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

Answer 1

The ground state of oxygen molecule is the triplet state, which has two unpaired valence electrons in its #pi^"*"# antibonding molecular orbitals:

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 #bb2# (why?). And this can be verified:

#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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Answer 2

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.

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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