Xe would certainly be your main atom, O is attached to it through a doublebond and also your 2 Fs space attached it is in a single bond.
There would certainly be 3 Lone Pair (LP) electrons approximately F, 2 about O,and 2 approximately Xe. Why ? there is no valance come this molecule for this reason youwant to meet the formal fee to be 0. Formal fee =(#electrons in ground state) - < (free electrons) - 1/2 (bondingelectrons)>.
Formal charge for F, would desire 0 so
0 = 7 - + 1/2 (2) > = 7 - (? +1) = 7 - ? - 1
? = 6, thus 3 lone pairs around fluorines.
Formal fee for O, would want 0 so
0 = 6 - + 1/2 (4) >...
? = 4, as such 2 LP roughly oxygen
Formal fee for Xe, again we desire it to be 0 so,
0 = 8 - + 1/2 (8)>.....
? = 4, because of this 2 LP around Xe.
Kind of a weird set up (since Xe does not follow the octectrule), but it works out =)
Wiki User∙ 9y ago
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