Sketch and label a qualitative molecular orbital (MO) energy level diagram for nitrogen monofluoride, NF.

Sketch and label a qualitative molecular orbital (MO) energy level diagram for nitrogen monofluoride, NF. (ii) Use your diagram to determine the bond order of NF, [NF]*, and [NF]. (iii) Use your diagram to determine whether each of NF, [NF]+, and [NF] is diamagnetic or paramagnetic. (b) (10 marks) The triboron cation (cyclo-B3+) has a triangular structure with each BB-B angle of 120 degrees. Construct a qualitative MO energy level diagram for the bonding in cyclo-B3+. Illustrate your answer by referencing clearly labeled sketches showing how the atomic orbitals interact to form the molecular orbitals.

The Correct Answer and Explanation is:

(a) Molecular Orbital (MO) Diagram for NF (Nitrogen Monofluoride)

(i) Qualitative MO Energy Level Diagram for NF:

1. Atomic Orbitals (AOs):
   • Nitrogen (N) has the electronic configuration 1s22s22p31s^2 2s^2 2p^31s22s22p3.
   • Fluorine (F) has the electronic configuration 1s22s22p51s^2 2s^2 2p^51s22s22p5.
2. MO Diagram:
   • The 2s2s2s orbitals combine to form a bonding σ2s\sigma_{2s}σ2s​ and an antibonding σ2s∗\sigma^*_{2s}σ2s∗​.
   • The 2p2p2p orbitals combine in two ways:
     • σ2pz\sigma_{2p_z}σ2pz​​ (bonding) and σ2pz∗\sigma^*_{2p_z}σ2pz​∗​ (antibonding).
     • π2px\pi_{2p_x}π2px​​, π2py\pi_{2p_y}π2py​​ (bonding), and π2px∗\pi^*_{2p_x}π2px​∗​, π2py∗\pi^*_{2p_y}π2py​∗​ (antibonding).
3. Electron Placement:
   • From the atomic orbitals of N and F, there are 10 electrons to place in the molecular orbitals.
   • Electrons fill from the lowest energy orbital upward, following the Pauli exclusion principle and Hund’s rule.
   The filling order for the molecular orbitals is as follows:
   • 2 electrons in σ2s\sigma_{2s}σ2s​
   • 2 electrons in σ2s∗\sigma^*_{2s}σ2s∗​
   • 2 electrons in π2px\pi_{2p_x}π2px​​ and π2py\pi_{2p_y}π2py​​
   • 2 electrons in σ2pz\sigma_{2p_z}σ2pz​​
   • 2 electrons in π2px∗\pi^*_{2p_x}π2px​∗​ and π2py∗\pi^*_{2p_y}π2py​∗​

(ii) Bond Order Calculation for NF:

The bond order is determined by the formula:Bond order=12(Number of electrons in bonding orbitals−Number of electrons in antibonding orbitals)\text{Bond order} = \frac{1}{2} \left( \text{Number of electrons in bonding orbitals} – \text{Number of electrons in antibonding orbitals} \right)Bond order=21​(Number of electrons in bonding orbitals−Number of electrons in antibonding orbitals)

• Bonding electrons: 2 (σ2s\sigma_{2s}σ2s​) + 2 (π2px\pi_{2p_x}π2px​​) + 2 (π2py\pi_{2p_y}π2py​​) + 2 (σ2pz\sigma_{2p_z}σ2pz​​) = 8 electrons.
• Antibonding electrons: 2 (σ2s∗\sigma^*_{2s}σ2s∗​) + 2 (π2px∗\pi^*_{2p_x}π2px​∗​) + 2 (π2py∗\pi^*_{2p_y}π2py​∗​) = 6 electrons.

Thus, the bond order for NF is:Bond order=12(8−6)=1\text{Bond order} = \frac{1}{2} \left( 8 – 6 \right) = 1Bond order=21​(8−6)=1

(iii) Magnetic Behavior of NF, [NF]*, and [NF]+:

• NF: Since there are no unpaired electrons in the molecular orbitals, NF is diamagnetic.
• [NF]*:
  • This ion has one additional electron, which will go into the π2px∗\pi^*_{2p_x}π2px​∗​ or π2py∗\pi^*_{2p_y}π2py​∗​ antibonding orbital.
  • With one unpaired electron, [NF]* is paramagnetic.
• [NF]+:
  • This ion has one fewer electron, which means an electron will be removed from one of the antibonding orbitals (likely π2px∗\pi^*_{2p_x}π2px​∗​ or π2py∗\pi^*_{2p_y}π2py​∗​).
  • With all electrons paired, [NF]+ is diamagnetic.

(b) Qualitative MO Energy Level Diagram for Cyclo-B3+ (Triboron Cation)

1. Atomic Orbitals (AOs):
   • Each boron atom has the electronic configuration 1s22s22p11s^2 2s^2 2p^11s22s22p1, so there are three 2p12p^12p1 orbitals from each boron atom.
2. MO Diagram:
   • Since there are three boron atoms in cyclo-B3+, the three 2p12p^12p1 orbitals combine to form three molecular orbitals:
     • One bonding π2p\pi_{2p}π2p​ orbital (lower energy).
     • One nonbonding π2p∗\pi^*_{2p}π2p∗​ orbital (higher energy).
     • One bonding σ2p\sigma_{2p}σ2p​ orbital.
3. Electron Placement:
   • The total number of electrons in cyclo-B3+ is 5 (since each boron contributes 1 electron and there is a +1 charge).
   • The 5 electrons fill the bonding π2p\pi_{2p}π2p​ and σ2p\sigma_{2p}σ2p​ molecular orbitals. Since the nonbonding orbital is empty, this makes the structure relatively stable.
4. Bonding and Geometry:
   • The bonding electrons form a molecular orbital that is symmetric about the center of the triangle.
   • This results in a triangular molecular geometry with equal BB−BBB-BBB−B bond angles of 120 degrees.

In summary, the bonding in cyclo-B3+ involves the overlap of the boron 2p2p2p orbitals to form both bonding and nonbonding molecular orbitals, with the structure adopting a triangular geometry due to the arrangement of the atoms in the molecule.