[airplanet56]

I have been limited in my learning in that the hybridization examples I typically see are only CH bonds.
In methane, the sp3 orbitals of carbon each form a sigma bond by overlapping with a 1s orbital of hydrogen.

When dealing with larger atoms, such as phosgene, or Cl₂CO I am confused about the specifics of what overlaps with what.

So the Lewis structure involves 2 chlorine atoms single-bonded to the carbon, with 3 lone pairs. The oxygen is double-bonded to the carbon, with 2 lone pairs. The carbon has no lone pairs, and has an electron/molecular geometry of trigonal planar.

So, is it true to say that the carbon's sp2 orbital overlaps with the sp3 orbital of chlorine? Unlike the hydrogen, there are now s and p orbitals. So, 3 of the sp3 orbitals would hold lone pairs, and the fourth would overlap with carbon's sp2. In oxygen, its sp2 orbital would overlap with carbons sp3. Oxygen's 2 other sp2 orbitals would hold the lone pairs. Carbon's unhybridized p orbital would form a pi bond with oxygen's unhybridized p orbital.

Am I making a huge error in my thinking? I thought the hybrid orbitals were for the central atom only initially, but where do chlorines electrons go? I was also initially assuming that the p orbitals of Cl overlapped with the sp3. But it has both p and s orbitals, and should, as a result, hybridize?

Thanks

[Hunter2]

Carbon is sp² hybrisized and chlorine sp³. So thwere is not much difference if you look on a formaldehyd molecule H₂C=O.  In this cas the two hydrogen s-orbitale overlap with the Sp² orbitale of carbon. Chlorine does the same compare it with acetone CH₃COCH₃. Each methyl group is sp³ so chlorine does. Cl₂C=O.

[airplanet56]

Thanks for the response!

I was actually told recently that the chlorine doesn't hybridize because it doesn't need to. It can, but it doesn't. One of its p orbitals overlap, because the s is filled first and possesses two of the lone pairs. The other 2 p orbitals also possess lone pairs.

So my new understanding is that terminal atoms can either be thought of as hybridized or un-hybridized. Does it matter? My textbook suggests that only the central atom hybridizes, unless a double or triple bond (pi bond) is involved. It all seems arbitrary based on the model or opinion.

[antimatter101]

Now, let us look at methane. Though the angle between the p orbitals is 90 degrees, the geometry of its bonds is different. How can this be?

Quantum mechanics explains this phenomenon by stating that all four valence orbitals of the carbon atom (2s, 2px, 2py, 2pz) are hybridised, where they merge together to form four degenerate hybrid orbitals. This phenomenon is called orbital hybridisation.

Figure 2.6 – The true structure of hybrid orbitals

 
Why do orbitals hybridise? There are three reasons for this phenomenon:

1) Unhybridised orbitals differ in energy. Recall that the s orbital is more stable than the p orbital, which is more stable than the d orbital, which is more stable than the f. As a result, bonding with the individual orbitals alone would cause differences in energy in each bond, thus unfavourable altering the shape of the molecule.

Figure 2.7

 

2) The electron shells of atoms repel each other. Not considering the effect of bonding, the hydrogen atoms in methane would prefer to be as far apart from each other as possible, in the shape of a tetrahedron. By hybridising, the molecule adopts a conformation which reduces repulsion between bonded atoms.

3) Unhybridised orbitals simply cannot bond effectively. The s orbital orbits the nucleus in the shape of a sphere, with no direction to bonded atoms; while the p orbitals have two equivalent ends which reduce the efficiency of bonding by half. However, hybridised orbitals (sp, sp2, sp3) have one end larger than the other, so that they can bond effectively.

You may be wondering about the exact structure of hybrid orbitals in relation to their bonding properties. Check out this interactive website for more information. http://www.uwosh.edu/faculty_staff/gutow/Orbitals/N/sp hybrid.shtml

 

When carbon forms four separate single bonds, three p orbitals and one s orbital hybridise to produce four equivalent sp3 orbitals. Each sp3 orbital extends in tetrahedron-shaped geometry outwards from the atomic nucleus.

When carbon forms three bonds (a double bond and two single bonds), it adopts the sp2 hybridisation. This means that one p orbital is left behind in an unhybridised state.

Figure 2.7

 

Why does one p orbital remain behind? This is because multiple bonds (double and triple bonds) exploit the overlap of unhybridised p (double and triple bonds). In such scenarios, each atom donates a p orbital to each other, which overlap to form a π bond. Likewise, hybridised orbitals bond together to form σ bonds.

Figure 2.8 – Structure of sigma and pi bonds

 

Likewise, in a triple bond, two pairs of pi bonds are located above and below the central sigma bond. This means that the atom is sp hybridised, where one s and one p orbital hybridise to produce 2 sp orbitals and two ground-state p orbitals.

(Note: Empty p orbitals are considered orbitals, NOT EMPTY SPACE, since electrons do occupy them after bonding.)

This also explains why quadruple bonds cannot form: If there are four bonds to a single atom, there must be three π bonds and one σ bond. Π bonds are formed only when unhybridised p orbitals overlap – meaning that there will be no more p orbitals to hybridise with the remaining s orbital. As a result, the sigma bond would be an overlap between a ground state s orbital, which cannot bond effectively due to its uniform structure. As a result, quadruple bonds do not exist.

Wait – what about other elements which do not need to form four bonds to achieve a stable octet/filled outer shell?

Figure 2.9 – The ammonia molecule
 

The answer is simple: Electron pairs which are not involved in bonding circulate around the nucleus in hybridised lone pairs. For example, in water (H2O), the oxygen atom is described as sp3 hybridised; since one s orbital and three p orbitals hybridise to form two sigma bonds and two lone pairs.

For a better understanding, refer to figure 2.4

(Copy and pasted from my own book, which I have been writing)