How do you draw the Lewis structure for the hypobromite ion, BrO⁻?

To draw the Lewis structure for the hypobromite ion (BrO⁻), follow these steps:

Count the Total Valence Electrons

First, determine the number of valence electrons:

• Bromine (Br) has 7 valence electrons.
• Oxygen (O) has 6 valence electrons.
• Since the ion has a negative charge (BrO⁻), add 1 more electron.

This gives a total of 7 + 6 + 1 = 14 valence electrons.

Arrange the Atoms

Next, place the bromine atom in the center and connect it to the oxygen atom:

• Br is less electronegative than O, so it goes in the center.
• Connect Br and O with a single bond.

Distribute Remaining Electrons

After forming the bond, you have 12 electrons left to distribute:

• Place 6 electrons (3 lone pairs) around the oxygen atom to satisfy its octet.
• This uses up 6 of the remaining electrons, leaving you with 6 electrons.

Complete the Octet for Bromine

Now, place the remaining electrons on the bromine atom:

• Add 3 lone pairs (6 electrons) to Br.
• This gives Br a total of 8 electrons, completing its octet.

Check Formal Charges

Finally, ensure that the formal charges are minimized:

• For Br: 7 (valence) - 0 (bonds) - 6 (lone pairs) = +1.
• For O: 6 (valence) - 2 (bonds) - 4 (lone pairs) = 0.

The overall charge of the ion is -1, which is consistent with the hypobromite ion.

Final Structure

The Lewis structure for the hypobromite ion shows a single bond between Br and O, with Br having three lone pairs and O having one lone pair:

Br: 3 lone pairs, O: 1 lone pair, and a single bond between them.