# What does a fluoride ion have in common with a neon atom and a sodium ion?

##### 1 Answer
Jan 9, 2016

The atomic number ($Z$) of the 3 elements $F$, $N e$, and $N a$, are $9$, $10$, and $11$.

#### Explanation:

Now $Z$ refers to the number of protons in the element's nucleus, and protons are POSITIVELY charged particles. So a fluoride ion, ${F}^{-}$, has 10 electrons rather than 9 (why?), a neutral neon atom has 10 electrons, and a sodium ion, $N {a}^{+}$, also has 10 electrons (why?).

So the 3 species are ISOELECTRONIC; they possess the same number of electrons.

You should look at the Periodic Table to confirm the electron number. Elements are (usually) electrically neutral (sometimes they can be ionic if they have lost or gained electrons). If there are 10 positively charged protons in the nucleus, there are NECESSARILY 10 electrons associated with the NEUTRAL atom. I don't know WHY I am capitalizing certain WORDS.

You might ask why sodium will form a positive ion, $N {a}^{+}$, whereas $F$ forms a negative ion, ${F}^{-}$. This again is a Periodic phenomenon, and explicable on the basis of the electronic structure that the Table formalizes.

Neutral metals tend to be electron-rich species, which have 1 or more electrons in a valence shell remote from the nuclear charge. On the other hand, neutral non-metals have valence electrons in incomplete shells, that do not effectively shield the nuclear charge. The demonstrable consequence is that metals lose electrons to form positive ions, whereas non-metals gain electrons to form negative ions.

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