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
Answers
Answered by
1
The decreasing order of oxygen in the given compounds is as given below
OF2>O3>KO2>BaO2
1) In OF2
1×(O) +2×(-1)= 0
O-2= 0
O = +2
2) In O3
oxidation number of O3 is zero
3) In KO2
1×(1) + 2×(O) = 0
1 + 2(O) = 0
O = -1/2
4) In BaO2
1×(2) + 2×(O) = 0
2+2(O) = 0
O= -1
so option (b) is the right answer
Answered by
1
OF₂ > O₃ > KO₂ > BaO₂ is the correct trend.
Explanation:
- In BaO2 the oxidation state of barium is +2.
- This leads to the oxidation state of oxygen to be - 1 each.
- In case of OF2, the oxidation state of fluorine is - 1.
- This leads to the oxidation state of oxygen to be +2.
- In case of KO2, the oxidation state of potassium is +1.
- So the oxidation state of oxygen is - 1/2.
- In case of O3 there's only oxygen atoms present in the molecule.
- So oxidation state of oxygen in this molecule is 0.
- So the trend of oxidation number among these compounds is OF₂ > O₃ > KO₂ > BaO₂.
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