Bonding orbitals in diatomic oxygen

Answer 1

I wanted explain it using an MO diagram but it's a little difficult to write in this format. I'll just go through with how to construct it and the results. Firstly, note that the atomic orbitals of the Oxygen elements are of the same energy levels so we can draw them parallel with respect to the diagram: Orbitals

O O We can then draw in the electron configuration with respect to each individual atom. It follows that: Orbitals

2p _|_|_

2s _

1s _

Orbitals

2p +-|+-|_

2s +-

1s +- *where +/- indicates sign of coefficient of electron spin number* Now that the two atoms are interacting, the probability regions defined by Schrodinger's wave equation are altered and new "molecular" solutions/orbitals are defined. Because of the nature and behavior of waves, and by application of the photoelectric effect, different interactions lead to different consequences. If a wave is in phase with another wave they result in an increased amplitude: this gives electrons the most ideal energy position and may result in a decrease of energy over atomic configurations.

However, it also means that there will be times of destructive interference as opposed waves collide. This collision leads to nodes of complete interference and oppose the bonding of the elements. Such regions are known as anti-bonding orbitals. Now we can draw in our molecular orbitals and fill them according to the aufbau principle, where electrons will fill according to the lowest possible energy state. Also note that the 2pσ bond in Oxygen is the lowest in energy as it is larger than nitrogen.Anti-Bonding Orbitals denoted by * 2P :-

2pσ* 2pπ* 2pπ 2pσ

2S :- 2sσ* 2sσ

1S : -

1sσ* 1sσ

From 1sσ, the energy increases in each molecular orbital. Using both Oxygen atom's electrons we fill the orbitals. The diagram will have these results:

2P :-

2pσ* 2pπ* + +

2pπ +- +-

2pσ + -

2S :- 2sσ* +-

2sσ +-

1S : -

1sσ* +-

1sσ + -

The idea is that the electrons left 2 half filled degenerate orbitals. This makes Oxygen a diradical. If we now calculate the bond order to see which state the elements favor. Bond order is the half the difference of bonding electrons to anti-bonding electrons. Here we have (10 bonding - 6 anti-bonding)/2. 4/2 = 2. Therefore:

By molecular orbital Theory, the gas O2 is energetically favored to the elemental Oxygen and is of Bond order 2. It will have a double bond.

Answer 2

The bond order of oxygen is 2 and molular orbital theory explain the paramagnetic character of oxygen by loosing two electrons aparamagnetic vanished and by adding two electrons paramagnetic vanished. it will be present only whn two unpaired electron anti bonding molecular orbitals

Answer 3

In diatomic oxygen, O2, there are two bonding orbitals: the sigma bonding orbital (σ2p) and the pi bonding orbital (π2p). These orbitals result from the overlap of the 2p atomic orbitals from each oxygen atom. The sigma bonding orbital is formed by head-on overlap of the p orbitals, while the pi bonding orbital is formed by sideways overlap.

Answer 4

The bonding orbitals in diatomic oxygen can easily be described. There is a sigma bond due to the overlap of two S orbitals. There is a PI bond due to the overlap of two PI bonds.

Answer 5

there is a sigma bond due to the overlap of 2 s orbitals and there is a pi bond due to the overlap of two pi bonds

Answer 6

1s [↓↑] 2s [↓↑] 2p [↓↑][↑][↑]

Answer 7

The molecule is neutral.

Answer 8

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