Understanding the Hybridization and Formation of Pi Bonds in Ammonia and Ethyne

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In ammonia (NH3), the nitrogen atom is sp3 hybridized due to the presence of four electron pairs, which includes one lone pair in addition to the three bonding pairs. This hybridization allows for a tetrahedral geometry, accommodating all electron pairs effectively. In ethyne (H-C≡C-H), the formation of the triple bond involves one sigma bond and two pi bonds. The sigma bond is formed by the overlap of one 2s electron from each carbon atom, while the pi bonds arise from the overlap of the unhybridized 2px and 2py orbitals. The 2pz orbital remains unhybridized and cannot participate in pi bonding due to its orientation along the internuclear axis, which would interfere with the sigma bond. The discussion clarifies that while orbitals in the same subshell are of equal energy in isolated atoms, hybridization alters their energy levels when forming molecules, leading to the observed bonding characteristics in NH3 and ethyne.
Clari
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In ammonia, why the N atom is sp3 hybridized and not sp2 hybridized...since there are only 3 bondings...

Also, in ethyne, that is H-C(triple bond)C-H...for the formation of 2 pi bonds, why it is the unhybridized 2px orbital and 2py orbital of each C atom overlap with each other? How about the 2pz orbital?


any help would be appreciated
 
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In NH3, there are 4 electron pairs about the N atom - you're omitting the lone pair.

The triple bond in H-C \equiv C-H is basically a sigma bond using one of the two 2s electrons (the other 2s electron is used for the sigma bond with H), and a pair of pi bonds using both of the 2p electrons. Since C has only 2 electrons in the 2p subshell, and since px and py are lower energy orbitals than pz, the pi bonds must involve an overlap of only these orbitals.
 
Gokul43201:

but in my textbook, it is said that orbitals in the same subshell is at the same energy level...I am confused now...
 
Clari said:
Gokul43201:

but in my textbook, it is said that orbitals in the same subshell is at the same energy level...I am confused now...

For isolated Carbon atoms they do:2p_{x},2p_{y} & 2p_{z} have all the same energy.But when the Carbon atom reacts with nonmetalic elements,like Hydrogen,those orbitals do not behave in the same manner.That's because 2 of them have one electron and the remaining one (2p_{z}) has none.The 2 orbitals with one electron they "hybridize" with the 2s orbital which has 2 electrons,resulting in 4 hybridized orbitals of smaller energy than the unhybridized 2p_{z}.Since it overlaps 3 hybrid orbitals (one \sigma and 2 \pi) with another C atom,it has a triple bond.Since 2 of the bonds are \pi,and another 2 are \sigma,it id hybridized sp.


Daniel.

PS.Maybe Gokul can put it in a more coherent way... :wink:
 
As Dexter mentioned, your textbook is right only when it talks about isolated atoms.

However, when a molecule is being formed, the direction along the line joining the atoms breaks the degeneracy (x, y and z are not symmetric anymore; the z direction is the internuclear direction and so, is different from x and y) between px, py and pz.
 
A "pz" pi molecular orbital would be along the same axis as the sigma, perhaps this clarifies things a bit more.
 

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