What do the carbon's valence electrons reveal about the bonding in CH4?
(Hint: since the dipole moment of CH4 is zero, the correct response is "not enough")
How can we use the fact that the orbital configuration of carbon is 2s22p2 to determine the arrangement of the orbitals in a simple organic molecule such as methane (CH4)?
Methane is found to be tetrahedral, with four equal bond angles of 109.5°, four equal bond lengths, and no dipole moment.
This raises two concerns. How do we know that CH4 is a tetrahedral structure? And secondly, how do we reconcile the fact that there are four equal C–H bonds with this electronic configuration (2s22p2)?
1. The Electronic Configuration Of The Valence Electrons Of Carbon Is 2s22p2
As shown below, our review of atomic orbitals revealed that the orbital configuration of carbon's valence electrons is 2s22p2.
As the 2s orbital has less energy than the 2p orbital, it is filled first. Thus, there are two electrons in the 2s orbital and one in each of the three 2p orbitals. Also present is an empty 2p orbital.
[Additionally, there are two unbondable electrons in the 1s "inner shell" orbital].
2. Can We Use This Information To Figure Out The Structure Of Methane (CH4)? (Spoiler: No)
So far so good. This is acceptable if we are only discussing isolated carbon atoms.
To be truly useful, however, the orbitals of carbon must be related to the structure and bonding of actual organic compounds.
CH4 methane is the most basic organic compound. So, let's add four hydrogen atoms and attempt to apply what we've learned to generate hypotheses about the bonding in this molecule.
Along the x, y, and z axes, the 3 p-orbitals of carbon are at right angles to one another.
Shouldn't the structure of methane consist of three C-H bonds for each of the p orbitals (at right angles to one another) and a fourth C-H bond attached to the 2s orbital? Since electron pairs repel, perhaps we should place this C-H bond as far away from the other C-H bonds as possible; this would result in a H–C–H bond angle of 135°.
Following this logic would result in the following structure:
As it turns out, this proposal can be shown to be incorrect.
Why?
Dipole moment.
Remember that due to the difference in electronegativity between C (2.5) and H (2.2), each C–H bond has a small dipole. We anticipate that C will be partially negative while H will be partially positive.
If the structure depicted above accurately represented the structure of methane, we would expect methane to have three longer C–H bonds (to the 2p orbitals) and one shorter C–H bond (to the 2s orbital, which is closer to the nucleus).
In addition, we would anticipate three H–C–H bond angles of 90° and one of 135°.
In this structure, the vector sums of the C–H dipoles do not cancel out when added together.
Therefore, we would anticipate observing a small but measurable dipole moment for CH4. [Note 1]
The measured dipole moment of CH4 is however zero. Consequently, this cannot be the proper structure.
This indicates that the bond lengths and angles of methane are identical.
3. Maybe Methane (CH4) Is Square Planar?
Alright, you say. If all C-H bonds have equal lengths and angles, why can't CH4 have the structure shown below, in which all bond angles are 90 degrees and the molecule is planar in the page's plane? (This structure is referred to as "square planar").
Until roughly 1880, this was the consensus regarding the arrangement of bonds around carbon. Extremely brilliant chemists such as Berzelius had no reason to doubt that methane was not flat when they died.
However, we now know that this is incorrect. Why?
4. Disproving The Square Planar Structure Of CH4 (1874) And Proposal Of A Tetrahedral Structure
If the central carbon of methane is attached to four different groups, the molecule can exist as two isomers that are non-superimposable mirror images (this phenomenon is known as "optical isomerism" and will be covered later in the course).
This is possible if the four groups surrounding the central carbon are arranged tetrahedrally, but not if the molecule has a square planar structure. For instance, the methane derivative bromochlorofluoromethane has four different groups surrounding the carbon atom and can be separated into two isomers that rotate plane-polarized light in opposite directions. [As we will see in the following section, these isomers are known as "enantiomers"]
This observation disqualifies the square planar shape. If carbon were square-planar, the molecule would be flat, superimposable on its own mirror image, and there would be only one possible isomer.
Fellow at the veterinary college in Utrecht, Jacobus Henricus van't Hoff was among the first to consider the possibility of three-dimensional carbon. In "La Chimie dans l'Espace" (1874), he pointed out that the arrangement of atoms in space has significant practical implications, a point that had previously been completely overlooked. van't Hoff demonstrated that a tetrahedral arrangement of four different groups around a carbon atom (which he termed a "asymmetric carbon") would result in two distinct isomers, and this would explain why tartaric acid (with two asymmetric carbon atoms) existed in three forms (+, –, and meso). [Source]
Noting that van't Hoff's work was purely theoretical, it was not well received in some circles.
5. Tetrahedral Carbons: Not A Popular Idea In 1874
In 1901, van't Hoff flew his Pegasus to Stockholm to receive the first Nobel Prize in Chemistry. [Note 2]
In 1913, Bragg used X-ray crystallography to determine the structure of diamond and found it to be a tetrahedral network of carbon atoms with C-C-C bond angles of 109.5°, providing irrefutable evidence of the tetrahedral arrangement of bonds around the carbon atom.
Therefore, what orbits are involved?
We now attribute the tetrahedral arrangement of atoms surrounding methane to the repulsion between bonding pairs of electrons (also known as the VSEPR theory).
However, this does not help us understand the orbitals involved in bonding.
Given what we know about the geometry of s and p orbitals, how can we describe the bonding orbitals of methane if we accept that the arrangement of hydrogens around methane is tetrahedral?
Despite the fact that the 2p orbitals are all aligned at 90 degrees, the bond angles in methane are 109.5 degrees
How do we account for the fact that each of the four bonds in methane have identical lengths? What happened, for instance, to the 2s orbital?
Are C-H bond electrons considered to be in p or s orbitals? Or something else?
As it turns out, the conventional method involves treating the bonds surrounding carbon as hybrid orbitals.
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