Typical oxidation states for chemical elements

How to count the electrons' transmission

Be­fore we dis­cuss the ox­i­da­tion state, let’s re­call the main rules of chem­istry and physics:

  • all sub­stances are formed from mol­e­cules, and mol­e­cules from atoms;
  • any atom is elec­tri­cal­ly neu­tral; i.e. it has a to­tal charge of zero;
  • the zero charge of an atom is de­ter­mined by the iden­ti­cal num­ber of pos­i­tive­ly and neg­a­tive­ly charged par­ti­cles in it;
  • neg­a­tive­ly charged par­ti­cles in­side the atom are called “elec­trons” – they move around the atom nu­cle­us. The charge of one elec­tron is -1;
  • the to­tal neg­a­tive charge of all the elec­trons of the atom is equal to their quan­ti­ty;
  • pos­i­tive par­ti­cles of the atom are called “pro­tons” and are lo­cat­ed in its nu­cle­us, and the charge of one pro­ton is (+1);
  • the to­tal pos­i­tive charge of the nu­cle­us is equal to the to­tal num­ber of pro­tons in it;
  • the pre­cise num­ber of pro­tons and elec­trons in the atom can be de­ter­mined by check­ing its num­ber in the pe­ri­od­ic sys­tem:

№ of el­e­ment = num­ber of pro­tons in the atom = num­ber of elec­trons in the atom.

Let us ex­am­ine the above state­ments based on the ex­am­ples of oxy­gen (О), hy­dro­gen (Н), cal­ci­um (Са) and alu­minum (Аl).

Oxy­gen (О) has the or­di­nal num­ber of eight in the pe­ri­od­ic sys­tem, and this means that there are eight pro­tons in its nu­cle­us, and eight elec­trons move around its nu­cle­us.

Atomic structure of oxygen [Deposit Photos]

Thus, the charge of the nu­cle­us of its atom is equal to (+8), and the to­tal charge of elec­trons mov­ing around its nu­cle­us is equal to (-8).

The to­tal charge of the atom for the chem­i­cal el­e­ment is de­ter­mined by adding all the pos­i­tive and neg­a­tive charges in the atom:

(+8)+(-8)=0.

Hy­dro­gen (Н) holds first place in the pe­ri­od­ic sys­tem, and ac­cord­ing­ly there is one pro­ton in its nu­cle­us, and one elec­tron moves around the nu­cle­us.

(+1)+(-1)=0.

Cal­ci­um (Са) is in 20th place in the pe­ri­od­ic sys­tem, and so its atom has 20 pro­tons and elec­trons, the to­tal charges of which are (+20) and (-20) re­spec­tive­ly:

(+20)+(-20)=0.

As for alu­minum (Аl), its 13th place in the pe­ri­od­ic sys­tem shows that it has 13 pro­tons and 13 elec­trons:

(+13)+(-13)=0.

A few words about the ox­i­da­tion state

As we know, in the earth’s crust chem­i­cal el­e­ments not only ex­ist in a free state. Their atoms also en­ter into chem­i­cal re­ac­tions, form­ing com­plex sub­stances. This can be eas­i­ly demon­strat­ed from the ex­am­ple of the for­ma­tion of ox­ides.

For ex­am­ple, oxy­gen (О) can in­ter­act with hy­dro­gen (Н). At the same time, hy­dro­gen gives oxy­gen its only elec­tron. Af­ter this, the hy­dro­gen atom no longer has any free elec­trons, and ac­cord­ing­ly, the pos­i­tive charge of the atom’s nu­cle­us of (+1) has noth­ing to neu­tral­ize it, and the en­tire hy­dro­gen atom gains a charge of (+1). Thus, the elec­tri­cal­ly neu­tral hy­dro­gen atom turns into a pos­i­tive­ly charged par­ti­cle – a pro­ton:

(+1) + (-1) - (-1)= (+1).

The oxy­gen atom, which in a free state also has a zero charge, can join two elec­trons to it­self at once. This means that it en­ters into a re­ac­tion with two hy­dro­gen atoms at once, each of which gives it its only elec­tron.

Thus, oxy­gen, which be­fore its re­ac­tion with hy­dro­gen has eight pro­tons and elec­trons, ac­quires an­oth­er two elec­trons in this chem­i­cal in­ter­ac­tion. And so its to­tal charge be­comes:

(+8)+(-8)+(-2)=(-2).

This ex­am­ple il­lus­trates a re­ac­tion in which an atom of one chem­i­cal el­e­ment gives its elec­trons to an atom of an­oth­er chem­i­cal el­e­ment. These re­ac­tions are called re­duc­tive-ox­ida­tive in chem­istry.

[Deposit Photos]

It is ac­cept­ed that the atom that gives its elec­trons ox­i­dizes, and the atom that re­ceives them is re­duced. In this case, hy­dro­gen ox­i­dizes, and oxy­gen re­duces. The charge which both atoms re­ceive as a re­sult of the re­ac­tion is writ­ten in the top right-hand cor­ner, above the sym­bols of their chem­i­cal el­e­ments.

