# How do you write the electron configuration for #"H"#, #"H"_2#, #"H"_2^-#, and #"H"_2^(2-)# and calculate their bond order?

##### 1 Answer

Hydrogen **atom**'s valence orbitals, ** before** bonding, include every orbital, and all are the same energy for a specific

**one**electron.

**HAVING MORE THAN ONE ELECTRON SPLITS ENERGY LEVELS**

When introducing **more electrons** into the system, i.e. with another hydrogen wanting to bond, the repulsion *splits* the energy levels of the AOs of hydrogen.

They start out where each

**MOLECULAR ORBITAL THEORY**

*The basic tenant of Molecular Orbital Theory (MO Theory) is that the number of MOs formed by a linear combination of atomic orbitals (LCAO) is equal to the number of AOs used.*

The energy splitting caused by *electron/electron repulsion* generates **two MOs** due to the one **per hydrogen** that is bonding. They are called the **bonding** **antibonding**

The

**MO DIAGRAM FOR DIATOMIC HYDROGEN MOLECULE & ION**

Now, let us draw the MO diagram for the **neutral molecule**.

Each hydrogen contributes one electron, which therefore fills the lower-in-energy

Add one electron, and you will get

**MO ELECTRON CONFIGURATIONS**

Writing the **MO electron configuration** is fairly straightforward; read off of the diagram and put superscripts for how many electrons there are, **analogous** to the AO electron configuration.

*The only difference is the symbol used to indicate the MO instead of the AO and some parentheses.*

- For
#"H"# **atom**, it would just be#1s^1# . - For
#"H"_2# **molecule**, it would be#(sigma_(1s))^2# . - For
#"H"_2^(-)# **ion**, it would be#(sigma_(1s))^2 (sigma_(1s)^"*")^1# . - For
#"H"_2^(2-)# **ion**, it would be#(sigma_(1s))^2 (sigma_(1s)^"*")^2# .

**BOND ORDER DIRECTLY RELATES TO BOND STRENGTH**

The **Bond Order**, then, is the number of bonding electrons minus the number of antibonding electrons, the total divided by

For

#color(blue)("Bond Order") = ("No. of Bonding e"^(-) - "No. of Antibonding e"^(-))/2#

#= (2 - 0)/2 = color(blue)("1")# which makes sense because

#"H"_2# has one single bond.

For

#color(blue)("Bond Order") = ("No. of Bonding e"^(-) - "No. of Antibonding e"^(-))/2#

#= (2 - 1)/2 = color(blue)("1/2")#

The contribution of one electron in the *antibonding* orbital ** decreases** the bond order, meaning that the bond in

**than the bond in**

*weaker*If you're following the pattern, you should draw the MO diagram for

#color(blue)("Bond Order") = ("No. of Bonding e"^(-) - "No. of Antibonding e"^(-))/2#

#= (2 - 2)/2 = color(blue)("0")#

The contribution of one electron in the *antibonding* orbital ** decreases** the bond order, meaning that the bond in

**than the bond in**

*even weaker*So weak, in fact, that since the bond order is