Are there any exceptions to this trend observed?

TL;DR: Does the observed trend in bond orders for different diatomic electronic species, based on molecular orbital theory, have any exceptions?

The trend being observed in the bond orders for different diatomic electronic species is related to the molecular orbital theory and the filling of molecular orbitals in diatomic molecules.

Bond Order (BO)

Bond order is calculated as:

$$ \text{BO} = \frac{{N_B} - {N_A}}{2} $$

where $N_B$ is the number of electrons in bonding orbitals, and $N_A$ is the number of electrons in antibonding orbitals.

For a molecule with $n$ electrons, you can arrange the electrons in molecular orbitals according to the increasing energy levels. These orbitals are filled following the Pauli exclusion principle and Hund's rule.

Molecular Orbitals for Homonuclear Diatomic Molecules

For diatomic molecules formed by atoms from the second period (like ( $\text{N}_2 $) or ( $\text{O}_2$ )), the order of molecular orbitals from lowest to highest energy is generally:

$$ \sigma(1s) < \sigma^*(1s) < \sigma(2s) < \sigma^*(2s) < \left(\pi(2p_x) = \pi(2p_y)\right) < \sigma(2p_z) < \left(\pi^*(2p_x) = \pi^*(2p_y)\right) < \sigma^*(2p_z) $$

However, for species with a different number of electrons, the pattern can be understood by considering how electrons fill these orbitals.

THE TREND for ($2 \times \text{BO}$):

$$ 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 0, ...$$

($n^{th}$ term of this trend is twice the BO of n-electronic diatomic species)

Notice that the ($1s$) orbital can hold 2 electrons. Therefore, the sequence increases from 1 to 2 and then decreases to 0. This pattern repeats with the ($2s$) orbital. For the ($2p$) orbital, which can accommodate 6 electrons, the sequence increases from 1 to 6 before decreasing back to 0, and this trend continues similarly for other orbitals.

These numbers correspond to:

$$2 \times \text{BO} $$

So, the bond order (BO) values are:

$$ \frac{1}{2}, 1, \frac{1}{2}, 0, \frac{1}{2}, 1, \frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{5}{2}, 2, \frac{3}{2}, 1, \frac{1}{2}, 0,...$$

Explanation of the Pattern

The trend in bond orders can be understood by looking at how molecular orbitals are filled and the resulting net bonding or antibonding nature of these orbitals. Here's a detailed breakdown:

  1. 1s Orbital Filling:

    • 1 electron: ($\frac{1}{2}$) (one electron in bonding orbital)
    • 2 electrons: 1 (both in bonding orbital)
    • 3 electrons: ($\frac{1}{2}$) (two in bonding, one in antibonding)
    • 4 electrons: 0 (both bonding and antibonding filled)
  2. 2s Orbital Filling:

    • Same trend repeats as above for the 2s orbital.
  3. 2p Orbital Filling:

    • Electrons start filling the ($\pi$) orbitals before the ($\sigma$) orbital.
    • Up to 6 electrons increase the bond order as more bonding orbitals are filled.
    • After 6 electrons, electrons start filling antibonding orbitals, decreasing the bond order.

Why Does This Work?

This pattern works due to the specific way electrons fill molecular orbitals. As you move from species with fewer electrons to those with more:

  • Initially, electrons fill lower-energy bonding molecular orbitals.
  • This increases the bond order up to a maximum.
  • Once bonding orbitals are filled, electrons start occupying antibonding orbitals.
  • Filling of antibonding orbitals reduces the bond order.

This is consistent with molecular orbital theory and explains the observed sequence of bond orders.

The Question

Are there any exceptions to this trend?

  • $\begingroup$ What's wrong with the question? Is it irrelevant? What could be improved? @TheOneWhoDownvoted? $\endgroup$ Commented May 24 at 13:32
  • $\begingroup$ You're gonna get more downvotes if you won't clearly say what this "Specific Trend" is. I think you're talking about second row elements, but it's far from clearly stated. $\endgroup$
    – Mithoron
    Commented May 24 at 13:45
  • $\begingroup$ The trend of 2*BO is explained in my question for a diatomic species having n electrons. $\endgroup$ Commented May 24 at 13:50
  • 3
    $\begingroup$ 1) You're talking only about the lightest ones. 2) What you think is the reason for lower BO orders later on is not true. 3) It all breaks completely for transition metals. $\endgroup$
    – Mithoron
    Commented May 24 at 14:08
  • $\begingroup$ @Mithoron, so upto what number of electrons does it work? $\endgroup$ Commented May 24 at 16:41


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.