Chirality of heteroatoms
Solution 1
Typical amines ($\ce{NR3}$) are roughly $\ce{sp^3}$ hybridized. A chiral amine must undergo nitrogen inversion in order to racemize. The transition state for nitrogen inversion is a planar $\ce{sp^2}$ structure. So in order to invert the ligands attached to nitrogen we must move from an $\ce{sp^3}$ ground state with ca. 109 degree bond angles to an $\ce{sp^2}$ transition state with 120 degree bond angles.
On the other hand typical $\ce{PR3}$ compounds are roughly unhybridized with 90 degree bond angles in the ground state. In order to racemize they must also invert at phosphorous which again requires an $\ce{sp^2}$ transition state with 120 degree bond angles. Changing the geometry from 90 to 120 degrees in the case of phosphorous requires more energy that a distortion from 109 to 120 degrees in the case of nitrogen.
P.S. Put a chiral nitrogen in a 3-membered ring and chirality can be preserved because you've now made it so much more difficult to expand the 60 degree bond angle in the ring to 120 degrees.
Solution 2
It's exactly as you say. Some heteroatoms can be chiral if they have 4 different substituents (or 3 different substituents and a lone pair).
The problem with nitrogen is that it rapidly interconverts at normal temperatures. The inversion barrier is ~24 kJ/mol and quantum effects also play a role in the conversion. The barrier for phosphorous is much, much higher. It's a larger atom so the bonds are usually longer, and the lone pair occupies a larger volume.
There's actually a nice Wikipedia article on nitrogen inversion.
Incidentally, molecules like helicenes are chiral (i.e., with a right-handed or left-handed helicity), but like nitrogen can interconvert between forms through inversion.
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EJC
Updated on August 01, 2022Comments
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EJC over 1 year
Why is a phosphorus atom with three different substituents chiral, but nitrogen isn't? Nitrogen inverts fast, while phosphorus obviously much slower. How is that explained?
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EJC about 9 yearsAlso a nitrogen atom that connects two condensed rings has a sufficiently high barrier.
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ado sar about 3 years@ron But in general can we prepare the R or S enantiomer of a $\ce{PR_3} compound? Because if it is impossible then the samples we work maybe be R or S and because tthere is no energy difference between the two types of enantiomers we expect a racemic mixture.