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In covalent bonding, as in metallic bonding, the lowering in energy relative to the free-atom state is achieved by the reduction in the kinetic energy of valence electrons due to their delocalization.

I understand that according to Heisenberg's uncertainty principle the delocalization of valence electrons lowers their momentum uncertainty (standard deviation), but I don't see why it should lower the momentum itself and so the kinetic energy..!

Thanks for your help !

In covalent bonding, as in metallic bonding, the lowering in energy relative to the free-atom state is achieved by the reduction in the kinetic energy of valence electrons due to their delocalization.

I understand that according to Heisenberg's uncertainty principle the delocalization of valence electrons lowers their momentum uncertainty (standard deviation), but I don't see why it should lower the momentum itself and so the kinetic energy..!

Thanks for your help !

In covalent bonding, as in metallic bonding, the lowering in energy relative to the free-atom state is achieved by the reduction in the kinetic energy of valence electrons due to their delocalization.

I understand that according to Heisenberg's uncertainty principle the delocalization of valence electrons lowers their momentum uncertainty (standard deviation), but I don't see why it should lower the momentum itself and so the kinetic energy..!

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How the electron delocalization can lower its kinetic energy?

In covalent bonding, as in metallic bonding, the lowering in energy relative to the free-atom state is achieved by the reduction in the kinetic energy of valence electrons due to their delocalization.

I understand that according to Heisenberg's uncertainty principle the delocalization of valence electrons lowers their momentum uncertainty (standard deviation), but I don't see why it should lower the momentum itself and so the kinetic energy..!

Thanks for your help !