Arrange the following compounds in increasing order of the oxidation number of oxygen.

K O_{2}

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O F_{2}

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O_{3}

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B a O_{2}

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Solution

In $O_{3}$ , O.N. of Oxygen is 0, elemental state.
Lets assume the O.N. of O is x in $O F_{2}$ :
$x + 2 (- 1) = 0$
$x - 2 = 0$
$x = + 2$
In $B a O_{2};$
$2 x + 1 (+ 2) = 0$
$2 x = - 2$
$x = - 1$
In $K O_{2}$ ;
$+ 1 + 2 (x) = 0$
$2 x = - 1$
$x = - \frac{1}{2}$
So, increasing order of O.N is
$B a O_{2} < K O_{2} < O_{3} < O F_{2}$

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In which of the following has the oxidation number of oxygen been arranged in increasing order?

N a C l O_{4}, N a C l O_{3}, N a C l O, K C l O_{2}, C l_{2} O_{7}, C l O_{3}, C l_{2} O, N a C l, C l_{2}, C l O_{2}

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