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QUESTION

# How do you draw the lewis structure with formal charges?

• Draw possible Lewis structures for carbon monoxide. Include all formal charges.

• We first draw a skeleton structure with only a single bond between the two atoms: This gives C-O.

• We now write a trial structure, in which we put enough electron pairs (two dots) around every atom to give it an octet of . This puts three electron pairs around C and three electron pairs around O.
• Count all the in the trial structure. We have twelve lone pair electrons plus two electrons in the C-O single bond. This makes 14 valence electrons.
• Now we use to figure out how many valence electrons we can actually have in our structure. We count as follows: C = 4; O = 6. This makes a total of only 10 valence electrons. Our trial structure is WRONG. It has four electrons too many.
• We start over. This time, for each two electrons in excess, we must insert a double bond. Since we have four electrons in excess, we must insert two double bonds or a triple bond.
• We write C≡O. Again, we give every atom enough electron pairs to complete its octet.
• We end up with :C≡O: as the Lewis structure for CO (Note: 10 valence electrons!). Now we calculate the formal charge on each atom. To do this, we assign electrons as follows: --> Shared electrons are shared equally. Each atom gets half of the electrons in the bond. --> Lone pair electrons belong entirely to the atom on which they reside.

Then we compare with the number of valence electrons in an isolated atom.

We see that in this structure C has five valence electrons — two from the lone pair and three from the C≡O triple bond, and O also has five valence electrons.

Since an isolated C atom has 4 valence electrons and here it has 5, it has gained an electron. It has a formal charge of -1.

Since an isolated O atom has 6 valence electrons and here it has 5, it has lost an electron. It has a formal charge of +1.

We might write the structure as :C⁻¹≡O:⁺¹. This structure satisfies the , but it contains formal charges.

We can avoid formal charges if we write the structure as :C=Ö: However, although this structure avoids formal charges, it does not give C an octet. There is just no single good Lewis structure for CO.

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