Editing Openai/69227557-0850-8003-9c19-8f26825e337d (section)

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Solving the Schrödinger equation for two atoms yields:
* a bonding orbital (lower energy)
* an antibonding orbital (higher energy)

Electrons occupy the bonding orbital, whose energy is lower because the wavefunction places electron probability between the nuclei, increasing negative–positive attraction.

This modifies VenV_{en}Ven favorably:

ΔVen<0\Delta V_{en} < 0ΔVen<0
That's the dominant stabilizing term.

===== For a Coulombic system in a stable bound state: =====

E=12Vwhere V<0E = \frac{1}{2} V
\quad\text{where } V<0E=21Vwhere V<0
So if the potential energy becomes more negative during orbital formation, the total energy drops.
The system must dispose of that energy → release.

===== Electrons in a molecular orbital are spatially more delocalized than in isolated atoms. =====

Heisenberg tells us:

Δx↑⇒Δp↓⇒Te↓\Delta x \uparrow \Rightarrow \Delta p \downarrow \Rightarrow T_e \downarrowΔx↑⇒Δp↓⇒Te↓
So electron delocalization reduces kinetic energy, contributing to stabilization.

This is not a chemical heuristic—it is straight quantum mechanics.

===== Plotting energy vs. internuclear separation produces a Lennard-Jones-type or Morse potential: =====

E(r)=Ar12−Br6(simplified)E(r) = \frac{A}{r^{12}} - \frac{B}{r^6} \quad (\text{simplified})E(r)=r12A−r6B(simplified)
Real quantum mechanical treatment yields a potential well:

r=r0⇒dEdr=0andd2Edr2>0r = r_0 \Rightarrow \frac{dE}{dr}=0 \quad \text{and} \quad \frac{d^2E}{dr^2} > 0r=r0⇒drdE=0anddr2d2E>0
The depth of that well is the bond energy.

The energy released upon formation equals the drop to that minimum.