Why can't Sp3 hybrid orbitals can form pi bonds?

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Why can't Sp3 hybrid orbitals can form pi bonds?

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Answer 1 · 2016-02-23T10:20:19

A single bond consists of a $σ$ $sigma$ bond, which is defined as the head-on overlap between two compatible orbitals.

Because hybrid orbitals are made to match the symmetry of the incoming (i.e. not-yet-bonded) atom's atomic orbital for the sake of being able to bond at all (which is usually if not always "totally symmetric about the internuclear axis"), they always have to overlap head-on.

Thus, a hybrid orbital has to make a $σ$ $sigma$ bond.

Not just $s p^{3}$ $sp^3$ , but any hybrid orbital. Even in a triple bond, like in acetylene ( $H - C \equiv C - H$ $"H"-"C"-="C"-"H"$ ), the $π$ $pi$ bonds are made by the $p_{x}$ $p_x$ and $p_{y}$ $p_y$ orbitals (or any qualified equivalent sidelong orbital overlap), while the $σ$ $sigma$ bonds are made with the hybrid orbitals, which consist of only the $p_{z}$ $p_z$ and $s$ $s$ orbitals.

![http://2012books.lardbucket.org/]( useruploads.socratic.org )

Here, the $z$ $z$ axis is defined as the internuclear axis, whereas the $x$ $x$ and $y$ $y$ axes are vertical and towards us, respectively.

As an alternative, consider ethene ( $H_{2} C = {CH}_{2}$ $"H"_2"C"="CH"_2$ ), where the double bonds lie on the $x y$ $xy$ -plane in the plane of the screen/paper, and the $z$ $z$ axis protrudes outward towards us.

If we consider the $s p^{2}$ $sp^2$ hybridization of carbon in ethene, carbon actually uses three $s p^{2}$ $sp^2$ hybrid orbitals: one each to $σ$ $sigma$ bond with the hydrogens, and one to $σ$ $sigma$ bond with the other carbon. (You would get that if you drew the molecule with all single bonds and no double bonds.)

Organic Chemistry, Bruice

It made each hybrid orbital by mixing the $2 s$ $2s$ , $2 p_{x}$ $2p_x$ , and $2 p_{y}$ $2p_y$ orbitals together and distributing their angular positions evenly on the $x y$ $xy$ -plane since the original $2 p$ $2p$ orbitals used to construct them were actually aligned in the $x$ $x$ and $y$ $y$ directions, respectively. (The notation $s p^{2}$ $sp^2$ comes from mixing one $s$ $s$ and two $p$ $p$ orbitals together to get $33 %$ $33%$ $s$ $s$ character and $66 %$ $66%$ $p$ $p$ character.)

Hence, the $\mathbf(sigma)$ bonds of ethene (shown in yellow) lie on a single plane.

In addition, we have the $\mathbf(2p_z)$ atomic orbital (perpendicular to the plane of the molecule, in purple) from each carbon that is used to $\mathbf(pi)$ bond (sidelong overlap) with the other carbon, thus constructing the second component of the double bond (one $sigma$ + one $pi$ bond = one double bond).

It was not used in the orbital hybridization, and it remains as a different, incompatible orbital (with respect to the $\mathbf(2p_x)$ and $\mathbf(2p_y)$ ) for $\mathbf(sigma)$ bonding within the molecule. The only thing it can do at this point is $pi$ bond because it is oriented precisely to do so.

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Why can't Sp3 hybrid orbitals can form pi bonds?

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