- Joined
- Nov 18, 2007
- Messages
- 18
- Reaction score
- 0
- Points
- 0
- Pre-Medical
Hi everyone, I'm looking for some help with the following question:
Question:
Molecular orbitals in hydrocarbons are formed between the 1s atomic orbital of hydrogen and the sp, sp2, or sp3 hybrid atomic orbitals of carbon. Which choice correctly lists the energy level of the C-H bonds, from lowest to highest?
A. C6H6, HC≡CH, CH4
B. H2C=CH2, CH4, C6H6
C. C6H6, CH4, H2C=CH2
D. HC≡CH, C6H6, CH4
---
Kaplan Answer:
To answer this question, you will have to remember the types of hybrid atomic orbitals formed by these simple organic molecules. In CH4, methane, the single 2s orbital and the three 2p orbitals hybridize to form an sp3 hybrid orbital. In C2H4, the 2s orbital hybridizes with only two of the p orbitals to form an sp2 hybrid. The third p orbital remains separate and forms the second carbon-carbon bond. In C6H6, benzene, the six carbons also form sp2 hybrid orbitals and the remaining p orbitals form the pi electron cloud above and below the plane of the ring. In C2H2, acetylene, only one p orbital hybridizes with the s orbital, and the two remaining p orbitals form the second and third carbon-carbon bonds.
The C-H bonds of each of these molecules are formed by the hybrid sp, sp2 or sp3 orbitals. Since the s subshell is lower energy than the p subshell, the greater the s character of an orbital, 10 the lower its energy will be. Thus, the energy level transition from lowest to highest will go from sp to sp2 to sp3, choice D.
--
My comments: I understand the explanation. However, if the CH bond in HC≡CH is lower energy than CH4, why is HC≡CH a stronger acid than CH4? What am I missing?
Question:
Molecular orbitals in hydrocarbons are formed between the 1s atomic orbital of hydrogen and the sp, sp2, or sp3 hybrid atomic orbitals of carbon. Which choice correctly lists the energy level of the C-H bonds, from lowest to highest?
A. C6H6, HC≡CH, CH4
B. H2C=CH2, CH4, C6H6
C. C6H6, CH4, H2C=CH2
D. HC≡CH, C6H6, CH4
---
Kaplan Answer:
To answer this question, you will have to remember the types of hybrid atomic orbitals formed by these simple organic molecules. In CH4, methane, the single 2s orbital and the three 2p orbitals hybridize to form an sp3 hybrid orbital. In C2H4, the 2s orbital hybridizes with only two of the p orbitals to form an sp2 hybrid. The third p orbital remains separate and forms the second carbon-carbon bond. In C6H6, benzene, the six carbons also form sp2 hybrid orbitals and the remaining p orbitals form the pi electron cloud above and below the plane of the ring. In C2H2, acetylene, only one p orbital hybridizes with the s orbital, and the two remaining p orbitals form the second and third carbon-carbon bonds.
The C-H bonds of each of these molecules are formed by the hybrid sp, sp2 or sp3 orbitals. Since the s subshell is lower energy than the p subshell, the greater the s character of an orbital, 10 the lower its energy will be. Thus, the energy level transition from lowest to highest will go from sp to sp2 to sp3, choice D.
--
My comments: I understand the explanation. However, if the CH bond in HC≡CH is lower energy than CH4, why is HC≡CH a stronger acid than CH4? What am I missing?
Last edited:
