Question

What is formal charge and how is it determined?

Answer 1

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 - 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_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`,`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.
  `rarr V=4`

• For `N`
  Carbon has all its valence electrons shared in bonds (1 with carbon and 3 with 3 hydrogen atoms)
  `rarr 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*(2)` electrons shared.
  `rarr B=4*(2)=8`

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

`"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

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

Explanation:

Take ethanol `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

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.

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

A useful example is the zig-zag-shaped `"H"_2"O"_2`, hydrogen peroxide:

`"H"-"O"-"O"-"H"`

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

`"H"-stackrel(..)"O"cdot + cdot stackrel(..)"O"-"H"`
`" "color(white)(a)^(..)" "" "color(white)(a)^(..)`

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

`"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?