MCAT® ExamChemical and Physical Foundations of Biological SystemsEasy
A research team is investigating the properties of a diatomic molecule. They determine that the molecule has a bond order of 2 and contains one net sigma bond and one net pi bond. Which of the following elements is most likely to form such a diatomic molecule?
- AFluorine (F)
- BCarbon (C)
- COxygen (O)
- DNitrogen (N)
Show answer & explanationAnswer & explanation
Correct answer: C. Oxygen (O)
Oxygen forms a diatomic molecule (O2) with a double bond. A double bond consists of one sigma bond and one pi bond, resulting in a bond order of 2. Nitrogen forms a triple bond (bond order 3), while fluorine forms a single bond (bond order 1). Carbon typically forms networks or more complex molecules, not simple diatomic molecules with a bond order of 2.
Why the other options are wrong
- A. Fluorine forms a single bond (F-F), which has a bond order of 1 (one sigma bond).
- B. Carbon typically forms more complex structures and does not form a stable diatomic molecule with a bond order of 2 under normal conditions.
- D. Nitrogen forms a triple bond (N≡N), which has a bond order of 3 (one sigma and two pi bonds).
Bond Order
Bond order is a measure of the number of chemical bonds between a pair of atoms, indicating the stability of a chemical bond.
- Single bond = bond order 1 (one sigma bond)
- Double bond = bond order 2 (one sigma, one pi bond)
- Triple bond = bond order 3 (one sigma, two pi bonds)
Memory trick: Sigma starts with S for Single, Pi is the Partner in a Double.