Answer:
Option E!
Explanation:
If we were to draw the lewis dot structure for IBr2 -, we would first count the total number of valence electrons ( " available electrons " ). Iodine has 7 valence electrons, and so does Bromine, but as Bromine exists in 2, the total number of valence electrons would be demonstrated below;
[tex]7 + 7 * ( 2 ) =\\7 + 14 + 1 =\\22 Electrons[/tex]
Don't forget the negative on the Bromine!
Now go through the procedure below;
1 ) Place Iodine in the middle and draw single bonds to each of the bromine.
2 ) Add three lone pairs on each of the Bromine's
3 ) Now we have 6 electrons left, if we were to exclude the electrons shared in the " single bonds. " This can be placed as three lone pairs on Iodine ( central atom )!
The molecular geometry can't be linear, as there are lone pairs on the atoms. This makes it bent.
