A) \[O{{F}_{2}}<K{{O}_{2}}<Ba{{O}_{2}}<{{O}_{3}}\]
B) \[Ba{{O}_{2}}<K{{O}_{2}}<{{O}_{3}}<O{{F}_{2}}\]
C) \[Ba{{O}_{2}}<{{O}_{3}}<O{{F}_{2}}<K{{O}_{2}}\]
D) \[K{{O}_{2}}<O{{F}_{2}}<{{O}_{3}}<Ba{{O}_{2}}\]
E) \[O{{F}_{2}}<{{O}_{3}}<K{{O}_{2}}<Ba{{O}_{2}}\]
Correct Answer: B
Solution :
Let the oxidation number of oxygen in following compounds is\[x\]. In \[O{{F}_{2}}\] \[x+(-1)2=0\] \[x=+2\] In \[K{{O}_{2}}\] \[+1+(x\times 2)=0\] \[2x=-1\] \[x=-\frac{1}{2}\] In \[Ba{{O}_{2}}\] \[+2+(x\times 2)=0\] \[2x=-2\] \[x=-1\] In\[{{O}_{3}},\]oxidation number of oxygen is zero because oxidation number of an element in Free State or in any of its allotropic form is always zero. Thus, the increasing order of oxidation number is \[\begin{align} & Ba{{O}_{2}}<K{{O}_{2}}<{{O}_{3}}<O{{F}_{2}} \\ & -1\,\,\,\,\,\,\,\,\,\,\,\,\,-\frac{1}{2}\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,+2 \\ \end{align}\]
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