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Why does a bond vibration frequency depend on the strength of the bond (does it relate to electrons shared in any way)?

And why does it depend on the atomic masses of the atoms in the bond? I don't see how the size of an atom determines its attraction to a pair of electrons.

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    $\begingroup$ A classical analog is a mass on a spring. Bond strength relates to the spring constant and atomic mass related to, well, the mass... $\endgroup$
    – Jon Custer
    Mar 26, 2020 at 18:50
  • $\begingroup$ @JonCuster why would a bond behave like this? $\endgroup$
    – XXb8
    Mar 26, 2020 at 18:54
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    $\begingroup$ I’m unclear what you mean by ‘why’ - there is a force between two atoms from the bond, that varies with separation, and the atoms have mass. $\endgroup$
    – Jon Custer
    Mar 26, 2020 at 19:03
  • $\begingroup$ @JonCuster sorry, it was unclear. Why would a bond vibrate in the first place let alone vibrate like a spring? $\endgroup$
    – XXb8
    Mar 26, 2020 at 19:10
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    $\begingroup$ Because there is a potential well for it to rattle around in. Give it a push and it can vibrate. Why wouldn’t it be able to oscillate? $\endgroup$
    – Jon Custer
    Mar 26, 2020 at 19:20

1 Answer 1

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Bonds are formed when the system is at a local/global minima in the potential. A typical potential vs separation between two masses is given by: enter image description here

And by Taylor expanding the potential at the minima we see that the leading order contribution is from the quadratic term in displacement from the equilibrium (minima). This is nothing but a harmonic oscillator potential. The vibration frequency then is related to the second derivative of the potential profile. And this depends on the local electronic distribution and the shape of the atoms.

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