Final answer:
The oxidation number of Cr in K2Cr2O7 is +6, S in S2O82- is +7, S in Na2S4O6 is +2.5, and P in H2P2O72- is +5. Calculations are based on known oxidation states of other elements in the compounds and the overall charge balance.
Step-by-step explanation:
Finding Oxidation Numbers
When finding the oxidation number of an element in a compound, keep in mind the rules of oxidation states such as the oxidation state of oxygen usually being -2, and the sum of the oxidation states in a neutral compound being zero, or equal to the charge for an ion. Here are how oxidation numbers are assigned to each query:
- For Cr in K2Cr2O7, knowing that O (oxygen) has an oxidation number of -2, and there are 7 oxygen atoms, that totals to -14. K (potassium) has an oxidation number of +1, there are 2 potassium atoms, totaling +2. The overall charge of the compound is 0, so the oxidation number for the two Cr (chromium) atoms must total +12. Therefore, each Cr has an oxidation number of +6.
- For S in S2O82-, each O (oxygen) has an oxidation number of -2 for a total of -16. Since the molecule has a charge of -2, the total of the two S (sulfur) atoms must be +14. Hence, each S has an oxidation number of +7.
- For S in Na2S4O6, O has an oxidation number of -2 (6 oxygens total -12). Na (sodium) has an oxidation number of +1 (2 sodiums total +2). The compound is neutral, so the total of the S atoms must be +10. With 4 sulfur atoms, each S has an oxidation number of +2.5.
- For P in H2P2O72-, O has an oxidation number of -2 (7 oxygens total -14). H (hydrogen) has an oxidation number of +1 (2 hydrogens total +2). With a -2 charge on the ion, the P atoms must total +10, so each P (phosphorus) has an oxidation number of +5.