equation containing equal no of moles of all the atoms of the reacting species and that of the products is known as the balanced chemical equation.

Complete answer:
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 $F:1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}.$

Note:
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 ${{F}^{-}}:[Ne]$