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What is the most logical and easy to remember step by step method for drawing Lewis dot structures? I'm having trouble drawing a simple one for $\ce{CO2}$.

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  1. Calculate the total number of valance shell electrons ($n_1$) in all constituent atoms or ions. For example, in $\ce{NH_4^+}$, $n_1$ will be 8 (5 of N, 4 from all H's and -1 because the molecule has +1 charge).
  2. Now, calculate the total number of electrons that are required to fill valance shell of every constituent atom or ion ($n_2$). In other words, calculate the total number of electrons that will be required to fill valance shell of each atom or ion individually. We can use following formula to find $n_2$
    $n_2 = [\text{# of Hydrogen atoms} \times 2] + [\text{# of all other atoms except Hydrogen} \times 8]$ In case of $\ce{NH_4^+}$, $n_2$ will be 16 ($[4 \times 2] + [1 \times 8]$).
  3. Now, calculate the total number of electrons that atoms or ions will need to share among each other in order to complete their octets ($n_3$). $n_3 = n_2 - n_1$ For $\ce{NH_4^+}$, $n_3$ will be 8.
  4. Now calculate the number of bonds that will be present in our molecule $n_4 = \frac{n_3}{2}$ For $\ce{NH_4^+}$, $n_4$ will be 4.
  5. Now let's find the number of unshared electrons, $n_5$ $n_5 = n_1 - n_3$
    For $\ce{NH_4^+}$, $n_5$ will be 0.
  6. Now find the number of lone pairs, $n_6$, which is nothing but $n_5$ divided by 2.
    _For $\ce{NH_4^+}$, $n_6$ will be 0.
  7. Now find the central atom. Central atom is the one which is either least in count or more electropositive. Always remember that Hydrogen and Fluorine can never be placed at center. So in our case, Nitrogen gets to be the central atom.
  8. After allocating the central atom, place the surrounding atoms such that the molecule is symmetric. In our case, geometry of molecule will be tetragonal.
  9. Distribute the bonds between atoms. Remember that the total number of bonds, that you distribute, should be equal to $n_4$. In our case, we simply place a bond between Nitrogen and each Hydrogen atom.
  10. Now use the lone pairs to complete the octet of atoms. Remember that, by now, every atom will already have 2 electrons in their valance shell because of the bond.
    In our case, which is $\ce{NH_4^+}$, we don't have any lone pairs. OOPS!
  11. Now comes the Formal Charge.

    Formal Charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in a chemical bond are shared equally between atoms, regardless of relative electronegativity. It is not an actual electrical charge but just a mathematical construct.

$\text{Formal Charge of an atom} = \text{# of valence shell electrons (in Ground state)} - \text{# of unshared/non-bonded electron pairs or lone pairs} - \text{# of bonds}$

Please note that you must place formal charge on each and every atom when it is a non-zero value for that atom. This step is little confusing so let's get back to our example. Nitrogen has 5 electrons in its valance shell in ground state. It made 4 bonds in our molecule and has none unshared electron pairs or lone pairs (please note that a single unpaired electron doesn't count here). So formal charge on Nitrogen will be 1 (5 - 0 - 4), whereas formal charge on each Hydrogen atom will be 0 as every Hydrogen atom has 1 electron in its valance shell and made one bond in our molecule.

In a sense, you actually obtain the structure in $10^{th}$ step but Formal charge, if a non-zero value, should be represented for each atom in molecule in order to complete its Lewis Dot Structure.

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  • $\begingroup$ I found this explanation confusing, does anyone have a simpler method? $\endgroup$ – jaykirby Jun 28 '13 at 3:09
  • $\begingroup$ this is the actual method! $\endgroup$ – ashu Jun 28 '13 at 4:55
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    $\begingroup$ Must be a simpler way.. $\endgroup$ – jaykirby Jun 28 '13 at 6:55
  • $\begingroup$ Thank @bon for the edit. I am using S.E. on mobile so its pretty hard to get things working in first try. :p $\endgroup$ – ashu Feb 15 '15 at 18:17
  • $\begingroup$ dont worry. there was just a single $ missing which was messing it up $\endgroup$ – bon Feb 15 '15 at 18:19
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I will give you a simple way, but if you need to study properly, you can follow Ashutosh Gangwar's answer.

C has 4 valance electrons. O has 6 valance electrons. O has 6 valance electrons.

First distribute these electrons to each element and indicate the covalent bonds.

So you know carbon and oxygen need to obey octed rule, so in this case carbon makes 4 bonds with oxygens;

enter image description here

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