Question #93d0a

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Question #93d0a

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Answer 1 · 2016-07-10T20:46:14

No, there is no need for hybridization in $^{-} : C \equiv O :^{+}$ $""^(-):"C"-="O":^(+)$ .

It's a diatomic linear molecule, and both atoms have all the necessary orbitals to make their bonds. So, no hybridization is required.

There are one sigma ( $σ$ $sigma$ ) and two pi ( $π$ $pi$ ) bonds made between $C$ $"C"$ and $O$ $"O"$ , since a triple bond contains one $σ$ $sigma$ and two $π$ $pi$ bonds.
The $σ$ $sigma$ bond is made via a $C_{2 p_{z}} - O_{2 p_{z}}$ $"C"_(2p_z)-"O"_(2p_z)$ head-on overlap. That is enough to make this bond, because each atom has one $2 p_{z}$ $2p_z$ atomic orbital.
The first $π$ $pi$ bond is made via a $C_{2 p_{x}} - O_{2 p_{x}}$ $"C"_(2p_x)-"O"_(2p_x)$ sidelong overlap. That is enough to make this bond, because each atom has one $2 p_{x}$ $2p_x$ atomic orbital.
The second $π$ $pi$ bond is made via a $C_{2 p_{y}} - O_{2 p_{y}}$ $"C"_(2p_y)-"O"_(2p_y)$ sidelong overlap. That is enough to make this bond, because each atom has one $2 p_{y}$ $2p_y$ atomic orbital.

Finally, the nonbonding electron pair on $C$ $"C"$ and that on $O$ $"O"$ (otherwise known as "lone pairs") are stored in nonbonding molecular orbitals, which were nonbonding because they:

could not overlap to form bonds because they weren't compatible with any other orbitals (incorrect symmetry).
did not hybridize (because they didn't need to)

Lastly, if it were ${CO}_{2}$ $"CO"_2$ , then yes, there would be hybridization, but on $^{-} C \equiv O :^{+}$ $""^(-)"C"-="O":^(+)$ , no. The molecule has to have more than two atoms to require hybridization.

( $: . . O = C = . . O :$ $:stackrel(..)"O"="C"=stackrel(..)"O":$ has $s p$ $sp$ hybridization on carbon and $s p^{2}$ $sp^2$ hybridization on oxygen.)

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