When calculating oxidation numbers, use '-2' for oxygen as a standard. It does vary in some oxygen molecules, but '-2' is a good starting point.
So in KNO3
The oxygen moiety is 3 x -2 = -6
Potassium(K) is always '+1' because it only loses one electron .
Hence the nitrgen os
+1 + N - 6 = 0
N- 5 = 0
N = +5 (oxidation state)
NB For neutrally charged molecule, the sum of the oxidation number is always 'zero'(0).
+1 for K -2 for each O +5 for N
In KMnO4, Mn has an oxidation number of +7, KNO2 has N with an oxidation number of +3, and H2SO4 has S with an oxidation number of +6. In MnSO4, Mn has an oxidation number of +2, H2O has O with an oxidation number of -2, KNO3 has N with an oxidation number of +5, and K2SO4 has S with an oxidation number of +6.
The oxidation number for Mn in MnO4^- is +7. This is found by balancing the charges on the formula MnO4^- where O has an oxidation number of -2.
In KNO3 the oxidation number of potassium (K) is 1+, for nitrogen (N) it is 5+, and for each oxygen (O) it is 2-
K has an oxidation number of +1 O has an oxidation number of (-2) x 4 So... the oxidation number for Mn is whatever is needed to make 1-8 equal to zero. Therefore, the oxidation number for Mn is +7
+1 for K -2 for each O +5 for N
In KMnO4, Mn has an oxidation number of +7, KNO2 has N with an oxidation number of +3, and H2SO4 has S with an oxidation number of +6. In MnSO4, Mn has an oxidation number of +2, H2O has O with an oxidation number of -2, KNO3 has N with an oxidation number of +5, and K2SO4 has S with an oxidation number of +6.
The oxidation number for Mn in MnO4^- is +7. This is found by balancing the charges on the formula MnO4^- where O has an oxidation number of -2.
In KNO3 the oxidation number of potassium (K) is 1+, for nitrogen (N) it is 5+, and for each oxygen (O) it is 2-
K has an oxidation number of +1 O has an oxidation number of (-2) x 4 So... the oxidation number for Mn is whatever is needed to make 1-8 equal to zero. Therefore, the oxidation number for Mn is +7
In chromium(III) oxide, the oxidation number of chromium is +3 and the oxidation number of oxygen is -2. This compound has the chemical formula Cr2O3.
The oxidation number for H is +1, and the oxidation number for O is -1.
The oxidation number of Cl in Cl2O7 is +7. In Cl2O7, there are 2 Cl atoms with an oxidation number of +14 total, and the oxidation number of O is -2 (since O usually has an oxidation number of -2 in compounds). Use the formula: 2x + 7(-2) = 0, which gives x = +7 for the oxidation number of Cl.
The correct chemical formula should be P2O5 or P4O10 (its dimer) and not P3O10 (as originally asked in the question). The oxidation numbers are -2 for each O +5 for each P
KNO3 has no oxidation number. However, the individual atoms do each have an oxidation number. Using oxygen as the 'yardstick' at '-2' ; NB There are exceptions this rule. There are 3 oxygens, so the oxygen component has an oxidation of 3 x -2 = -6. So for it to be a neutrally charge molecule the potassium and the nitrogen must sum to +6. Since potassium only loses one electron its oxidation number is '+1'. It follows that nitrogen must be in oxidation state '+5' Summarising;- Potassium(K) = +1 Nitrogen(N) = +5 Oxygen(O) = 3 x -2 = -6 +1+5-6 = 0 NB Nitrogen , like sulphur and phosphorus exhibit various oxidation states. =
Oxidation number is 4, formula C3O2O=C=C=C=O
It is not possible to know the oxidation number in an formula with more then 2 unknowns such as OCN- you only know O is -2.