Using the Hybridisation theory, please account for the observed bonding pattern in the following molecules:
Cl Cl H
| | |
C - C H-C-C(triple bond)C-H
| | |
Cl Cl H
To account for the observed bonding pattern in the given molecules, we need to apply the concept of hybridization. Hybridization theory helps us explain the way atomic orbitals combine to form molecular orbitals during chemical bonding.
Let's analyze each molecule individually:
1. For the molecule with the carbon-carbon double bond (C=C), we have:
- Each carbon atom is connected to one chlorine atom and one hydrogen atom.
- The carbon atoms are connected by a double bond (C=C), indicating the presence of a sigma bond and a pi bond.
- To explain this bonding pattern, we will consider the hybridization of carbon.
- Carbon needs to form four bonds in total to satisfy its valency.
Step-by-step explanation:
a) Start by counting the number of valence electrons for each atom.
- Carbon has 4 valence electrons.
- Chlorine has 7 valence electrons.
- Hydrogen has 1 valence electron.
b) Calculate the total number of valence electrons by adding up the valence electrons of each atom in the molecule:
2 carbon atoms * 4 valence electrons = 8 valence electrons
4 chlorine atoms * 7 valence electrons = 28 valence electrons
2 hydrogen atoms * 1 valence electron = 2 valence electrons
Total = 8 + 28 + 2 = 38 valence electrons
c) Determine the central atom by checking which atom appears least frequently. In this case, it is carbon.
d) Use the remaining valence electrons to form bonds around the central atom. Remember, each bond (single, double, or triple) requires two electrons.
4 chlorine atoms require 4 * 2 = 8 electrons
2 hydrogen atoms require 2 * 2 = 4 electrons
Total electrons used for bonding = 8 + 4 = 12 electrons
Remaining valence electrons = Total valence electrons - Total electrons used for bonding
Remaining valence electrons = 38 - 12 = 26 electrons
e) Assign the hybridization of the central atom based on the number of electron groups around it.
In this case, carbon is bonded to three other atoms (two chlorine and one hydrogen), indicating that it has three electron groups.
Therefore, the carbon atom undergoes sp2 hybridization, leading to the formation of three sp2 hybrid orbitals.
f) Form sigma bonds using the hybrid orbitals of the central atom and the atomic orbitals of the surrounding atoms. In this case, the carbon atom forms sigma bonds with the two chlorine atoms and the hydrogen atom using its three hybrid orbitals.
Additionally, there is a pi bond formed between the two carbon atoms using the unhybridized p orbitals.
The resulting bonding pattern in the given molecule is consistent with an sp2 hybridization of carbon.
2. For the molecule with the carbon-carbon triple bond (C≡C), we have:
- Each carbon atom is connected to one chlorine atom and one hydrogen atom.
- The carbon atoms are connected by a triple bond (C≡C), indicating the presence of a sigma bond and two pi bonds.
- To explain this bonding pattern, we will again consider the hybridization of carbon.
Following the same steps as explained earlier, we find that the carbon atoms undergo sp hybridization. Each carbon atom now has two sp hybrid orbitals and two unhybridized p orbitals.
The sigma bond in this case is formed using sp hybrid orbitals, and the two pi bonds are formed using the unhybridized p orbitals. The resulting bonding pattern is consistent with an sp hybridization of carbon.
In summary, based on the bonding patterns observed in the given molecules, we can account for them using the hybridization theory. The first molecule exhibits sp2 hybridization of carbon to form a double bond (C=C), while the second molecule shows sp hybridization of carbon to form a triple bond (C≡C).