The bond graph summarizes these connectivities. The bond orders (also called bond valences) sum up to oxidation states according to the attached sign of the bond's ionic approximation (there are no formal charges in bond graphs). Determination of oxidation states from a bond graph can be illustrated on ilmenite, . We may ask whether the mineral contains and , or and . Its crystal structure has each metal atom bonded to six oxygens and each of the equivalent oxygens to two irons and two titaniums, as in the bond graph below. Experimental data show that three metal-oxygen bonds in the octahedron are short and three are long (the metals are off-center). The bond orders (valences), obtained from the bond lengths by the bond valence method, sum up to 2.01 at Fe and 3.99 at Ti; which can be rounded off to oxidation states +2 and +4, respectively: Balancing redox Oxidation states can be useful for balancing chemical equations for oxidation-reduction (or redox) reactions, because the changes in the oxidized atoms have to be balanced by the changes in the reduced atoms. For example, in the reaction of acetaldehyde with Tollens' reagent to form acetic acid (shown below), the carbonyl carbon atom changes its oxidation state from +1 to +3 (loses two electrons). This oxidation is balanced by reducing two cations to (gaining two electrons in total). An inorganic example is the Bettendorf reaction using tin dichloride () to prove the presence of arsenite ions in a concentrated HCl extract. When arsenic(III) is present, a brown coloration appears forming a dark precipitate of arsenic, according to the following simplified reaction: 2 As^3+ + 3 Sn^2+ -> 2 As^0 + 3 Sn^4+ Here three tin atoms are oxidized from oxidation state +2 to +4, yielding six electrons that reduce two arsenic atoms from oxidation state +3 to 0. The simple one-line balancing goes as follows: the two redox couples are written down as they react; As^3+ + Sn^2+ <=> As^0 + Sn^4+ One tin is oxidized from oxidation state +2 to +4, a two-electron step, hence 2 is written in front of the two arsenic partners. One arsenic is reduced from +3 to 0, a three-electron step, hence 3 goes in front of the two tin partners. An alternative three-line procedure is to write separately the half-reactions for oxidation and reduction, each balanced with electrons, and then to sum them up such