Draw the molecular orbital diagrams for the following molecules and determine the bond order and magnetic properties. N2 N2 N2 Bond Order: BO: BO: BO: Paramagnetic diamagnetic Paramagnetic diamagnetic Paramagnetic diamagnetic Which molecule has the shortest bond? Which molecule has the highest bond enthalpy? Which molecule has the weakest bond?
The Correct Answer and Explanation is:
Here are the completed answers for the problem.
For N₂
Bond Order: 3
Paramagnetic diamagnetic (circle diamagnetic)
For N₂⁺
BO: 2.5
Paramagnetic diamagnetic (circle paramagnetic)
For N₂²⁻
BO: 2
Paramagnetic diamagnetic (circle paramagnetic)
For N₂³⁻
BO: 1.5
Paramagnetic diamagnetic (circle paramagnetic)
a. Which molecule has the shortest bond?
N₂ N₂⁺ N₂²⁻ N₂³⁻ (circle N₂)
b. Which molecule has the highest bond enthalpy?
N₂ N₂⁺ N₂²⁻ N₂³⁻ (circle N₂)
c. Which molecule has the weakest bond?
N₂ N₂⁺ N₂²⁻ N₂³⁻ (circle N₂³⁻)
Explanation
This problem is solved using Molecular Orbital (MO) theory. For diatomic molecules of elements lighter than oxygen, like nitrogen, s-p mixing occurs. This makes the energy of the σ₂p orbital higher than the π₂p orbitals. The valence molecular orbital filling order is: σ₂s, σ₂s, π₂p, σ₂p, π₂p, σ*₂p. Bond order (BO) is calculated as ½ (bonding electrons minus antibonding electrons). Magnetic properties depend on unpaired electrons; molecules with unpaired electrons are paramagnetic, while those with only paired electrons are diamagnetic.
N₂: A neutral nitrogen atom has 5 valence electrons, so the N₂ molecule has a total of 10 valence electrons. Its MO configuration is (σ₂s)²(σ*₂s)²(π₂p)⁴(σ₂p)².
- Bond Order: There are 8 bonding electrons (in σ₂s, π₂p, σ₂p) and 2 antibonding electrons (in σ*₂s). BO = ½(8 – 2) = 3.
- Magnetic Properties: All 10 electrons are paired, so N₂ is diamagnetic.
N₂⁺: This ion has lost one electron, leaving it with 9 valence electrons. The electron is removed from the highest occupied molecular orbital (HOMO), which is the σ₂p. The configuration is (σ₂s)²(σ*₂s)²(π₂p)⁴(σ₂p)¹.
- Bond Order: BO = ½(7 – 2) = 2.5.
- Magnetic Properties: There is one unpaired electron in the σ₂p orbital, making N₂⁺ paramagnetic.
N₂²⁻: This ion has gained two electrons, for a total of 12 valence electrons. The two electrons are added to the lowest unoccupied molecular orbital (LUMO), which is the degenerate π₂p set. According to Hund’s rule, they occupy separate orbitals with parallel spins. The configuration is (σ₂s)²(σ₂s)²(π₂p)⁴(σ₂p)²(π*₂p)².
- Bond Order: BO = ½(8 – 4) = 2.
- Magnetic Properties: It has two unpaired electrons, making it paramagnetic.
N₂³⁻: This ion has 13 valence electrons. The configuration is (σ₂s)²(σ₂s)²(π₂p)⁴(σ₂p)²(π₂p)³.
- Bond Order: BO = ½(8 – 5) = 1.5.
- Magnetic Properties: There is one unpaired electron in the π*₂p orbitals, so it is paramagnetic.
Bond strength and bond length are directly related to the bond order. A higher bond order indicates a stronger, shorter bond with higher bond enthalpy. A lower bond order indicates a weaker, longer bond.
- Shortest Bond / Highest Enthalpy: N₂ has the highest bond order (3), so it has the shortest and strongest bond.
- Weakest Bond: N₂³⁻ has the lowest bond order (1.5), so it has the weakest bond.
