Hybridization of Phosphorus

[DariushS]

When Phosphorus (P) is bonded to three elements, and double bonded to one, how is P sp^3 hybridized? sp^3 hybridization only offers 4 orbitals, yet there are 5 bonds. McMurry's Organic Chemistry textbook says the bond angles are similar to a tetrahedral configuration, and that is why P is sp^3 hybridized in that situation. I understand P is capable of forming a different number of bonds depending on what molecule it is a part of.

 

How can I determine the hybridization of P in a molecule?

Edited June 2, 2014 by DariushS

[studiot]

Phosphorus, ground state.

 

1s²2s²2p⁶3s²3p³3d⁰

 

So phosphorus has 5 electrons in the third shell. This makes the empty 3d orbital available for hybridsation.

 

This is called the expanded octet. In the case of POCl₃ the octet is expanded to 10 this way.

 

Note that resonance is possible with alternate partially polar forms of bonding and an octet.

[hypervalent_iodine]

Phosphorus, ground state.

 

1s²2s²2p⁶3s²3p³3d⁰

 

So phosphorus has 5 electrons in the third shell. This makes the empty 3d orbital available for hybridsation.

 

This is called the expanded octet. In the case of POCl₃ the octet is expanded to 10 this way.

 

Note that resonance is possible with alternate partially polar forms of bonding and an octet.

 

This is a little pedantic of me, but it is a pet peeve. The d orbitals have very little / nothing to do with hypervalent bonding. The ability for certain atoms to possess an 'expanded octet' is accounted for by applications of molecular orbital theory and the associated quantum mechanical calculations (for example, 3c-4e bonds). The competing theory of d-orbital hybridisation is a nice and easy to understand concept for undergrads (and honestly, the distinction is probably not going to be relevant to anyone except for the poor souls such as myself who go on to do research involving hypervalent compounds), but it has been outdated since about the 90's or earlier.