The electron configuration of a particular diatomic species is:
[core electrons]( $\sigma$ _2s) ²( $\sigma$ *_2s) ²( $\sigma$ _2p) ²( $\pi$ _2p) ²( $\pi$ *_2p) ⁰
What is the bond order for this species?

Question

The electron configuration of a particular diatomic species is:
[core electrons](

\sigma

_2s) ²(

\sigma

*_2s) ²(

\sigma

_2p) ²(

\pi

_2p) ²(

\pi

*_2p) ⁰
What is the bond order for this species?

Quizplus · Accepted Answer

To determine the bond order, we need to count the total number of bonding electrons and divide by the number of bonding pairs. The bonding electrons are the ones in the molecular orbitals, which are shown in the configuration as (σ and π) orbitals.

Counting the electrons in these bonding orbitals, we find:

Bonding electrons = 2 + 2 + 2 + 2 + 2 = 10

Next, we need to count the number of bonding pairs. To do this, we need to look at the Lewis structure for this species. From the configuration, we can see that there are two unpaired electrons in the π*4p orbital of each fluorine atom. These two electrons can form one σ bond with the unpaired electrons of the other fluorine atom, resulting in a double bond. Thus, the Lewis structure for this species is:

F≡F

There is only one bonding pair, so the bond order is:

Bond order = 10/1 = 2

Therefore, the best choice is C) 2 because this is the bond order that corresponds to the given electron configuration.

The Electron Configuration of a Particular Diatomic Species Is: [Core σ\sigmaσ

The Electron Configuration of a Particular Diatomic Species Is:
[Core $\sigma$