XH3 Optimization

From WikidChem

XH3 Optimization

Boron- 3 valence electrons

Carbon- 4 valence electrons

Nitrogen- 5 valence electrons

There is a relationship between Hybridization and Structure

The angle made in the structure depends on hybridization.

With this equation we can figure out what the angle will be between any bonds within a structue. The m and n are the hybridization numbers on the p values that correspond with specific bonds made.

For example if there is two sp bonds made with a central C atom (let's say propane) then the angle would be 180 (linear). And so on for the rest of the molecule.

The geometry will tell about the hybridization and vice versa.

How to optimize hybridization of X atom in XH3?

Atom wants to get the best overlap while maximizing s-orbital occupancy.

The Boron atom can make 3 sp^2 with one vacant p orbitals because it only has 3 valence electrons.

Think about electron configurations:

1s^2 2s^1 2px^1 2py^1 (and no 2pz) valence electrons Thus we have sp^2 hybridization

Here, in making three sp^2 hybrids, we have used all of the s character to stabilize these bonds. The molecule is happy! There is no problem because it wants to be planar!

The Nitrogen, however cannot use the sp^2 because then there will be 2 electrons in the p orbital (which is less stable than the s orbital). Since we hypothsize that N would want to maximize s-orbital occupancy by putting 2 electrons in its s orbital and one 1 in each p. Think about electron configuration:

1s^2 2s^2 2px^1 2py^1 2pz^1 valence electrons

Howeve, N is conflicted. To increase overlap, it would want to hybridize to sp^3. If it were to use sp^3 hybridization, then the three bonds would be more stable than the lone pair becuase they would all have the same amout of s character. To fix this, we imagine the molecule would devote more s character to the lone pair. We see then that NH3 is in conflict between forces working for overlap and forces working for stability of s character. Only through ESR are we able to see that NH3 gives in and creates hybrids that are slightly less than sp^3 and creates a bend molecule.

We cannot predict how CH3 will act because if it hybridized with 4 sp^3 hybrids, it would use the same amount of energy and s character as it would if it hybridized to 3sp^2 hybrids and 1 p orbital. The only way we can tell what will happen is through IR and ESR. IR reveals that the potential energy is a single minimum, so the lowest energy structure it planar. However, the curve is shallower than BH3's curve, so it is less strongly planar. ESR reaveals that there is some s character in the valence electrons, though, and becuase of this we can see that CH3 is easilly bent out of its preferrable planar state easilly.

last modified by Hayley Israel (hpi2) on Oct 23, 2006 4:33:22 PM