what is the oxidation state of XeOF4, XeO4,XeO2F2,XeOF2
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XeOF4=x-2-4=6
XeO4=. x+4(-2)=x=8
XeO2F2=x+2(-2)+2(-1)=x-4-2=x=6
XeOF2=x-2-2=4
i hope this will help u
XeO4=. x+4(-2)=x=8
XeO2F2=x+2(-2)+2(-1)=x-4-2=x=6
XeOF2=x-2-2=4
i hope this will help u
pratiksha01:
Thanks a lot
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