# What is the MO diagram of #"O"_2# in its ground state, and how do I calculate the bond order?

The ground state of oxygen molecule is the triplet state, which has two unpaired valence electrons in its

From this, you should be able to write the electronic configuration by inspection:

#overbrace(color(blue)((sigma_(1s))^2 (sigma_(1s)^"*")^2))^"deep core orbitals " overbrace(color(blue)((sigma_(2s))^2 (sigma_(2s)^"*")^2))^"outer core orbitals" underbrace(color(blue)((sigma_(2p_z))^2 (pi_(2p_x))^2(pi_(2p_y))^2 (pi_(2p_x)^"*")^1 (pi_(2p_y)^"*")^1))_"valence orbitals"#

*It's exactly how you would approach the atomic configuration, with changed notation. Instead of notating atomic orbitals, we notate molecular orbitals.
*

*As you (should) have been taught, #sigma# orbitals are made from head-on overlap, and #pi# orbitals are made from side-long overlap.*

By conservation of orbitals, there must be the same number of molecular orbitals as atomic orbitals.

Its bond order is intuitive. It has a double bond, so its bond order is anticipated to be

#color(blue)("Bond Order") = 1/2 ("Bonding e"^(-) - "Antibonding e"^(-))#

#= 1/2 [underbrace(overbrace(2)^(sigma_(1s))+overbrace(2)^(sigma_(2s))+overbrace(2)^(sigma_(2p_z))+overbrace((2 xx 2))^(pi_(2p_(x//y))))_("Bonding e"^(-)'s) - (underbrace(overbrace(2)^(sigma_(1s)^"*")+overbrace(2)^(sigma_(2s)^"*")+overbrace((2 xx 1))^(pi_(2p_(x//y)^"*")))_("Antibonding e"^(-)'s))]#

#= color(blue)(2)#

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The MO diagram of ( \text{O}_2 ) in its ground state consists of two electrons each occupying the ( \sigma_1s ), ( \sigma_1s^* ), ( \sigma_2s ), and ( \sigma_2s^* ) molecular orbitals.

To calculate the bond order, use the formula:

[ \text{Bond Order} = \frac{1}{2} \times (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) ]

For ( \text{O}_2 ), the number of bonding electrons is 8 and the number of antibonding electrons is 4.

[ \text{Bond Order} = \frac{1}{2} \times (8 - 4) ] [ \text{Bond Order} = \frac{1}{2} \times 4 ] [ \text{Bond Order} = 2 ]

Therefore, the bond order of ( \text{O}_2 ) in its ground state is 2.

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*Answer from HIX Tutor*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

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