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

The `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` and `s"/"p_z` combinations, where let's say the `z` axis is the horizontal, internuclear axis.

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

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

The lewis structure for `"CO"` is `""^((-)):"C"-="O":^((+))`

The MO diagram is:

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

• `sigma_(2s)` bonding MO, from the `2s(C)"/"2s(O)` head-on combination
• `sigma_(2s)^"*"` antibonding MO, from the `2s(C)"/"2s(O)` head-on combination
• `pi_(2p_x)` and `pi_(2p_y)` bonding MOs, from the `2p_x(C)"/"2p_x(O)` and `2p_y(C)"/"2p_y(O)` sidelong combination
• `sigma_(2p_z)` bonding MO, from the `2p_z(C)"/"2p_z(O)` head-on combination. This is the HOMO.
• `pi_(2p_x)^"*"` and `pi_(2p_y)^"*"` antibonding MOs (empty), from the `2p_x(C)"/"2p_x(O)` and `2p_y(C)"/"2p_y(O)` sidelong combination. This is the LUMO.
• `sigma_(2p_z)^"*"` antibonding MO (empty), from the `2p_z(C)"/"2p_z(O)` head-on combination

The `sigma_(2s)` and `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 `sigma_(2p_z)` contributes two bonding electrons, and the `pi_(2px)` and `pi_(2p_y)` contribute two bonding electrons each. That is 6 bonding.

That gives a bond order of `([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.