How to find oxidation number of an element

The oxidation state of an element in a compound is the charge left on the atom of interest when all the This definition is, I grant, a mouthful, but let's see how it is applied. We start with the water molecule, #OH_2#, which of course is neutral. The #O-H# bond is of course composed of 2 electrons (this goes back to the idea of covalent bonds as the result of the sharing of 2 electrons between adjacent atoms), and when we break this bond, we are left with #O^(2-)# and #2xxH^+# (because oxygen is more electronegative than hydrogen, the 2 electrons in the bond devolve to oxygen). Thus given the definition, the oxidation state of #O# #=# #-II#, and that of #H# #=# #+I#. #0# oxidation states, because in their elemental state they neither accepted not donated electrons: cf oxidation of carbon: #C(s)+O_2(g)rarrCO_2(g)# Elemental carbon is conceived to have been oxidized, i.e. it has lost 4 electrons to give #C(IV+)#, and dioxygen has accepted those same electrons to give #2xx(-II)#. Now of course, these ideas of electron loss and electron gain are formalisms in that they have no actual reality, but often they allow us to balance #CO_3^(2-), C(+IV)#; #SO_3^(2-), S(+IV)#; #SO_4^(2-), S(+VI)#; #ClO_4^(-), Cl(+VII)#; #ClO_3^(-), Cl(+V)#. Anyway, I hope this spray hasn't managed to confuse you totally. Remember, these are simple ideas, and they use very simple arithmetic.

Name Symbol Oxidation number hydrogen H +1 +1 lithium Li +1 +1 sodium Na +1 +1 potassium K +1 +1 rubidium Rb +1 +1 cesium Cs +1 +1 copper (I) Cu +1 +1 gold (I) Au +1 +1 silver Ag +1 +1 thallium (I) Tl +1 +1 mercury (I) Hg 2+2 +1 beryllium Be +2 +2 calcium Ca +2 +2 magnesium Mg +2 +2 strontium Sr +2 +2 barium Ba +2 +2 radium Ra +2 +2 cadmium Cd +2 +2 mercury (II) Hg +2 +2 cobalt (II) Co +2 +2 copper (II) Cu +2 +2 iron (II) Fe +2 +2 lead (II) Pb +2 +2 tin (II) Sn +2 +2 chromium (II) Cr +2 +2 nickel (II) Ni +2 +2 zinc Zn +2 +2 manganese (II) Mn +2 +2 boron B +3 +3 aluminum Al +3 +3 gallium Ga +3 +3 indium In +3 +3 thallium (III) Tl +3 +3 scandium Sc +3 +3 iron (III) Fe +3 +3 chromium (III) Cr +3 +3 nickel (III) Ni +3 +3 cobalt (III) Co +3 +3 gold (III) Au +3 +3 titanium (III) Ti +3 +3 tin (IV) Sn +4 +4 lead (IV) Pb +4 +4 titanium (IV) Ti +4 +4 carbon C +4 +4 silicon Si +4 +4 germanium Ge +4 +4 manganese (IV) Mn +4 +4 fluoride F -1 -1 chloride Cl -1 -1 bromide Br -1 -1 iodide I -1 -1 hydride H -1 -1 oxide O -2 -2 sulfide S -2 -2 selenide Se -2 -2 telluride Te -2 -2 nitride N -3 -3 phosphide P -3 -3 carbide C -4 -4

\( \newcommand\) • • • • • • • • Formal Charge On the page discussing Formal Charge is a somewhat artificial device that exists in the minds of chemists (not within the molecules, themselves) to help keep track of electrons in their bonding configurations. The Formal Charge is the charge an atom in a molecule or polyatomic ion would have if all of the bonding electrons were divided equally between atoms in the bond. Formal Charge "Rules" Here are some rules for determining the Formal Charge on each atom in a molecule or polyatomic ion: • Electrons within a Lone Pair on an atom are assigned exclusively to that atom. • Half of the electrons in each bond around an atom are assigned to that atom. • The Formal Charges on all atoms in a molecule must sum to zero; for a polyatomic ion the Formal Charges must sum to the charge on the ion (which may be positive or negative). The Formal Charge is defined by the relationship: Formal Charge = [ number of valence electrons in an isolated atom] - [( number of lone pair electrons) + ½ ( number of bonding electrons)] With the definitions above, we can calculate the Formal Charge on the thiocyanate Ion, SCN -: Table \(\PageIndex\) S C N Valence electrons 6 4 5 Lone pair electrons 6 0 2 Shared electrons (bonds) 2 (1) 8 (4) 6 (3) Formal Charge = 6 - (6 + ½·2) 4 - (0 + ½·8) 5 - (2 + ½·6) Formal Charge -1 0 0 Notice how the sum of all of the formal charges adds up to the charge of the thiocyanate ion (-1). When drawing Lewis Structures, we use...

