Explanation
A. In $$BaO_2$$, oxygen is in a peroxide state. therefore has a charge of $$-1$$.
B. In $$KO_2$$, oxygen is in a superoxide state. therefore has charge of $$\frac {-1}{2}$$.
C. In $$O_3$$, as there is $$3$$ oxygen. First one has charged $$-1$$, second one has a charge of $$0$$ and third one has therefore had a charge of $$+1$$.
Hence, overall charge is $$0$$.
In $$OF_2$$, oxygen is less electronegative than fluorine. Therefore has a charge of $$-2$$.
As shown above, the calculated oxidation state of compounds is as follows.
A. $$BaO_2$$ $$\rightarrow$$ $$-1$$
B. $$KO_2$$ $$\rightarrow$$ $$\frac {-1}{2}$$
C. $$O_3$$ $$\rightarrow$$ $$0$$
D. $$OF_2$$ $$\rightarrow$$ $$+2$$
Which of the following shows highest oxidation number in combined state?
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