Throwback to Chapter One!

In lieu of reviewing for exams, I wanted to go back to basics and look at a problem from chapter one. This question in number 2 for sapling and focuses on the octet rule and formal charges. The octet rule is when an atom has all eight of it's electrons in its valance shell. An atom is more stable when this happens which is why we try so hard to obtain an octet. If an atom doesn't have eight valance electrons filled it has the possibility of getting a formal charge. A formal charge is pretty much saying how negative or positive the charge on the atom is. To find the formal charge you have to subtract it's valance electrons from its lone pair electrons then add the bonding electrons divided by to. The formula looks like this: valance electrons - [lone pairs electrons + (bonding electrons/2)].

*I say that there is a possibility of getting a formal charge because there are cases when the atom can not have an octet and not have a formal charge! For example, BH^3 is one of these cases.*

The first part of the problem gives us three different examples to assign formal charges: BH^3, BH^4, and I. For BH^3 the formal charge is 0 because boron has 3 valance electrons, 0 lone pairs, and 6 bonding electrons. So the formula would be 3- [0 + (6/2)] = 0.

*Don't forget to count each bond counts as 2 electrons! So for BH^3, it has three bonds to three hydrogens which means it has 6 bonding electrons.*

For BH^4 the formal charge is -1 because boron has 3 valance electrons, 0 lone pairs, and 8 bonding electrons. So the formula would be 3- [0 + (8/2)] = -1.

For BH^4 the formal charge is -1 because boron has 7 valance electrons, 6 lone pairs, and 0 bonding electrons. So the formula would be 7- [6 + (0/2)] = +1.

The second part of the problem asks us to identify which atom has the complete octet. The answer is boron in BH^4 become all 8 of its valance electrons are filled. The answer can't be BH^3 or Iodine because only have 6 valance electrons.