# On average, how many #pi# bonds does #"O"_2^(+)# have?

##### 2 Answers

As mo theory there should be 1.5 pi bond

#### Explanation:

MO configuration of

You would have a bond order of *average*.

**BOND ORDER OF HOMONUCLEAR DIATOMICS**

**Bond order** is a measure of bond strength, and suggests stability. It is *half the number of bonding minus the number of antibonding molecular orbitals*.

#"Bond Order" = ("Bonding e"^(-) - "Antibonding e"^(-))/2#

You can have *optimal* overlap, *less than optimal* overlap, or *no* overlap. ** Poorer** bonding overlaps correspond with

**values of bond order. Or, less antibonding overlap corresponds with higher values of bond order (which is what applies here).**

*lower*The structures of

**DIATOMIC OXYGEN IS PARAMAGNETIC**

While **Valence Bond Theory** suggests oxygen is diamagnetic, **Molecular Orbital Theory** correctly demonstrates that oxygen, *paramagnetic*.

That means it has unpaired electrons. Specifically, two unpaired electrons, one in each **antibonding** orbital (

The **MO diagram** for *neutral*

When you take away one electron, you take it away from the **highest-occupied molecular orbital**. Since either the **lose** an electron when we wish to form

**DETERMINING BOND ORDER**

Naturally,

Two bonding electrons each come from the

#(10 - color(red)(6))/2 = color(blue)(2)#

When taking away one *antibonding* electron to form

#(10 - color(red)(5))/2 = color(blue)(2.5)#

Since *one of two* antibonding *less weak* by *half*. So, instead of going from **bonds**.