VSEPR geometry of Bromate Ion
There’s a simple way to calculate how many bonds you need to form a Lewis structure, and how many lone pairs remain. It is four simple steps:


How many valence electrons do we have? (add up)

For bromate, the calculation is:

$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$
How many valence electrons are required for full octets/hydrogen dublets?

This equates to $2m + 8n$, where $m$ is the number of hydrogens and $n$ the number of other elements.

$$4\times 8 = 32\tag{2}$$
How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$

$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$
How many electrons are remaining? These will form lone pairs. $(1) - (3)$

$$26 - 6 = 20 = 10~\text{pairs}\tag{4}$$


Thankfully, knowing the general structure is ‘halogen in the centre, oxygen atoms around it’ and knowing that there are three oxygen atoms, this clearly shows us that all $\ce{Br-O}$ bonds are single bonds. The halogen would then need one more lone pair to be satisfied, the remaining nine lone pairs go to the three oxygen atoms (3 each). This leads us to the following, final Lewis structure:


(Unfortunately, the $2+$ on iodine and its remaining lone pair intersect in the image. Originally taken from an answer about iodate hence the central iodine. Replace it with bromine to get bromate.)