Molecular Orbital Diagram of O2, F2, and Ne2 Molecules.

Описание к видео Molecular Orbital Diagram of O2, F2, and Ne2 Molecules.

0:21 Molecular Orbital Diagram of Oxygen Molecule
3:30 Molecular Orbital Diagram of Florine Molecule
5:25 Molecular Orbital Diagram of Neon Molecule

So as we discussed in our previous video how to create molecular orbital diagrams of simple diatomic molecules. Now lets take molecules with p orbitals involved. So first example is Oxygen, here oxygen atom have 8 electrons with electronic configuration 1s2, 2s2, and 2p4. These electrons are filled as per the afbaou principle in their respective atomic orbitals.

For formation of oxygen molecule, two oxygen atoms come close and their atomic orbitals overlap to make molecular orbitals. In this case 1s orbital is inner sell orbital. So they are very far from each other and overlap is not possible. This generates nonbonding situation. 2s orbitals overlap to create sigma 2s and sigma star 2s orbitals. 2pz orbitals are along the axis or overlap so they combine to give sigma 2pz and sigma star 2pz orbitals. While 2px and 2py orbitals overlap is lateral to the axis of overlap so they form pi bonds as pi 2p and pi star 2p orbitals. As the energy of 2px and 2py are same they are represented on the same level.

Now oxygen molecule have total 16 electrons out of these only 12 electrons have to be filled in molecular orbitals. So filling of electrons takes place as per afbaus principle. Incase of pi 2p orbital, first single electrons are placed and afterwords pairing takes place.
The electronic configurations of oxygen molecule will be KK…
So from the molecular orbital diagram there are two unpaired electrons thus the molecule is paramagnetic in nature. For calculation of bond order we use the formula Bond order = no of electrons in bonding - no of electrons in anti bonding mo divide by 2. In this case there are total 8 electron in bonding and only 4 electrons in anti bonding, so bond order becomes 2. Therefore two bonds are formed between oxygen atoms.
Similarly we can also construct molecular orbital diagram for Florine molecule. Florine atom contains 9 electrons so electronic configuration is 1s2, 2s2 and 2p5. This means in fluorine molecule we will have total 18 electrons but 4 electrons from 1s orbital will not be involved as this is non bonding situation, remaining 14 electrons can be accommodated in molecular orbitals. Here also 2s orbitals of both Florine atoms overlap to give sigma 2s and sigma star 2s orbitals. Similarly 2pz orbitals being at the axis of overlap produce sigma 2pz and sigma star 2pz orbitals, while 2px and 2py orbitals being lateral to overlap axis produce pi and pi star orbitals.

Filling of electrons takes place from lower energy orbitals as from sigma 2s, then sigma star 2s then sigma 2pz then pi 2p orbitals and finally in pi star 2p orbitals
So as we see from the diagram as all the electrons are paired up the molecule should be diamagnetic in nature and the bond order can be calculated from the same formula where number of bonding electrons are 8 and number of anti bonding electrons are 6. So bond order becomes 1. Therefore single bond is formed between two Florine atoms.
Now lets try to find how many bonds are formed between two Neon atom to make a Neon molecule. In this case Neon atom is having 10 elections with electronic configuration as 1s2 2s2 and 2p6. So in diatomic neon molecule there must be 20 electrons, but 4 elections from 1s orbitals will not be involved in bonding as they are in nonbonding situation.

So lets fill remaining 16 electrons in molecular orbitals, from lower energy to higher energy as sigma 2s, sigma star 2s, sigma 2pz, pi 2px and 2py then pi star 2px and 2py and finally sigma star 2pz.

Here as we can see all electrons are again paired up, the molecule should be diamagnetic, and if we calculate the bond order it comes as zero. This means zero bonds are formed between two neon atoms. Or no bond is formed, thats why diatomic Neon molecule does not exist.

So thats all about these molecules, in next lecture we will study the formation of molecule in Boron, carbon and nitrogen.

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