Draw out the Lewis structure for the given ion IBr2

To draw the Lewis structure for the ion IBr₂^–, we need to consider the valence electrons for each atom involved.

Iodine (I) is in Group 17 and has 7 valence electrons. Each Bromine (Br) atom, also in Group 17, has 7 valence electrons as well. Therefore, for IBr₂, we have:

• 1 Iodine: 7 electrons
• 2 Bromines: 2 x 7 electrons = 14 electrons

In total, we start with 7 + 14 = 21 valence electrons. Since we have a -1 charge on the ion, we add an additional electron, bringing the total to 22 valence electrons.

Next, we’ll place the Iodine atom in the center and surround it with the two Bromine atoms:

  Br
   |
   I
   |
  Br

Now, we start bonding the atoms. Each bond between Iodine and Bromine uses 2 electrons. With 2 Br atoms bonded to I, we use 4 electrons, leaving us with 22 – 4 = 18 electrons remaining.

Next, we complete the octets of the Bromine atoms. Each Bromine requires 6 more electrons to complete their octets:

   :Br:
  ..|
   I
  ..|
   :Br:

Now, we have used 4 electrons for the bonds and 12 for the lone pairs on the 2 Bromine atoms, totaling 16 electrons used. This leaves us with 2 electrons to place on the Iodine atom. These 2 electrons will be set as lone pairs:

   :Br:
  ..|
 ..I..
  ..|
   :Br:

Finally, we ensure that the overall structure has the right charge. The Iodine atom has 2 lone pairs and is bonded to two other atoms, maintaining the overall charge of -1 for the ion. This concludes the Lewis structure for IBr₂^–.