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It has both +3 and +2 oxidation states
+1 for H, -2 for each O, +3 for I
oxygen is 2- so four oxygen have 8- charges therefore two Sb should have 8+ and therefore the oxidation number of Sb is 4+
First O has an oxidation number of -2; K will an oxidation number of +1 as it is an alkali metal. Therfore the Mn is +6. note that while the manganate ion exists, the question may relate to the permanganate ion MnO4-, so KMnO4, where Mn has oxidation number of +7
The oxidation number determines how much an element is oxidated, so the oxidation number of...1. elements is always 0.2. of simple ions is always the charge, e.g. in Cu2+ the oxidation number of copper is +2.3. hydrogen is usually +1, oxygen usually -2, alkali metals +1, etc.In molecules without a charge, the sum of the oxidation numbers has to be 0. This way you can calculate the oxidation number of its compounds. For example in KMnO4, the oxidation number of oxygen is -2, of K is +1, so if the sum is zero then the oxidation number of Mn has to be +7.In complex ions (OH-, MnO4-, ...) the sum of the oxidation numbers has to be the charge of the ion. (so in OH- and MnO4- it's -1).
It has both +3 and +2 oxidation states
In a compound the sum of oxidation states of the elements contained is zero.E1 + E2 + ... = 0If you know the oxidation states of the elements E1... you can calculate the oxidation state of the element E2.
+1 for H, -2 for each O, +3 for I
To calculate the oxidation state of fluorine in O2F2, first draw a diagram of the molecule: F-O-O-F Determine the most electronegative atom(s), which are fluorines -- the most electronegative atom there is. Being in the group 7A, a fluoride ion would gain an electron to a -1 charge, so each has an oxidation number of -1. The oxygens, therefore, have an oxidation number of +1 each.
oxygen is 2- so four oxygen have 8- charges therefore two Sb should have 8+ and therefore the oxidation number of Sb is 4+
First O has an oxidation number of -2; K will an oxidation number of +1 as it is an alkali metal. Therfore the Mn is +6. note that while the manganate ion exists, the question may relate to the permanganate ion MnO4-, so KMnO4, where Mn has oxidation number of +7
The oxidation number determines how much an element is oxidated, so the oxidation number of...1. elements is always 0.2. of simple ions is always the charge, e.g. in Cu2+ the oxidation number of copper is +2.3. hydrogen is usually +1, oxygen usually -2, alkali metals +1, etc.In molecules without a charge, the sum of the oxidation numbers has to be 0. This way you can calculate the oxidation number of its compounds. For example in KMnO4, the oxidation number of oxygen is -2, of K is +1, so if the sum is zero then the oxidation number of Mn has to be +7.In complex ions (OH-, MnO4-, ...) the sum of the oxidation numbers has to be the charge of the ion. (so in OH- and MnO4- it's -1).
Co = +2 oxidation C = +4 oxidation O = -2 oxidation
Oxidation number is usually defined along the lines of the charge on the central atom if all "ligands" are removed along with the electron pairs.Now C2H4 is H2C=CH2 so we remove H as H- and leave a "charge" on the carbon atoms of 2+. the oxidation number of carbon is therefore (+)2.There is always debate about this term and the closely related term oxidation state. Confusingly some people use the term oxidation number to mean oxidation state.To calculate oxidation state which uses the relative electronegativity to determine which atom "keeps" the electrons, the hydrogens would be treated not as H- but as H+ leaving the carbon in an oxidation state of -2.My advice would be to check your notes and see which definition your teacher is using.
O = -2 oxidation state H = +1 oxidation state
Oxidation number is oxidation states of an element. It can be positive or negative.
oxidation number of I is -1. oxidation number of F is +1.