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Lewis Dot Diagrams
In Lewis Dot diagrams a dot represents and electron. Thus H indicates a hydrogen atom with one electron. Another H can get together with the first hydrogen so that the two
together share two electrons between them. This can be written H:H. Each hydrogen
then has a share in the two electrons and (somewhat) reaches the stable He configuration. Hydrogen is said to have a duet of electrons. The two electrons shared are counted as a single bond and the Lewis Dot Diagram (LDD) is written H─H.
Follow ―Steps for Drawing Lewis Dot Diagrams‖ to draw the LDD for an an O 2 molecule. The structure (step 1) must be O─O. Each oxygen atom brings in 6 valence
electrons for a total of 12 electrons (step 2). Apply step 3 to get O
─O
: Application of
step 4a requires that the completed structure have octets on both oxygen atoms. The
oxygen on the right has eight electrons around it (─O
: ). That oxygen is said to have an
octet. The other oxygen has just 6 electrons around it (O
─ ). That is not an octet, so
electrons must be moved to form octets on both oxygen atoms. There is no point to taking electrons from the right oxygen and placing them on the left oxygen, because then the right oxygen would not have an octet. But if a pair of electrons on the right oxygen is rotated into the bond so that four electrons are shared between oxygen atoms, the bonding
could be represented asO
::O
, where the four shared electrons are counted as a double
bond and the LDD is written asO
═O
Now each oxygen (O
═ and ═O
) has an octet,
(4 unshared electrons and 4 shared electrons). Notice bonding electrons are counted twice to determine octets.
Follow ―Steps‖ to draw the LDD for NOF. The structure must be O−N−F according to step 1a, because nitrogen with electronegativity 3.0 is less than oxygen with 3.5 and fluorine with 4.0, so it must be the central atom. The number of electrons must be 5 (for
N) +6 (for O) + 7 (for F) = 18. Applying step 3 gives :O
−N
−F
: , in which the N does
not have an octet. The octet rule (step 4a) can be satisfied either by moving electrons
around the oxygen into the N−O bond to giveO
═N
−F
: or by moving electrons around
the fluorine into the N−F bond to give :O
−N
═F
. To choose the better of these we
need to apply step 5a. To apply step 5 we need formal charges , which are calculated as follows:
Formal Charges Each atom in a molecule, ion or radical is assigned a charge
using the LDD. To assign charges in an LDD, the electrons in each bond are divided equally between the atoms sharing them. From a single bond each atom gets one of the electrons. From a double bond each atom gets two electrons and from a triple bond each atom gets 3 electrons. These electrons are added with the lone pair (unshared) electrons and the sum is compared to the valence electrons the atom brought into the LDD. In
O
═N
−F
: oxygen has 6 electrons (4 unshared + 2 from the bond). This is exactly equal
to the 6 electrons oxygen brought into the LDD so formally it has a charge of zero. It has neither gained nor lost electrons in the bonding process. Nitrogen formally has 5 electrons ( 2 unshared + 2 of the electrons bonded with O +1 of the electrons bonded with F) for a formal charge of zero. Likewise fluorine formally has 7 electrons (6 unshared + 1 from the bond) for a formal charge of zero. A formula for formal charge would be:
Formal charge = valence electrons – bonds – lone pair electrons
In :O
−N
═F
oxygen has a formal charge of −1 ( = 6 – 1 – 6) Nitrogen has formal
charge 0 (= 5 – 3 – 2). The fluorine, however, has a formal charge of +1 (=7 – 2 – 4).
So according to step 5aO
═N
−F
: with all formal charges 0, 0, 0 is better than
:O
−N
═F
with formal charges −1, 0 ,+1. Also formal charges are useful for choosing
the best skeletal structures (step 1C). Notice thatO
═N
−F
: is better than N
═O
−F
which has formal charges −1, +1, 0. LDDs with fluorine in the middle are worse yet.
For example, N
═F
−O
: has formal charges −1,+2, −1. Usually the step 1b picks the
correct skeletal structure without the necessity of drawing LDDs for each possible skeletal structure, but occasionally step 1c will pick a better skeletal structure than 1b.
Sometimes it is not possible to draw an LLD with octets having all formal charges equal to zero. The best structure has octets and formal charges as close to zero as possible.
Consider the molecule ozone, O 3 , with skeletal structure O–O–O and 18 electrons (3×6).
To get octets on each oxygen we could draw the LLD as O
═O
−O
: with formal
charges 0,+1,–1 or as :O
−O
═O
with formal charges – 1,+1,0. It is not possible to
draw the molecule with octets and formal charges 0, 0, 0. The two Lewis Dot structures are equivalent but they are not the same structure. It is possible to distinguish the different ends of the ozone molecule by putting a different isotope of oxygen on one end, maybe oxygen-17. Then in one of the diagrams the molecule would have a double bond to the oxygen-17 and the other has the single to the oxygen-17.
than H–O
═N
−O
: , which has formal charges 0,+1,0,–1. So these two LLDs are not
equivalent and no resonance diagram is required. The ON and NO bond lengths are expected to be different in this molecule and experimentally they are, 143.3 pm and 117. pm, respectively. And this is consistent with the best structure—the one having the single bond (longer bond) with the O bonded to H.
