Determine the number of bonding electrons and the number of nonbonding electrons in the structure of of2.

OF2 - 
O has 6 electrons in outer shell and F has 7 in its outer shell 
Therefore, you have to account for 20 electrons total in the 
structure (7+7+6 = 20) 
therefore draw it linear first. F OF 
The two bonds take care of 4 electrons now you have to add another 16. 
Therefore 3 lone pairs on each F and 2 lone pair on O. 
If you check for formal charges, all the atoms are neutral 
F will have 3 lone pairs + 1 bond = 7 electrons (bond = 1/2 electron for formal charge distribution) therefore both the F's are neutral 
Now look at the O: it should have 6.. it has two lone pair and 2 bonds = 4 electrons and 2 bonds = 1 electron each = 2 electrons from bonds = 6 total electrons for formal charge which is exactly the # it should have. There is no need for any double bond in this as there are no charges to be separated. 
Now if u look at the # of domains around O you will see if you include the lone pairs it has a sp3 hybridization (4 domains) therefore a tetrahedron which has 2 lone pairs and 2 bonds.. since there are two lone pairs, the lone pair/bond pair repulsion is so high it is going to repel the two Fluorines and form a bent structure, looks a lot like H2O. 

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