Chemistry · General chemistry I · Worked example
Draw the Lewis structure of CH₂O
Draw the Lewis structure of formaldehyde, CH₂O, and check the formal charges.
Count the valence electrons
Carbon brings 4, each hydrogen 1 and oxygen 6.
Connect the atoms
Carbon goes in the center, bonded to both hydrogens and to oxygen. Three single bonds use 6 electrons, leaving 6.
Complete the octets
The 6 remaining electrons go on oxygen as three lone pairs, giving it an octet. Carbon now has only 6 electrons, in its three bonds.
Form a double bond
Move one of oxygen’s lone pairs into the carbon–oxygen bond. Now carbon has 8 electrons, and oxygen still has 8: two shared pairs and two lone pairs.
Check the formal charges
Carbon: 4 − 0 − 8/2 = 0. Oxygen: 6 − 4 − 4/2 = 0. Each hydrogen: 1 − 0 − 2/2 = 0. They add up to 0, the charge of the neutral molecule.
| Atom | Valence V | Nonbonding N | Bonding B | V − N − B/2 |
|---|---|---|---|---|
| C | 4 | 0 | 8 | 0 |
| O | 6 | 4 | 4 | 0 |
| H (each) | 1 | 0 | 2 | 0 |
Result
Carbon has single bonds to both hydrogens and a double bond to oxygen, which keeps two lone pairs. Every formal charge is zero.
Your turn
Draw the Lewis structure of hydrogen cyanide, HCN.
Show the answer and explanation
H–C≡N, with one lone pair on nitrogen; every formal charge is zero.
1 + 4 + 5 = 10 electrons. The single bonds H–C and C–N use 4; putting the other 6 on nitrogen leaves carbon with only 4, so two of nitrogen’s lone pairs become bonds: C≡N. Carbon then has 8 electrons and nitrogen 8. Formal charges: C = 4 − 0 − 4 = 0 and N = 5 − 2 − 3 = 0.
Keep exploring
In Lewis electrons & resonance, change the carbon–oxygen double bond to a single bond and give oxygen six nonbonding electrons. The count is still 12, but carbon has only 6 shell electrons and the formal charges become +1 on carbon and −1 on oxygen.
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