Untitled - Kelly Walsh High School

Chemical Bonding 139
Quick Tip
involve the hydrogen since it already has the one and only bond that it can have.
The oxygen next to the hydrogen is not a likely candidate since it already has two
bonds. The most likely candidate for another bond is the other oxygen atom. We
can add this bond, which changes the N-O bond to a double bond. The bonding
arrangement is now H-O-N=O. The four bonds, at two electrons each, account
for eight of the available electrons (A). To finish the structure, we need to add the
remaining 10 electrons (A 8). Usually electrons come in pairs, so our 10 electrons
will appear as 5 pairs. We will begin with the oxygen atoms since oxygen is
the most electronegative element in this compound. (The most electronegative
elements will get its octet.) Each oxygen atom has 2 bonds (4 electrons), so each
needs 4 electrons (2 pairs). We need to add two separate pairs to each oxygen
atom. It is not too important where you place the pairs as long as you do not place
the electrons between the oxygen and another atom. These pairs account for 8
more electrons. The remaining 2 electrons constitute one more pair. This pair will
probably go on the nitrogen, but you should check to make sure. The nitrogen
atom has 3 bonds (6 electrons), and it needs 2 more to achieve an octet.
Therefore, we can add our last 2 electrons as a pair on the nitrogen. The positioning
of the pair is not important. Do not add the two electrons separately.
All the available electrons, A, must appear in the final Lewis structure. They will
normally appear in pairs unless there are an odd number of electrons. If the number
of electrons is odd, there will be only one unpaired electron.
The final Lewis structure for nitrous acid is:
Η Ο Ν
In this structure, we have two central atoms. The nitrogen is one, and the oxygen
with the hydrogen is the other.
Now we will begin drawing the Lewis structure of XeF 4. Xenon will be the central
atom, and we will arrange fluorine atoms around it. In this way, we avoid
attaching identical atoms to each other. We will need a bond between the central
xenon and each of the fluorine atoms. This arrangement means there will
be at least four bonds.
If you wish to apply the S N A rule, you will need to assign values to N and
A. To find N, we use 8 for xenon and 8 for each of the fluorine atoms. This gives
O