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Report (any type)/Brief report
22/09/2021
Ammonia (##”NH”_3)##, or, more accurately, the central atom in ammonia, is ##”sp”^3## . Here’s how you’d go about determining this.
First, start with ##”NH”_3##’s , which must account for 8 – 5 from nitrogen and 1 from each hydrogen atom.
As you can see, all the are indeed accounted for – 2 for each between nitrogen and hydrogen, and 2 from the lone pair present on the nitrogen atom.
Now, here is where it gets interesting. Nitrogen’s energy levels look like this
Looking at this energy diagram, one could see that each of the three p-orbitals is available for , so why would the atom need to be hybridized? Here’s where stability and geometry come into play.
If the three hydrogen atoms would bond with nitrogen using the available p-orbitals, the bond angles would be ##”90″^@##. However, this cannot take place since electron-rich regions must be located as far away from each other as possible in ##”3D”## space – this is why the bond angle in ammonia is approximately ##”107″^@##.
Moreover, the hybrid orbitals would ensure the formation of a stronger bond with the hydrogen atoms, since hybrid orbitals formed from s and p-orbitals have a greater electron on one side of the lobe – the side that bonds with the hydrogen atom.
Molecule stability also come into play, since the ##”sp”^3## hybrid orbitals will be lower in energy than the three unhybridized p-orbitals.
So, in order to determine hybridization, you must determine the central atom’s steric number, which represents the number of electron-rich regions around the atom.
Since it forms 3 covalent bonds and has 1 lone pair, nitrogen’s steric number will be equal to 4, which implies that one s and three p-orbitals will combine for a total of 4 hybridized orbitals.
Here’s a video on this subject:
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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