To draw the Lewis structure for the bromate ion (BrO3–), we start by calculating the total number of valence electrons. Bromine (Br) has 7 valence electrons, and each oxygen (O) has 6; since there are three oxygen atoms, that gives us a total of 18 electrons from the oxygen. We also have one additional electron because the ion has a -1 charge. So, the total number of valence electrons is:
7 (Br) + 3 × 6 (O) + 1 (charge) = 7 + 18 + 1 = 26 electrons.
Next, we place Br in the center and surround it with the three O atoms. Connect each oxygen to bromine with a single bond. This accounts for 6 of the 26 electrons (2 electrons per bond). We then distribute the remaining electrons to satisfy the octet rule for each oxygen atom. After distributing, each oxygen will have 6 electrons (3 lone pairs) and Br will have 6 total electrons (2 from single bonds). To satisfy the octet and minimize formal charges, we can form a double bond between Br and one of the O atoms.
Here’s the final structure obeying the octet rule:
Now, if we consider a scenario where we break the octet rule while still aiming to minimize formal charges, we might involve a structure where bromine exceeds the octet by forming fewer double bonds or solely single bonds. In this case, we can allow bromine to have 10 electrons total. To illustrate this, we can have Br bonded to two O atoms with double bonds and one O atom with a single bond:
With this structure, bromine has 10 electrons (4 from the double bonds and 2 from the single bond), and the formal charges on the atoms are minimized, leading to a stable form of the ion.