Determine the oxidation number of the underlined element in NaBrO₄.

Question

Quizplus · Accepted Answer

The oxidation number of sulfur (S) in NaBrO3 is +5. This is determined by considering the oxidation states of the other atoms in the compound: Na (+1), Br (as a halogen in a compound like this, typically -1), and O (almost always -2). The sum of the oxidation states in a neutral compound must equal zero. With three oxygens, the total oxidation from oxygen is -6. To balance the +1 from Na and the -1 from Br, leaving a net of 0, S must be +5 to ensure the overall charge of the compound sums to zero. However, the compound provided seems to be incorrectly notated (NaBrO₄), which doesn't correspond to a recognizable chemical formula and likely represents a typographical or formatting error. The correct interpretation based on a common compound that fits part of the provided notation (NaBrO3) leads to sulfur having an oxidation state of +5, but since +5 is not an option and considering possible confusion or error in the question, the closest available choice reflecting a common oxidation state for sulfur in such compounds is +7 (A), which is a common oxidation state for sulfur in other compounds, but not correctly applicable here based on the provided (and likely incorrect) formula.

Determine the Oxidation Number of the Underlined Element in NaBrO4

Determine the Oxidation Number of the Underlined Element in NaBrO₄