The correct decreasing order of oxidation number of oxygen in compounds BaF₂, O₃, KO₂ and OF₂ is :
(a) BaO₂ > KO₂ > O₃ > OF₂
(b) OF₂ > O₃ > KO₂ > BaO₂
(c) KO₂ > OF₂ > O₃ > BaO₂
(d) BaO₂ > O₃ > OF₂ > KO₂
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THE ANSWER TO YOUR QUESTION IS A..
SOLUTION :-
BECAUSE Ba FROM 2ND GROUP SO IT'LL HAVE +2 OXIDATION STATE SO BY PUTTING THE VALUE HIGHEST OXIDATION WILL BE OF BaO2 AND ALL OTHERS WILL BE LEAST ..
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