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The oxidation number of potassium in \mathrm{K_2O , K_2O_2 \: and\: KO_2} respectively is :
Option: 1 +1,\: +1\: and\: +1
Option: 2 +2,\: +1\: and\: +\frac{1}{2}
Option: 3 +1,\: +2\: and\: +4
Option: 4 +1,\: +4\: and\: +2
 

Answers (1)

best_answer

\mathrm{For \: K_2O:} 2x - 2 = 0\\\: \: \:

                        x = +1

\mathrm{For\ K_2O_2:} 2x - 2 = 0

(In peroxide, the oxidation state of oxygen is -1)
                        x = +1

\mathrm{For\ KO_2:} x - 2\times \frac{1}{2} = 0

(In superoxide, the oxidation state of oxygen is -1/2)
                      x = +1

Therefore, Option(1) is correct.

Posted by

Kuldeep Maurya

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