It should also be re­mem­bered that oxy­gen and hy­dro­gen are gas­es, and so there are two iden­ti­cal atoms in their mol­e­cules. Ac­cord­ing­ly, the full re­ac­tion of the in­ter­ac­tion of oxy­gen with hy­dro­gen looks like this:

2Н₂⁰ + О₂⁰ → 2Н₂⁺¹О⁻²

In this case, this con­cerns the for­ma­tion of com­pounds of the type X₂O, in which in or­der to re­ceive a mol­e­cule of a com­plex sub­stance, two iden­ti­cal atoms of an­oth­er el­e­ment at­tach to one oxy­gen atom. The ox­i­da­tion state (+1) is char­ac­ter­is­tic for el­e­ments of the first group of the pe­ri­od­ic sys­tem be­long­ing to the main sub-group.

Ox­i­da­tion state in XO

The sec­ond group of the pe­ri­od­ic sys­tem, name­ly its main sub­group, con­tains chem­i­cal el­e­ments in which each atom can give two elec­trons to oxy­gen. This atom, its re­duc­tive-ox­ida­tive re­ac­tion, gains a charge (+2), and oxy­gen, as al­ways, re­ceives the charge (-2). For ex­am­ple, the re­ac­tion of the ox­i­da­tion of cal­ci­um:

2Са⁰ + О₂⁰→2Са⁺²О⁻².

Zinc (Zn), lo­cat­ed in the sub­sidiary sub­group of the sec­ond group, shows the same ox­i­da­tion state as cal­ci­um, name­ly XO:

2Zn⁰ + О₂⁰→2Zn⁺²О⁻²

Ox­i­da­tion state in X₂O₃

The fea­ture of the el­e­ments of the main sub­group of the third group of the pe­ri­od­ic sys­tem is that each atom can eas­i­ly give three elec­trons to the oxy­gen atom. But one oxy­gen atom can only ac­cept two elec­trons.

Ac­cord­ing­ly, the ra­tio of atoms in the mol­e­cule of ox­ides for el­e­ments of third group, on the ex­am­ple of alu­minum ox­ide, is the fol­low­ing:

  • If one alu­minum atom can give three elec­trons, two alu­minum atoms give six elec­trons – three each;
  • one oxy­gen atom can take only two elec­trons, and so each oxy­gen atom takes two elec­trons from the alu­minum atoms:

4Al⁰ + 3O₂⁰ → 2Al₂⁺³O₃⁻²

Thus, in this chem­i­cal re­ac­tion four alu­minum atoms take part, which give 12 elec­trons to six atoms (or three mol­e­cules) of oxy­gen.

As a re­sult of the re­ac­tion, each alu­minum atom will lack three elec­trons for a zero charge, and thus the pos­i­tive charge of the nu­cle­us will pre­dom­i­nate over the neg­a­tive charge of the elec­trons:

+13 (the charge of the Al atom has not changed) -10 (elec­trons left af­ter the re­ac­tion)= (+3).

Ox­i­da­tion states in XO₂

This ox­i­da­tion state is shown by chem­i­cal el­e­ments lo­cat­ed in the main sub­group of the fourth group of the pe­ri­od­ic sys­tem. Each atom can give four elec­trons at once, and as the oxy­gen mol­e­cule is di­atom­ic, each of the oxy­gen atoms takes two elec­trons.

Let us ex­am­ine this re­duc­tive-ox­ida­tive re­ac­tion on the ex­am­ple of the re­ac­tion of oxy­gen with car­bon:

С⁰ + О₂⁰ → С⁺⁴О₂⁻²

This is a re­ac­tion that il­lus­trates the com­bus­tion of a sol­id, car­bon, in the pres­ence of oxy­gen gas. So the oxy­gen mol­e­cule is di­atom­ic, and the car­bon mol­e­cule is monoatom­ic.

Click here for find­ing out how dif­fer­ent met­als can be ox­i­dized.

Ox­i­da­tion state in в X₂O₅ and XO₃

For some el­e­ments of the fifth group, in the main sub­group, the ox­i­da­tion state of (+5) is char­ac­ter­is­tic, i.e. they can give an oxy­gen atom five elec­trons. For ex­am­ple, the re­ac­tion of the com­bus­tion of phos­pho­rous in the pres­ence of oxy­gen:

4Р⁰ + 5О₂⁰ → 2Р₂⁺⁵О₅⁻².

Some el­e­ments of the sixth group can give six elec­trons, af­ter which their ox­i­da­tion state be­comes (+6). For ex­am­ple, the re­ac­tion of sul­fur with oxy­gen:

2S⁰ + 3O₂⁰ → 2S⁺⁶O₃⁻²