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..!
 A: The (non-relativistic) kinetic energy expectation value of a particle moving in $\mathbb{R}^d$ is proportional to $\lvert \nabla \psi \rvert^2$, so if you delocalize it, you make the gradients — and hence, the kinetic energy — smaller. So if you rescale a wave function $\psi_{\lambda}(x) = \lambda^{d/2} \, \psi(\lambda x)$ by $\lambda$, then you see that the kinetic energy scales with $\lambda^2$, i. e. if $\lambda$ is small, then the kinetic energy expectation value with respect to $\psi_{\lambda}$ is $\lambda^2$ times the expectation value with respect to $\psi$.
Of course, usually there is a price you pay by delocalising because decreasing the kinetic energy means you eventually increase the potential energy expectation value. Try minimizing the total energy expectation value for $H = \frac{1}{2m} (- \mathrm{i} \partial_r)^2 - \frac{e}{r}$ by scaling $\psi(r)$. You will see that there is an optimal point between $\lambda = 0$ (completely delocalized) and $\lambda = \infty$ (localized in a single point).
