What is formal charge and how is it determined?

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What is formal charge and how is it determined?

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Answer 1 · 2017-09-09T12:45:44

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:

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

Where,

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

Let us take for example ethane molecule $C_{2} H_{6}$ $C_2H_6$
![https://useruploads.socratic.org/eJ8DZZp9RcuRzGDl9pV4_C2H6.png)

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

For $V$ $V$ :
The number of valence electron on carbon is 4.
You can read more about finding the number of valence electrons here.
$\to V = 4$ $rarr V=4$
For $N$ $N$
Carbon has all its valence electrons shared in bonds (1 with carbon and 3 with 3 hydrogen atoms)
$\to N = 0$ $rarr N=0$
For $B$ $B$ :
Each carbon has 4 single covalent bonds in ethane and each single covalent bond is made of 2 electrons, thus each carbon atom has $4 \cdot (2)$ $4*(2)$ electrons shared.
$\to B = 4 \cdot (2) = 8$ $rarr B=4*(2)=8$

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

$formal charge = V - N - \frac{B}{2} = (4) - (0) - \frac{8}{2} = 4 - 4 = 0$ $"formal charge"= V - N - B/2=(4)-(0)-(8)/2=4-4=0$

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

Answer 2 · 2017-09-10T10:07:54

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

Explanation:

Take ethanol $C H_{3} - C H_{2} - O H$ $CH_3-CH_2-OH$

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.

Answer 3 · 2017-09-12T00:59:46

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.

$Formal charge$ $"Formal charge"$ $=$ $=$ $valence electrons - owned electrons$ $"valence electrons - owned electrons"$

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

$H - O - O - H$ $"H"-"O"-"O"-"H"$

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

$H - . . O \cdot + \cdot . . O - H$ $"H"-stackrel(..)"O"cdot + cdot stackrel(..)"O"-"H"$
$a^{. .} a^{. .}$ $" "color(white)(a)^(..)" "" "color(white)(a)^(..)$

And further cleaving the $O - H$ $"O"-"H"$ bond gives:

$H \cdot + \cdot . . O \cdot + \cdot . . O \cdot + \cdot H$ $"H"cdot + cdot stackrel(..)"O"cdot + cdot stackrel(..)"O"cdot + cdot "H"$
$" "color(white)(....)color(white)(icdot)^(..)" "" "color(white)(ii)^(..)$

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 $"H"$ and $"O"$ in $"H"_2"O"_2$ ? Why are they not zero?

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What is formal charge and how is it determined?

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