Question

what is the order of bond length of KO2,H2O2,Na2O2,O2BF4,O2..if anyone no the correct answer only those can answer plz​

Answer 1

Answer:

In compound Na2O2, H2O2 peroxide are present i.e single bond between oxygen.

Here oxygen is represented as O2^-2

In KO2 superoxide is present and represented as O2^-.

The bond order of O2^-2 is 1,and

Bond order of O2^- is 1.5

As there is inverse relation between bond order and bond length. Then,

Na2O2 = H2O2 > KO2.

Hope will help you….

Explanation:

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Answer 2

Answer:

2 nd Leter on body blood

Explanation:

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