# Why is ethyl alcohol neutral?

Jan 13, 2016

Because ethyl alcohol is composed of $C$, $O$, and $H$ atoms. Associated with each atom are sufficient electrons to complement (i.e. balance) the nuclear charge.

#### Explanation:

The atomic number, $Z$, of $H$, $C$, and $O$, are $1$, $6$, and $8$ respectively. These atomic numbers necessitate that there are $1$, $6$, and $8$ protons (positively charged nuclear particles) in the respective nuclei.

Thus for each hydrogen atom, there must be 1 electron for neutrality, and it gets one of the two electron that comprise the $O - H$ or $C - H$ bonds (yes, a covalent bond is comprised of 2 electrons, one is owned by the hydrogen, and one is owned by $C$). So hydrogen is neutral.

For $C$, $Z$$=$ $6$. There are 2 inner core electrons, so there must be 4 electrons from the covalent bonds; carbon here has 4 covalent bonds, so each carbon possesses (or owns) 4 bonding electrons. Since each carbon thus has 6 electrons, the carbons are also neutral.

For $O$, $Z$$=$ $8$. There are 2 inner core electrons; it gets 1 electron each from its 2 covalent bonds, and there are 2 lone pairs of electrons (these lone pairs devolve solely to oxygen). Thus there are 8 electrons in total, that precisely balance the charge of the 8 nuclear protons.

You can do this with any molecule and assign formal charge. Covalent bonds are shared equally (i.e. each atom "owns" 1 electron); lone pairs and radicals devolve solely to the atom of interest.

Can you use this treatment to designate formal charge in methyl radical, ${H}_{3} C \cdot$; $C {a}^{2 +} C \equiv {C}^{2 -}$, calcium carbide; methyl anion ${H}_{3} {C}^{-}$, and sodium amide ${H}_{2} {N}^{-} N {a}^{+}$? Where does the formal charge reside?