# Formal Charge

## Key Questions

Well, the modern concept of the chemical bond relies on the distribution of electrons between atoms....

#### Explanation:

Electrons can be transferred between species such that a cation, and an anion result....and the charged particles can bind electrostatically in an extended array to give a salt...

$N a \left(g\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow N {a}^{+} C {l}^{-} \left(s\right) \downarrow$

...and typically this interaction occurs between an electron-rich species, a metal, and an electron-poor species, a non-metal...

Alternatively, electrons can be shared to give covalent bonds, a region of high electron-density between two positively charged nuclei, such that internuclear repulsion is negated, and a net attractive force between the nuclei results...

In either case we utilize formal charges to keep track of the electrons..the which acts as the glue between the positively charged nuclear particles.

In either case, it is useful to introduce the concept of formal charge with respect to the ATOM, which may be altered by the loss or gain or electrons...and in every chemical reaction both mass and charge are conserved.

• The formula for calculating the formal charge on an atom is simple.

Formal charge = [ of valence electrons] – [electrons in lone pairs + 1/2 the number of bonding electrons]

Since the number of bonding electrons divided by 2 is equal to the number of bonds surrounding the atom, this formula can be shortened to:
Formal Charge = [ of valence electrons on atom] – [non-bonded electrons + number of bonds].

Let's look at an example. Take the compound $B {H}^{4}$, or tetrahydrdoborate. Boron, $\left(B\right)$ has 3 valence electrons, zero non-bonded electrons, and 4 bonds around it. This means that the formula becomes $3 - \left(0 + 4\right)$, giving an answer of $- 1$.

Next, let's look at the hydrogen atoms in $B {H}^{4}$. Hydrogen has one valence electron, zero non bonded electrons, and one bond. So the formal charge of hydrogen in $B {H}^{4}$ is $1 - \left(0 + 1\right)$, which gives a formal charge of 0.

Formal charge is the charge left on the central atom when all the bonding pairs (of electrons) are removed sequentially.

#### Explanation:

Let's consider the simple case of ammonia, $N {H}_{3}$, versus its ammonium salt, $N {H}_{4}^{+}$. Now ammonia is a neutral molecule, and there is a non-bonding pair, a lone pair, of electrons localized to the nitrogen centre. Its reaction with ${H}^{+}$ is very simply represented:

$N {H}_{3} + {H}^{+} \rightarrow N {H}_{4}^{+}$

The ammonium is now quaternized. Why? Well, any chemical reaction conserves mass and charge, and this one does as well. But why do we write ammonium as $N {H}_{4}^{+}$, with a formal positive charge on the nitrogen nucleus?

The nitrogen in ammonia, $N {H}_{3}$ is neutral because it shares 3 electrons from the 6 electrons that comprise the $N - H$ bonds, and gets a full contribution from its lone pair, i.e. 5 electrons + 2 inner electrons electrostatically balance the 7 protons in the nucleus of a nitrogen atom. When $N$ is quaternized as ammonium, $N {H}_{4}^{+}$, it is conceived to have a $\frac{1}{2}$ share only of the 8 electrons of the $N - H$ bonds, $4 e$ in total, and is therefore written as $N {H}_{4}^{+}$, with the positive charge formally associated with the nitrogen atom.