# VSEPR geometry of Bromate Ion [duplicate]

After doing the Lewis structure of Bromate Ion and in order to find the molecular geometry using VESPR method, we have:

central atom: Br

Electrons of the central atom: 7

Electrons that contribute the 3 Oxygens : 3

Electrons that contribute the central atom for the π bonds: -2

Charge of ion (negative) : 1

Total electrons : 9

But if we divide 9 with 2 in order to find the σ-bonding pairs, we get 4,5. what is wrong? Because after searching the molecular geometry of BrO3- it says that it is trigonal pyramid.

• I'm not sure what is your difficulty exactly. If you've already done the Lewis structure, then you know the Bromate ion looks like this. Three terminal atoms + a lone pair indicate that this polyatomic ion is of the type AX3E1. According to this table, this is a trigonal pyramid. So what exactly is the problem? What purpose do these calculations serve? – Don_S Mar 1 '17 at 13:21
• @Jan I am a bit hesitant to agree that this is a duplicate. The reason for this is that the target question is convoluted with needless information of a pupil-teacher e-mail back and forth and it is hard to find the actual question. I like your answer there and would prefer to see it here. – Martin - マーチン Mar 1 '17 at 13:54
• @Martin-マーチン Fairy ’nuff ;) – Jan Mar 1 '17 at 13:57

## 1 Answer

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:

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

For bromate, the calculation is:

$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$

2. 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.

$$4\times 8 = 32\tag{2}$$

3. How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$

$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$

4. 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.)

• Thank you, that was very helpful! According to this, the molecular geometry of ClO2- will be linear? valence electrons=20, electrons we need to form octets: 3x8=24 , electrons we're missing : 24-20=4 or 2 pairs and electrons that will form lone pairs: 20-4=16 or 8 pairs – user40808 Mar 1 '17 at 15:22
• @MariaP No, it will be bent; you will have two lone pairs on chlorine (and three per oxygen). Linear compounds are very rare for purely single-bonded atoms; to the best of my knowledge, they only occur with beryllium. Multiple bonds are a different story. – Jan Mar 1 '17 at 15:33