Draw Lewis Structure for BrO3^-
The textbook answer involves a central Br atom with one lone electron pair and five shared pairs (two double bonds and one single bonds).
This gives Br twelve electrons rather than the typical eight. I understand that some atoms like Br have expanded valence shells, but how would I know exactly how much it has expanded? I don't see how I could have come up with this structure by myself without a supplied answer.
thanks!
I can't draw an example on this forum but you can tell by counting the electrons as they are introduced to the atom in question, in this case Br. Consider PF5.
Count the valence electrons in P and F.
P = 1 atom x 5e = 5electrons
F = 5 atoms x 7e each = 35 electrons.
Total = 40 electrons
Making P the central atom, start adding F atoms around the P, each F atom takes 8 electrons. That will take up the 40 electrons (8*5 = 40) but that places 10 electrons around P. I don't think there is another way of doing it. In practice, I start placing the atoms. When I've added 4 F atoms, I have the octet for P BUT I still have another F to add so add that to the central atom.
To determine the Lewis structure for the BrO3^- ion, it is helpful to know the number of valence electrons for each atom involved. In this case, Br is in Group 7A or 17 of the periodic table, and oxygen is in Group 6A or 16. Hence, the Br atom has 7 valence electrons, while each oxygen atom has 6 valence electrons. The negative charge in the BrO3^- ion indicates the addition of an extra electron, bringing the total number of valence electrons to account for to 26 (= 7 + 6 + 6 + 7).
To begin constructing the Lewis structure, we typically place the atom with the lowest electronegativity (in this case, Br) in the center. Placing the Br atom as the central atom, we then draw single bonds between the Br atom and each oxygen atom, using up 4 electrons (2 from the Br atom and 2 from each oxygen atom).
At this point, we have used up 4 electrons and have 22 remaining. We notice that Br still has an incomplete octet (it only has 6 electrons around it), so we need to be creative with the remaining electrons. One way to achieve a total electron count of 26 is by converting one of the oxygen-bronze single bonds into a double bond. This requires two additional electrons, reducing the remaining electron count to 20.
To maintain the overall charge of -1, we then place the remaining 20 electrons on oxygen atoms as lone pairs, giving each oxygen atom a complete octet. Oxygen can accommodate 3 lone pairs, which requires 6 electrons per oxygen atom. Assigning 6 electrons to each oxygen atom uses up the remaining 18 electrons, leaving us with 2 unassigned electrons.
To distribute these remaining electrons, we place them as a lone pair on the central Br atom, giving it a total of 12 electrons. This helps accommodate the expanded valence shell of bromine in this particular ion.
To summarize, the Lewis structure for BrO3^- is as follows:
O
/
Br=O
\
O^-
By following the general rules for constructing Lewis structures, we can determine that Br can have an expanded valence shell with 12 electrons in this particular ion. It is important to note that the formal charges on the atoms in the Lewis structure also need to be considered to ensure the overall charge is balanced.