Fluorine in compounds is always assigned an oxidation number of -1. For HSO3- it is the negative sign, indicating a -1 charge for the bisulfite ion. The oxidation number of a monatomic ion equals the charge of the ion. Which best identifies why the rusting of an iron nail in the presence of water and oxygen is an oxidation-reduction reaction? Add . Since the zinc ion's charge is +2, so is its oxidation number (rule 2). Oxidation State of Sulfur + (3 Oxidation state of Oxygen) = -2 Oxidation state of sulfur + (3 (-2 view the full answer Previous question Next question The oxidation number of a free element is always 0. Don't get the two confused, they may both be written with out the charge, but if SO3 is (aq) it has a charge of -2. PLEASE HELp In the redox conversion of SO3 to SO−, S is ? Because the sulfite ion (SO3) has a -2 charge (and you'd have to consult a list of polyatomic ions to know that) and the zinc ion MUST be +2 in order to balance that out. 2 Answers. The sum of the oxidation numbers of all of the atoms in a neutral compound is zero. Favorite Answer. So, in SO3, in order to create a net neutral, or zero charge, the S must have an oxidation number 6+ to cancel out the 3*(-2)= -6 of the oxygen in the compound. The sulphite anion is SO3^2- Using '-2' for oxygen as the yardstick . The oxidation number of a Group 17 element in a binary compound is -1. Oxygen has an oxidation number of 2-. Then there are 3 oxygens hence 3 x -2 = -6 . However in SO3^2- (aq) the Oxidation states are: Sulfur (+4) & Oxygen (-2). Lv 7. Consider the following reaction. That would be the oxidation number. Answer (g) :- Fe2(SO3)3 species contain SO3(2-) ion. The alkali metals (group I) always have an oxidation number … The given compound is, Let the oxidation state of S be, 'x' Hence, the oxidation state of S is, (+4) Rules for assigning oxidation numbers. Relevance. Electrons are transferred. To answer this question, let's first look at the atoms in the compounds for which we know the oxidation number. It can be determined using the following rules: 1. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. (Recall that O has an oxidation number of -2.) d. Which identifies an oxidation-reduction reaction? Explanation: let oxidation number of S be x. oxidation number of O is -2 . c. ... Why? Mg + Cl2 mc005-1.jpg Mg2+ + 2Clmc005-2.jpg. to ? Commonly, sulfur (s) forms 2- oxidation state, but in bisulfite it forms 4+. hydrogen. What is the oxidation number for S in the compound SO3? The Oxidation states in in SO3(g) are: Sulfur (+6) & Oxygen (-2), b/c SO3(g) has no charge. Roger the Mole. S is reduced and its oxidation number goes from 6+ … + indicates +1 charge, sum of charges of compounds always equals zero. Sum of charges in this case, for these polyatomic ions, equals the charge on the ion. so, x+3(-2)= -2. x-6=-2. The oxidation number for SULFUR is +4, and the oxidation number for oxygen is -2. x= +4 Answer Save. What is the oxidation number for S in the compound SO3? S + - 6 = -2 . and its oxidation number goes from ? c. Given the reaction below, which is the oxidized substance? Since the anion has an oerall charge of '-2 ' , then we create a sum . (Recall that O has an oxidation number of -2.) 4 years ago. Answer: oxidation number of S in SO3-2 is +4. The oxidation number (ON) of an element details the number of electrons lost or gained by the element in order to achieve its current state. S = -2 + 6 . S = 4 The oxid'n number of sulphur. The easiest way is to remember certain common numbers: O (in most cases): -II, H: +I For oxidation numbers we use Roman numerals! Hydrogen has an oxidation number of 1+.