Final answer:
The oxidation number of iodine in sodium periodate (NaIO4) is +7. This is found by applying the general rules of oxidation numbers to the compound and considering the well-known oxidation states of sodium (+1) and oxygen (-2).
Step-by-step explanation:
The oxidation number of iodine in sodium periodate (NaIO4) can be determined by knowing the general rules for oxidation numbers. For ionic compounds, it's common to assign oxidation numbers for the cation and anion separately. Here, sodium (Na) has a well-known oxidation number of +1, as it is an alkali metal, and oxygen generally has an oxidation number of -2, except in peroxides or when bonded to fluorine.
To find the oxidation number of iodine, we set up the equation considering the oxidation numbers and the overall charge of the compound, which is neutral (0):
- +1 (for Na)
- Unknown oxidation number of iodine (I)
- 4 x (-2) (for four oxygen atoms)
Adding them together and setting them equal to 0 gives us the equation: +1 + I + (4 x -2) = 0. Simplifying, we get +1 + I - 8 = 0, which further simplifies to I - 7 = 0. Solving for the oxidation number of iodine (I) gives us I = +7. Therefore, the oxidation number of iodine in sodium periodate (NaIO4) is +7, which corresponds to option d).