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I am reading the paper "Twenty five years of Finnis-Sinclair potentials" by Graeme Ackland, Adrian Sutton, and Vasek Vitek, Philosophical Magazine 2009, 89, 3111-3116. It is a review-type article describing the notion that pair potentials are inadequate for describing bonding in metals. But it turns out that if the attractive interaction is "described by the square root of a sum of pair interactions to neighbors, centered on each atom in the solid," metallic bonding can be described reasonably well. The authors point out:

The novelty is entirely in the square root. The square root captures the dependence of atomic interactions on the local density: as the number of neighbors of an atom decreases, the strength of the remaining bonds increases. This immediately predicts an inward relaxation at metallic free surfaces with a tensile surface stress, both of which are widely observed but not predicted by models in which the cohesive energy is just a sum of pair potentials.

My question is, is it true for metals that "as the number of neighbors of an atom decreases, the strength of the remaining bonds increases"? Is there any way that I can understand this, either in a qualitative or quantitative way? Also, when it says "the strength of the remaining bonds increases," is this referring to the energy per bond?

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I'm not sure, does this help at all,en.wikipedia.org/wiki/Bond_order_potential – George Herold Nov 5 '15 at 20:57

I'm not sure if it's correct. However assuming that it is, keep in mind atoms in a metal are positively charged relative to the "electron sea". That being the case, intuition says if you have more atoms near eachother in a metal, you have larger positive charge density, which diminishes the electrostatic attraction (potential due to separation of charges) felt per atom against the electron sea.

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Question: when it says "the strength of the remaining bonds increases," is this referring to the energy per bond?
Answer: Yes.
Example: If a metal atom bonded to 6 other metal atoms with each bond having energy of 0.8 eV then when you remove one atom, for the remaining 5 bonds energy increases to say 0.85 eV each. The total bonding energy decreases while energy for each bond increases. Otherwise we would never have more than two metal atoms coming together to form a solid. This is somehow connected to the definition of cohesive energy, which is the energy required to remove an atom from the bulk of a metal to the free atomic state. .
Question: Is there any way that I can understand this, either in a qualitative or quantitative way?
Answer: One way to understand this, as discussed in the quoted part from the paper, is to think in terms of surfaces. To create a surface one break bonds and the atoms at the vacuum-solid interface loose one neighbor from above. These atoms feel an extra pull from the atoms underneath since the balancing force normal to the surface disappeared. One way to rationalize this downward extra pull is to think as if the strength of the bonds with the remaining atoms has increased. This phenomenon is called surface relaxation and observed almost for all metals.

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