Why does basicity go down along the group 15 hydrides? [duplicate]








**This question already has answers here**:
 
 




[Why does basicity of group 15 hydrides decrease down the group?](/questions/16215/why-does-basicity-of-group-15-hydrides-decrease-down-the-group)

 (3 answers)
 

Closed 5 years ago.



As we go down Group 15, the sizes of atoms increase and these are the central atoms in their corresponding hydrides. Shouldn't the large size of central atom (in case of $\ce{Bi}$) actually help the lone pair inside the $\ce{sp^3}$ hybrid orbital to easily be donated and thus increase the basic character down the group?


$\ce{N}$ is small and shouldn't it hold lone pair with much more strength than others, but $\ce{NH3}$ seems to be strong base while $\ce{BiH3}$ is the weakest one.




Going down this group the proton affinity decreases, so basicity decreases:


$$\mathrm{PA}(\ce{X})=\mathrm{D}(\ce{X+-H})-\mathrm{IE}(\ce{X})$$


where $\mathrm{D}$ is the bond disassociation energy, and $\mathrm{IE}$ is the ionization energy. Down a column this expression is dominated by the $\mathrm{D}(\ce{X^+-H})$ term, and since $\ce{N}$ has the highest bond strength with hydrogen, it is the most basic.


The structural (or hybridization) way to look at it is that the lone pair in $\ce{NH3}$ is very, very different than the lone pair of a similar compound of $\ce{P}$, $\ce{As}$, etc. $\ce{NH3}$ is almost perfectly $\mathrm{sp^3}$ hybridized, but the others have little to no hybridization, so the lone pair is actually held in a orbital that has mainly $\mathrm{s}$ character. This makes the lone pair far less directional (the charge is distributed over a larger area), making it less reactive and harder to form bonds, so less basic.


Finally, $\ce{N}$ is the only one on the series where the HOMO (which contains the lone pair) is concentrated on the $\ce{N}$. For the rest of the series, the HOMO is very delocalized and significant electron density is held on the $\ce{H}$ atoms, so it will not react well as a base as well.


 1

The very idea of $\ce{sp^3}$ hybridization you are referring to is pretty much *not there*, starting from $\ce{PH3}$. Look at its bond angles, they are almost exactly $90^\circ$. This means that phosphorus contributes to the $\ce{P-H}$ bonds almost exclusively with its $\ce p$ orbitals, and the lone pair just sits there hidden on $\ce s$ orbital, rather than stick out as $\ce{sp^3}$.


The same is even more true for the following hydrides.


Also, with nitrogen being the most electronegative of all, its partial negative charge in $\ce{NH3}$ attracts $\ce{H+}$, which is not the case for $\ce{PH3}$ and the rest.


 1