The formal charge (CF) is the one that is assigned to an atom of a molecule or ion, which allows to explain its structures and chemical properties based on it. This concept implies the consideration of the maximum character of covalence in the bond AB; that is, the pair of electrons is shared equally between A and B.
To understand the above, the image below shows two linked atoms: one designated with the letter A and the other with the letter B. As can be seen, in the intercept of the circles a bond is formed with the pair “:”. In this heteronuclear molecule, if A and B have equal electronegativities, the pair “:” remains equidistant from both A and B.
However, since two different atoms cannot have identical properties, the “:” pair is attracted to the one that is more electronegative. In this case, if A is more electronegative than B, the pair “:” is closer to A than to B. The opposite occurs when B is more electronegative than A, now approaching “:” to B.
So, to assign the formal charges to both A and B, it is necessary to consider the first case (the one at the top of the image). If the purely covalent bond AB were to break, a homolytic break would occur, generating the free radicals A · and · B.
Qualitative benefits of the use of formal cargo
The electrons are not fixed, as in the previous example, but travel and are lost through the atoms of the molecule or ion. If it is a diatomic molecule, it is known that the pair “:” must be shared or wandering between both atoms; the same occurs in a molecule of the ABC type, but with greater complexity.
However, when studying an atom and assuming a covalence of one hundred percent in its bonds, it is easier to establish whether it gains or loses electrons within the compound. To determine this gain or loss, your basal or free state must be compared to your electronic environment.
In this way, it is possible to assign a positive charge (+) if the atom loses an electron, or a negative charge (-) when, on the contrary, it gains an electron (the signs must be written inside a circle).
Thus, although the electrons cannot be located exactly, these formal (+) and (-) charges on the structures conform in most cases to the expected chemical properties.
That is, the formal charge of an atom is closely related to the molecular geometry of its environment and its reactivity within the compound.
Formula and how to calculate it
Are formal charges assigned arbitrarily? The answer is no. For this, the gain or loss of electrons must be calculated assuming purely covalent bonds, and this is achieved through the following formula:
CF = (group number of the atom) – (number of bonds it forms) – (number of unshared electrons)
If the atom has a CF with a value of +1, it is assigned a positive charge (+); whereas if you have a CF with a value of -1, then it is assigned a negative charge (-).
To correctly calculate the CF, the following steps must be followed:
- Locate in which group the atom is in the periodic table.
- Count the number of bonds it forms with its neighbors: double bonds (=) are worth two and triple bonds are worth three (≡).
- Finally, count the number of unshared electrons, which can be easily observed with Lewis structures.
Calculation variations according to the structure
Given the linear molecule ABCD, the formal charges for each atom can vary if the structure, for example, is now written as: BCAD, CABD, ACDB, etc. This is because there are atoms that, by sharing more electrons (forming more bonds), acquire positive or negative CFs.
So which of the three possible molecular structures corresponds to compound ABCD? The answer is: the one that generally has the lowest CF values; likewise, the one that assigns negative charges (-) to the most electronegative atoms.
If C and D are more electronegative than A and B, then by sharing more electrons they consequently acquire formal positive charges (seen from a mnemonic rule).
Thus, the most stable structure, and the most energetically favored, is CABD, since in this both C and B form only one bond. On the other hand, the ABCD structure and those that have C or B forming two bonds (–C– or –D–), are more unstable.
Which of all the structures is the most unstable? ACDB, because not only C and D form two bonds, but also their formal negative charges (-) are adjacent to each other, further destabilizing the structure.
Examples of formal load calculations
BF 4 – (tetrafluoroborate ion)
The boron atom is surrounded by four fluorine atoms. Since B belongs to group IIIA (13), it lacks unshared electrons and forms four covalent bonds, its CF is (3-4-0 = -1). On the other hand, for F, an element of group VIIA (17), its CF is (7-6-1 = 0).
To determine the charge of the ion or molecule, it is enough to add the individual CFs of the atoms that compose it: (1 (-1) + 4 (0) = -1).
However, the CF for B has no real meaning; that is, the highest electron density does not reside on it . In reality, this electron density is distributed towards the four atoms of F, an element much more electronegative than B.
BeH 2 (beryllium hydride)
The beryllium atom belongs to group IIA (2), forms two bonds and lacks, again, unshared electrons. Thus, the CFs for Be and H are:
CF Be = 2-2-0 = 0
CF H = 1-1-0 = 0
Load BeH 2 = 1 (0) + 2 (0) = 0
CO (carbon monoxide)
Its Lewis structure can be represented as: C≡O: (although it has other resonance structures). Repeating the CF calculation, this time for C (of group IVA) and O (of group VIA), we have:
CF C = 4-3-2 = -1
CF O = 6-3-2 = +1
This is an example where formal charges do not conform to the nature of the elements. O is more electronegative than C and therefore should not carry a positive.
The other structures (C = O and (+) CO (-) ), although they comply with the coherent assignment of charges, do not comply with the octet rule (C has less than eight valence electrons).
NH 4 + (ammonium ion), NH 3 and NH 2 – (amide ion)
The more electrons N shares, the more positive its CF is (even the ammonium ion, since it has no energy availability to form five bonds).
Equally applying the calculations for the N in the ammonium ion, ammonia and amide ion, we have then:
CF = 5-4-0 = +1 (NH 4 + )
CF = 5-3-2 = 0 (NH 3 )
CF = 5-2-4 = -1 (NH 2 – )
That is, in NH 2 – N has four unshared electrons, and it shares all of them when it forms NH 4 + . The CF for the H is equal to 0 and therefore your calculation is saved.