# Is this analogy right? [closed]

Is my logic correct in this case ?

Since the formal charge represents the charge the element possess in a covalent compounds, is it right to say that formal charge is the equivalent term for covalent bonds as is oxidation number is to ionic bonds ?

• Oxidation number applies to covalent compounds just as well. Mar 22, 2021 at 12:32
• But that is an assumption, right ? Covalent compounds don't generally carry charges to possess oxidation number. Mar 22, 2021 at 12:51
• No, that's a definition. We define oxidation numbers so as to be meaningful for covalent compounds too. They do not necessarily have any physical meaning, though. In particular, they are not real charges. Mar 22, 2021 at 13:13

## 2 Answers

The oxidation number is the charge that one atom would have if all the covalent bonds of the molecule (or of the ion) were transformed into ionic bonds, with a negative charge on the most electronegative atom of the bond, and a positive charge on the least electronegative atom.

• So if I'm correct formal charge is the actual charge possesed by the atom in a covalent compound ? Mar 22, 2021 at 13:44
• @Parvathy neither this is correct Mar 22, 2021 at 14:53
• @Parvathy. Not really. For example, in $\ce{CH4}$, no atoms are charged. The bonds are all covalent. But the electronegativities are $2.1$ for $\ce{H}$, and $2.5$ for $\ce{C}$. So there is a partial (or formal) positive charge $\delta +$ on the $\ce{H}$ atom, and a negative $\delta -$ on the Carbon atom. When calculating the oxidation number of C you add all the partial (or formal) charge, as if they were ionic charges. And you find $-4$ for the Carbon atom, and $+1$ for each hydrogen atom. Mar 22, 2021 at 14:53

Your analogy isn't correct. As @Ivan pointed out in the comments, oxidation numbers apply to covalent compounds as well as they do to ionic ones.

You calculate oxidation numbers assuming that the more electronegative atom takes up both the electrons in a bond. Formal charge, on the other hand, is determined assuming no difference in electronegativity.

Take the carbon atom in methane as an example. It forms four $$\ce{C-H}$$ bonds; none of them are polar enough to be called ionic. Yet we consider this bond to be perfectly ionic and assign a "charge" of $$-4$$ to carbon. This is its oxidation number.

When you calculate its formal charge, though, you ignore the (slight) difference in electronegativity between carbon and hydrogen. Carbon, then, happily ends up with the four electrons it started with. Its formal charge, therefore, is zero.

Note that neither of these numbers has any physical meaning.

Can you figure out the formal charge of each atom in the nitrate ion and ozone?

• Thank you for clearing my doubt ! It's 1 for O₃ and 0 for the oxygen of NO₃⁻ and -1/3 for the Nitrogen, if I'm not wrong. Mar 22, 2021 at 15:20
• Mar 22, 2021 at 15:32