How to Draw a Lewis Structure for BrO3⁻ That Obeys the Octet Rule and Assign Oxidation Numbers

To draw a Lewis structure for BrO₃^– (bromate ion) that obeys the octet rule, follow these steps:

1. Count the total number of valence electrons:
   • Bromine (Br) has 7 valence electrons.
   • Each oxygen (O) has 6 valence electrons.
   • There are 3 oxygen atoms, so 3 × 6 = 18 electrons.
   • Add 1 electron for the negative charge.
   • Total valence electrons = 7 + 18 + 1 = 26 electrons.
2. Place the least electronegative atom in the center:
   • Bromine is less electronegative than oxygen, so it will be the central atom.
3. Connect the central atom to the surrounding atoms with single bonds:
   • Draw single bonds between Br and each O atom. This uses 6 electrons (3 bonds × 2 electrons).
4. Distribute the remaining electrons to satisfy the octet rule:
   • After forming the bonds, 20 electrons remain (26 – 6 = 20).
   • Place lone pairs on the oxygen atoms first. Each oxygen needs 6 more electrons to complete its octet.
   • Each oxygen will get 6 electrons (3 lone pairs), using 18 electrons (3 O atoms × 6 electrons).
   • This leaves 2 electrons, which will be placed on the central bromine atom as a lone pair.
5. Check the formal charges:
   • Formal charge = (Valence electrons) – (Non-bonding electrons) – (Bonding electrons/2).
   • For Br: 7 – 2 – (6/2) = 0.
   • For each O: 6 – 6 – (2/2) = -1.
   • The overall charge of the ion is -1, which matches the sum of the formal charges.

To assign oxidation numbers:

1. Bromine (Br): The oxidation number is +5.
2. Oxygen (O): Each oxygen has an oxidation number of -2.

This structure obeys the octet rule and correctly assigns oxidation numbers to each atom.