\( \newcommand\) • • • For example, in NO 3 – the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. This arbitrary assignment corresponds to the nitrogen’s having lost its original five valence electrons to the electronegative oxygens. In NO 2, on the other hand, the nitrogen has an oxidation number of + 4 and may be thought of as having one valence electron for itself, that is, one more electron than it had in NO 3 –. This arbitrarily assigned gain of one electron corresponds to reduction of the nitrogen atom on going from NO 3 – to NO 2. As a general rule, reduction corresponds to a lowering of the oxidation number of some atom. Oxidation corresponds to increasing the oxidation number of some atom. Applying the oxidation number rules to the following equation, we have Since the oxidation number of copper increased from 0 to +2, we say that copper was oxidized and lost two negatively charged electrons. The oxidation number of nitrogen went down from 5 to 4, and so the nitrogen (or nitrate ion) was reduced. Each nitrogen gained one electron, so 2 e – were needed for the 2 NO 3 –. The nitrogen was reduced by electrons donated by copper, and so copper was the reducing agent. Copper was oxidized because its electrons were accepted by an oxidizing agent, nitrogen (or nitrate ion). Although they are useful and necessary for recognizing redox reactions, oxidation numbers are a highly artificial device. The nitrogen atom in NO 3 – does not ...

Oxidation–reduction or redox reactions are reactions that involve the transfer of electrons between chemical species (check out and charge, which can make them challenging to balance by inspection alone. In this article, we’ll learn about the half-reaction method of balancing, a helpful procedure for balancing the equations of redox reactions occurring in aqueous solution. To balance a redox equation using the half-reaction method, the equation is first divided into two half-reactions, one representing oxidation and one representing reduction. The equations for the half-reactions are then balanced for mass and charge and, if necessary, adjusted so that the number of electrons transferred in each equation is the same. Finally, the half-reaction equations are added together, giving the balanced overall equation for the reaction. Is this equation balanced? It appears to be balanced with respect to mass since there is one C o \ce N i N, i atom on each side of the equation. However, it is not balanced for charge: the net charge on the left side of the equation is 3 + 3+ 3 + 3, plus , while the net charge on the right side is 4 + 4+ 4 + 4, plus . To help us balance the equation for charge, we’ll use the half-reaction method. Reduction half-reaction: The reduction half-reaction shows the reactants and products participating in the reduction step. In this case, our equation should show C o X 3 + \ce C o X 2 + . It should also include an electron on the left side of the equation fo...

Iridium is the only element known (and this was only recently demonstrated) to achieve the +9 oxidation state (in the cation IrO₄⁺). There are seven elements that can achieve +8 (Ru, Xe, Os, Ir, Pu, Cm and Hs). Since the results for the +9 oxidation state in Ir were first published just a few weeks ago (October 23, 2014) most sources you see will say that +8 is the highest oxidation state. The lowest known oxidation state is −4, which only Group 14 elements are known to achieve. First, keep in mind that oxidation numbers are NOT charges. The oxidation numbers are statements about what the charge on the atom would be if all of its bonds were 100% ionic. Thus, in Mg(OH)₂ you have two separate things going on. First you have O and H covalently bonded to each other with a negative charge (taken from Mg) and you have two sets of O and H. So, we have two OH⁻ anions. They got their extra electrons from the Mg, so Mg has a charge of +2, so it is a cation. Thus, the second thing occurring is that we have an ionic bond between Mg⁺² cation and the two OH⁻ anions. That is why we have Mg(OH)₂ Thus, the oxidation states in Mg(OH)₂ are Mg +2 O -2 H +1 My Chemistry teacher taught us to assume that Oxygen is 2-, and Hydrogen is 1+. You always know the Charge of the compound, such as HO 's charge is 1-. So if Oxygen is 2-, and Hydrogen is 1+, then the charge is 1-. For something like MgO, where the charge is neutral. You can assume that Oxygen is 2-, and so Mg must be 2+ to keep the compoun...