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When sodium reacts with excess of oxygen, oxidation number of oxygen changes fromA. `0-1`B. `0-2`C. `-1` to `-2`D. `+1` to `-1`. |
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Answer» Correct Answer - A `2Na+O_(2)rarrNa_(2)O_(2)` Oxidation number of oxygen is 0 in `O_(2)` and `-1` in `Na_(2)O_(2)`. |
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