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Oxidation and Reduction

In O2[PtF6] w...

Question

In O2[PtF6] what will be charge on O2 and how to find it in case if we have to find the bond order of it

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Q. What is bond order in case of

O_{2}^{+}

?

Q. Bond order is associated with the strength of bond and bond length. The higher the bond order, the smaller will be the bond length and stronger will be the bond.

O_{2}, O_{2}^{+}, O_{2}^{-}

O_{2}^{2 -}

in order of increasing bond order.

Q. What will be the bond order stability magnetic character of

O_{2}^{+}, O_{2}^{-}

N_{2}^{+}

N_{2}^{2 -}

Q. Oxygen reacts with platinum (VI) fluoride,

P t F_{6}

O_{2} + P t F_{6} \to O_{2}^{+} P t F_{6}^{-}

It was suggested that xenon should react similarly and, in this way,the first noble gas compound was produced.

X e + P t F_{6} \to X e^{+} P t F_{6}^{-}

The most likely reason for the suggestion being made, is:

The first ionization energy of Na, NO, Xe, and O

_{2}

follows the order

N a < N O < X e = O_{2} . O_{2}

reacts with the powerful oxidizing agent

P t F_{6}

to yield a compound

O_{2} [P t F_{6}] .

P t F_{6}

is allowed to react with other mentioned species, then the product is :

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