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S and p character with hybridisation

A nucleophile will tend to donate electrons which are at the highest energy level. Generally:


Which makes sense. For example, we already expect ammonia to donate its lone pair rather than a hydride ion (H-). 

There is a related pattern in the hybridisation of s and p orbitals. In ethyne, a carbon atom has two p orbitals used in pi bonding, and one s and one p orbital used in sigma bonding. The sigma bonds are therefore 50% s character.

For a carbon atom in ethene, one p orbital is used in a pi bond and one s and two p orbitals are used in sigma bonds, sigma bonds in ethene are therefore 33% s character.

In ethane, sigma bonds are 25% s character.

A s orbital is lower in energy than a p orbital, so molecular orbitals with high s character are lower in energy.

A visible result of this is the structure of CH3+, which has three sp2 orbitals and one empty pi orbital:


If someone asked "why isn't it sp3 hybridized with an empty sp3 orbital. Just remember that lower-energy orbitals only affect stability when electrons are contained in them. sp2 hybridisation rather than sp3 in this molecule lowers the energy of the occupied orbitals and increases the energy of the unoccupied orbital.. 

In other words, it won't waste s character on an empty p orbital.

And obviously if you added a lone pair to make CH3- then it would sp3 hybridise and adopt a tetrahedral shape.

3 comments:

  1. If in case of PBr3 the bond angle is 101° the S character is 16%.
    How it is possible?

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