show Louis dot structure and hybridization for co2
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You must first draw the Lewis structure for "CO"_2.
CO2
CO2
According to VSEPR theory, we can use the steric number ("SN") to determine the hybridization of an atom.
"SN" = number of lone pairs + number of atoms directly attached to the atom.
"SN = 2" corresponds to sp hybridization.
"SN"= 3" corresponds to sp^2 hybridization.
We see that the "C" atom has "SN = 2". It has no lone pairs, but it is attached to two other atoms.
It has sphybridization.
Each "O" atom has "SN = 3". It has 2 lone pairs and is attached to 1 "C" atom.
Just as the carbon atom hybridized to form the best bonds, so do the oxygen atoms.
The valence electron configuration of "O" is ["He"] 2s^2 2p^4.
To accommodate the two lone pairs and the bonding pair, it will also form three equivalent sp^2 hybrid orbitals.
Two of the sp^2 orbitals contain lone pairs, while the remaining sp^2 orbital and the unhybridized p orbital have one electron each.
We can see this arrangement in the "C=O" bond of formaldehyde, which is equivalent to the right-hand side of the "O=C=O" molecule.
Hope it will help...
CO2
CO2
According to VSEPR theory, we can use the steric number ("SN") to determine the hybridization of an atom.
"SN" = number of lone pairs + number of atoms directly attached to the atom.
"SN = 2" corresponds to sp hybridization.
"SN"= 3" corresponds to sp^2 hybridization.
We see that the "C" atom has "SN = 2". It has no lone pairs, but it is attached to two other atoms.
It has sphybridization.
Each "O" atom has "SN = 3". It has 2 lone pairs and is attached to 1 "C" atom.
Just as the carbon atom hybridized to form the best bonds, so do the oxygen atoms.
The valence electron configuration of "O" is ["He"] 2s^2 2p^4.
To accommodate the two lone pairs and the bonding pair, it will also form three equivalent sp^2 hybrid orbitals.
Two of the sp^2 orbitals contain lone pairs, while the remaining sp^2 orbital and the unhybridized p orbital have one electron each.
We can see this arrangement in the "C=O" bond of formaldehyde, which is equivalent to the right-hand side of the "O=C=O" molecule.
Hope it will help...
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