Show the molecular orbital diagram for BN (boron nitride), calculate its bond order and compare its stability to BC (boron carbide) – which is paramagnetic? Which is diamagnetic
The Correct Answer and Explanation is:
To solve this, we analyze the molecular orbital (MO) diagrams for BN (boron nitride) and BC (boron carbide) and use electron configurations to determine bond order, magnetism, and stability.
1. MO Diagram Overview
- Left MO diagram (BN): Resembles that of diatomic molecules like CO or NO, where nitrogen is more electronegative.
- Right MO diagram (BC): Resembles earlier-period diatomic molecules like B₂ or C₂.
The key difference lies in the energy order of molecular orbitals:
- For BN (left): the ordering is affected by nitrogen’s higher electronegativity, placing σ2p lower than π2p.
- For BC (right): due to smaller electronegativity difference, π2p < σ2p.
2. Electron Count
- Boron (B) = 5 electrons
- Nitrogen (N) = 7 electrons
- Carbon (C) = 6 electrons
So:
- BN = 5 + 7 = 12 electrons
- BC = 5 + 6 = 11 electrons
3. Bond Order Calculation
Bond order = (Number of bonding electrons – Number of antibonding electrons) / 2
BN (12 electrons):
Filling the MOs:
σ1s², σ1s², σ2s², σ2s², π2p⁴
- Bonding electrons: 2 (σ2s) + 4 (π2p) = 6
- Antibonding electrons: 2 (σ2s*) = 2
Bond order = (8 – 2)/2 = 3
→ BN has bond order = 3
→ All electrons paired → Diamagnetic
BC (11 electrons):
Same orbitals, but one fewer electron.
σ1s², σ1s², σ2s², σ2s², π2p³
- Bonding electrons: 2 (σ2s) + 3 (π2p) = 5
- Antibonding: 2 (σ2s*) = 2
Bond order = (7 – 2)/2 = 2.5
→ BC has bond order = 2.5
→ Unpaired electron in π2p → Paramagnetic
4. Stability Comparison
- BN (bond order 3) is more stable than BC (bond order 2.5).
- Diamagnetic (BN) indicates all electrons are paired → typically more stable.
- Paramagnetic (BC) has unpaired electrons → less stable.
Final Answer Summary:
- BN: Bond order = 3, Diamagnetic, More stable
- BC: Bond order = 2.5, Paramagnetic, Less stable
