# What is formal charge and how is it determined?

Sep 9, 2017

Charge assigned to a certain atom in a molecule.

#### Explanation:

Formal charge is the charge assigned to a certain atom in a molecule.

It can be calculated using the following formula:

$\text{formal charge} = V - N - \frac{B}{2}$

Where,

• $V$ is the number of valence electrons on an atom in its ground state.
• $N$ is the number of none bonding valence electrons.
• $B$ is the number of electrons shared in bonds.

Let us take for example ethane molecule ${C}_{2} {H}_{6}$

The formal charge for both carbon atoms will be the same, since they are surrounded by the same atoms (3 hydrogen and 1 carbon atoms)

We will first find $V$,$N$, and $B$ and then put their values in the formula to get the formal charge.

• For $V$:
The number of valence electron on carbon is 4.
You can read more about finding the number of valence electrons here.
$\rightarrow V = 4$

• For $N$
Carbon has all its valence electrons shared in bonds (1 with carbon and 3 with 3 hydrogen atoms)
$\rightarrow N = 0$

• For $B$:
Each carbon has 4 single covalent bonds in ethane and each single covalent bond is made of 2 electrons, thus each carbon atom has $4 \cdot \left(2\right)$ electrons shared.
$\rightarrow B = 4 \cdot \left(2\right) = 8$

Now that we have calculated $V , N \mathmr{and} B$ we put their values in the formula:

$\text{formal charge} = V - N - \frac{B}{2} = \left(4\right) - \left(0\right) - \frac{8}{2} = 4 - 4 = 0$

Therefore, the formal charge of each carbon atom is zero.

Sep 10, 2017

The charge (if any) on an atom in a molecule, leaving aside any effects of polarity caused by electronegativity differences.

#### Explanation:

Take ethanol $C {H}_{3} - C {H}_{2} - O H$

In the -OH group there is a large difference between the electronegativity of H and that of O, meaning that the O atom pulls the charge density towards itself, creating a partial negative charge on O (and therefore a partial positive charge on H). However, the formal charge on each atom is still zero.

Why? Because whilst O may have pulled a greater share of the charge density towards itself, it hasn't actually exchanged any electrons. It still continues to share electron orbitals with H as in any covalent bonded molecule.

So the formal charge is zero, even though some partial charges may still occur due to uneven electron sharing (polarity) of the bond.

Sep 12, 2017

Formal charge is the charge assigned to each atom in a molecule, assuming that all the bonds are 100% covalent, i.e. a 50/50 share of valence electrons.

$\text{Formal charge}$ $=$ $\text{valence electrons - owned electrons}$

A useful example is the zig-zag-shaped ${\text{H"_2"O}}_{2}$, hydrogen peroxide:

$\text{H"-"O"-"O"-"H}$

As the two oxygen atoms are equivalent, cleaving the molecule in two gives:

$\text{H"-stackrel(..)"O"cdot + cdot stackrel(..)"O"-"H}$
$\text{ "color(white)(a)^(..)" "" } {\textcolor{w h i t e}{a}}^{. .}$

And further cleaving the $\text{O"-"H}$ bond gives:

$\text{H"cdot + cdot stackrel(..)"O"cdot + cdot stackrel(..)"O"cdot + cdot "H}$
$\text{ "color(white)(....)color(white)(icdot)^(..)" "" } {\textcolor{w h i t e}{i i}}^{. .}$

And the formal charges for each of these would be $0$, as we would hope for the most stable structure, since the number of valence electrons is equal to the number of owned electrons under the assumption of 100% covalent character in these bonds.

What are the oxidation states of $\text{H}$ and $\text{O}$ in ${\text{H"_2"O}}_{2}$? Why are they not zero?