How many sigma and pi bonds are generally part of a triple bond?

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How many sigma and pi bonds are generally part of a triple bond?

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Answer 1 · 2016-02-19T19:19:52

The $σ$ $sigma$ bond is made by a head-on overlap between two compatible atomic orbitals that are symmetric about the internuclear axis. Two common examples are $s / s$ $s"/"s$ and $s / p_{z}$ $s"/"p_z$ combinations, where let's say the $z$ $z$ axis is the horizontal, internuclear axis.

The $π$ $pi$ bonds are made by the sidelong overlap between two compatible atomic orbitals, which are specifically the $p_{x} / p_{x}$ $p_x"/"p_x$ and $p_{y} / p_{y}$ $p_y"/"p_y$ combinations. Let's say that $x$ $x$ is vertical and $y$ $y$ is towards or away from you.

Let us consider a diatomic molecule for simplicity. How about $CO$ $"CO"$ ?

The lewis structure for $CO$ $"CO"$ is $^{(-)} : C \equiv O :^{(+)}$ $""^((-)):"C"-="O":^((+))$

The MO diagram is:

Inorganic Chemistry, Miessler et al. Ch. 5.3, Figure 5.13

From bottom to top, where again, $z$ $z$ is horizontal, $x$ $x$ is vertical, and $y$ $y$ is in or out of the screen/paper, we have:

$σ_{2 s}$ $sigma_(2s)$ bonding MO, from the $2 s (C) / 2 s (O)$ $2s(C)"/"2s(O)$ head-on combination
$σ_{2 s}^{*}$ $sigma_(2s)^"*"$ antibonding MO, from the $2 s (C) / 2 s (O)$ $2s(C)"/"2s(O)$ head-on combination
$π_{2 p_{x}}$ $pi_(2p_x)$ and $π_{2 p_{y}}$ $pi_(2p_y)$ bonding MOs, from the $2 p_{x} (C) / 2 p_{x} (O)$ $2p_x(C)"/"2p_x(O)$ and $2 p_{y} (C) / 2 p_{y} (O)$ $2p_y(C)"/"2p_y(O)$ sidelong combination
$σ_{2 p_{z}}$ $sigma_(2p_z)$ bonding MO, from the $2 p_{z} (C) / 2 p_{z} (O)$ $2p_z(C)"/"2p_z(O)$ head-on combination. This is the HOMO.
$π_{2 p_{x}}^{*}$ $pi_(2p_x)^"*"$ and $π_{2 p_{y}}^{*}$ $pi_(2p_y)^"*"$ antibonding MOs (empty), from the $2 p_{x} (C) / 2 p_{x} (O)$ $2p_x(C)"/"2p_x(O)$ and $2 p_{y} (C) / 2 p_{y} (O)$ $2p_y(C)"/"2p_y(O)$ sidelong combination. This is the LUMO.
$σ_{2 p_{z}}^{*}$ $sigma_(2p_z)^"*"$ antibonding MO (empty), from the $2 p_{z} (C) / 2 p_{z} (O)$ $2p_z(C)"/"2p_z(O)$ head-on combination

The $σ_{2 s}$ $sigma_(2s)$ and $σ_{2 s}^{*}$ $sigma_(2s)^"*"$ cancel out and contribute zero to the bond order because they contribute two bonding, but also two antibonding electrons. That is 2 bonding and 2 antibonding.

The $σ_{2 p_{z}}$ $sigma_(2p_z)$ contributes two bonding electrons, and the $π_{2 p x}$ $pi_(2px)$ and $π_{2 p_{y}}$ $pi_(2p_y)$ contribute two bonding electrons each. That is 6 bonding.

That gives a bond order of $\frac{[2 + 6] - 2}{2} = 3$ $([2+6] - 2)/2 = 3$ , as expected.

For a diatomic molecule whose bond order is $3$ , we can conclude that there is one $sigma$ bond and two $pi$ bonds.

CHALLENGE: If $NO^(+)$ ion is isoelectronic with $CO$ , what about neutral $NO$ ; what is the bond order on that, and from that, how many $pi$ bonds on average does it have? Hint: It still has one $sigma$ bond.

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How many sigma and pi bonds are generally part of a triple bond?

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