What orbitals form sigma bonds?

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What orbitals form sigma bonds?

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Answer 1 · 2016-02-27T00:36:04

$σ$ $sigma$ bonds must be made by orbitals that overlap head-on.

POSSIBLE ORBITAL COMBINATIONS TO GENERATE SIGMA MOLECULAR ORBITALS

For simplicity, if we examine only the $s$ $s$ , $p$ $p$ , and $d$ $d$ orbitals, let's suppose that all orbitals we are examining are similar enough in energy to interact.

Let's also suppose that we are ignoring $d$ $d$ - $d$ $d$ interactions, since we know those should work (they are the same $l$ $l$ so it's not so interesting).

Then, orbitals that can capably overlap with each other to form $σ$ $sigma$ bonds include the following linear combinations:

$s + p_{z} \to σ$ $s + p_z -> sigma$ (bonding)
$s - p_{z} \to σ^{*}$ $s - p_z -> sigma^"*"$ (antibonding)
$s + d_{z^{2}}^{2} \to σ$ $s + d_(z^2) -> sigma$ (bonding)
$s - d_{z^{2}}^{2} \to σ^{*}$ $s - d_(z^2) -> sigma^"*"$ (antibonding)
$s + d_{x^{2} - y^{2}}^{2} \to σ$ $s + d_(x^2-y^2) -> sigma$ (bonding)
$s - d_{x^{2} - y^{2}}^{2} \to σ^{*}$ $s - d_(x^2-y^2) -> sigma^"*"$ (antibonding)
$p_{z} + d_{z^{2}}^{2} \to σ$ $p_z + d_(z^2) -> sigma$ (bonding)
$p_{z} - d_{z^{2}}^{2} \to σ^{*}$ $p_z - d_(z^2) -> sigma^"*"$ (antibonding)
$p_{x} + d_{x^{2} - y^{2}}^{2} \to σ$ $p_x + d_(x^2 - y^2) -> sigma$ (bonding)
$p_{x} - d_{x^{2} - y^{2}}^{2} \to σ^{*}$ $p_x - d_(x^2 - y^2) -> sigma^"*"$ (antibonding)
$p_{y} + d_{x^{2} - y^{2}}^{2} \to σ$ $p_y + d_(x^2 - y^2) -> sigma$ (bonding)
$p_{y} - d_{x^{2} - y^{2}}^{2} \to σ^{*}$ $p_y - d_(x^2 - y^2) -> sigma^"*"$ (antibonding)

where the $z$ $z$ axis is the internuclear axis (i.e. the axis along which the single bond---which is also a $σ$ $sigma$ bond---is made), and the $x$ $x$ and $y$ $y$ axes are where you should expect them to be for the Cartesian coordinate system.

We would also suppose that the antibonding molecular orbitals are unoccupied so that the bond is a standard single bond.

HOW TO DEPICT/IMAGINE THESE ORBITAL OVERLAPS

When you sketch these orbital overlaps:

All $s$ $s$ orbitals are spheres. The only way these can change sign is if the whole thing changes sign.
The $p_{z}$ $p_z$ orbitals can be approximated as dumbbells, regardless of their $n$ $n$ , without losing the essence of the $σ$ $sigma$ MOs generated (head-on overlap). One lobe is the opposite sign to the other.
The $d_{z^{2}}^{2}$ $d_(z^2)$ look almost like $p_{z}$ $p_z$ orbitals, except there is a donut in the middle. You can also approximate these as dumbbells, regardless of their $n$ $n$ , without losing the essence of the $σ$ $sigma$ MOs generated (head-on overlap). Both lobes are the same sign.
The $d_{x^{2} - y^{2}}^{2}$ $d_(x^2-y^2)$ can be approximated as four-leaf clovers, essentially, on the $x y$ $xy$ -plane, with the lobes aligned along the $x$ $x$ and $y$ $y$ axes. The opposite lobes along each axis are the same sign. Therefore, they overlap with the $p_{x}$ $p_x$ and $p_{y}$ $p_y$ , which also lie long those axes.

Since they are all aligned along the same axis ( $p_{z}$ $p_z$ with $d_{z^{2}}^{2}$ $d_(z^2)$ , $p_{x}$ $p_x$ with $d_{x^{2} - y^{2}}^{2}$ $d_(x^2-y^2)$ , and $p_{y}$ $p_y$ with $d_{x^{2} - y^{2}}^{2}$ $d_(x^2-y^2)$ ) AND they are compatible ( $s$ $s$ with $p_{z}$ $p_z$ , $s$ $s$ with $d_{z^{2}}^{2}$ $d_(z^2)$ , and $s$ $s$ with $d_{x^{2} - y^{2}}^{2}$ $d_(x^2-y^2)$ ), they form $σ$ $sigma$ bonding and $σ^{*}$ $sigma^"*"$ antibonding MOs. Since we supposed that only the $σ$ $sigma$ bonding MO is occupied, we have a single bond.

(Since $s$ $s$ orbitals are spheres, it doesn't matter along which axis they bond.)

Of course, there exist $f$ $f$ orbitals of some sort that can overlap with $s$ $s$ , $p_{z}$ $p_z$ , $d_{z^{2}}^{2}$ $d_(z^2)$ , and $d_{x^{2} - y^{2}}^{2}$ $d_(x^2-y^2)$ in a $σ$ $sigma$ fashion, but that's left up to the really motivated chemist to figure out.

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What orbitals form sigma bonds?

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