1.

A compound of Xe and F is found to have 53.3% Xe (atomic weight=133). Oxidation number of Xe in this compound is

Answer»

`+2`
0
`+4`
`+6`

Solution :`XE :F=( 53.3)/(133):(46 .7 )/(19)`
`= 0.4 : 2.45= 1 :6`
thusE.Fofxenonfluorideis `XeF_6`
` thereforeO.Nof XeinXeF_6` is + 6


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