Question

write the formal charge and lewis structure of H2CO3 ?​

Answer 1

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Explanation:

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his is the H2CO3 Lewis structure: carbonic acid. Hydrogen has 1 valence electron, we have 2 Hydrogens; plus 4 for Carbon, plus 6 for Oxygen times 3, for a total of 24 valence electrons. Whenever you see Hydrogens in front of a polyatomic ion like CO3, NO3, or SO4, it's going to be an acid and you're going to need to put those Hydrogens attached to the outside Oxygens. So we'll put the Carbon at the center. We have three Oxygens, we'll put three Oxygens around it. And then we'll put the two H's around the outside of it.

So we have a total of 24 valence electrons. We'll put 2 electrons between atoms to form chemical bonds. So we've used 2, 4, 6, 8, 10. Then we'll form octets for each atom: 12, 14, 16, 18, 20, 22, and 24. At this point, the Hydrogens have 2 valence electrons each so their outer shells are full. And the Oxygens have 8 valence electrons each, so they have octets. Their outer shells are full, as well. However, the Carbon only has 6 valence electrons, so it needs 2 more to form an octet.

What we can do is take 2 valence electrons from this Oxygen and move them to the center and share them in a double bond. By sharing the valence electrons in that double bond, Oxygen has 8 still, but now the Carbon has 8 and we're still only using 24 valence electrons. So we've used all 24 valence electrons and each of the atoms in H2CO3 has a full outer shell. So that's the Lewis structure for H2CO3. We could check our formal charges.

If we did, we'd find that the formal charge for each atom in H2CO3 is 0, making this the Lewis structure for H2CO3.

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Answer 2

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This is the$H$₂$C$$O$₃ Lewis structure: carbonic acid. Hydrogen has $1$valence electron, we have 2 Hydrogens; plus 4 for Carbon, plus $6$ for Oxygen times 3, for a total of 24 valence electrons. Whenever you see Hydrogens in front of a polyatomic ion like$CO$₃, $NO$₃, or $SO$₄, it's going to be an acid and you're going to need to put those Hydrogens attached to the outside Oxygens. So we'll put the Carbon at the center. We have three Oxygens, we'll put three Oxygens around it. And then we'll put the two H's around the outside of it.

So we have a total of $24$ valence electrons. We'll put $2$ electrons between atoms to form chemical bonds. So we've used 2, 4, 6, 8, 10. Then we'll form octets for each atom: $12, 14, 16, 18, 20, 22, and \:24$. At this point, the Hydrogens have $2$ valence electrons each so their outer shells are full. And the Oxygens have$8$ valence electrons each, so they have octets. Their outer shells are full, as well. However, the Carbon only has $6$  valence electrons, so it needs $2$ more to form an octet.

What we can do is take $2$valence electrons from this Oxygen and move them to the center and share them in a double bond. By sharing the valence electrons in that double bond, Oxygen has 8 still, but now the Carbon has $8$and we're still only using 24 valence electrons. So we've used all $24$ valence electrons and each of the atoms in$H$₂$C$$O$₃ has a full outer shell. So that's the Lewis structure for$H$₂$C$$O$₃. We could check our formal charges.

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