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In $\ce{H2O}$, I am able to understand that the enthalpy needed to break two $\ce{OH}$ bonds are not the same. But I don't understand why is the bond energy for breaking the first $\ce{O-H}$ bond more than the second one.

My logic for second bond energy more than the first one is, when a $\ce{H}$ is removed from $\ce{H2O}$, The remaining $\ce{O}$ gets a negative charge and that ensures strong bonding with other $\ce{H}$ which will imply more energy for breaking the second bond.

First bond enthalpy is 502 kJ/mol and second bond enthalpy is 427 kJ/mol.

I would appreciate someone who could correct me

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  • $\begingroup$ After dissociation of the first bond, the remaining hydroxyl radical does not get a negative charge. $\endgroup$
    – Loong
    Nov 7 at 9:25
  • $\begingroup$ How can that happen? $\endgroup$
    – Shashaank
    Nov 7 at 9:40
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    $\begingroup$ You seem to think that water is made of ions, because it dissociates to ions (which it does). This is wrong. $\endgroup$ Nov 7 at 10:00
  • $\begingroup$ Oh OK. But what's the reason for such trend, first bond enthalpy being more than second one $\endgroup$
    – Shashaank
    Nov 7 at 10:02
  • $\begingroup$ There is no reason to expect anything. O in OH is in a pathological state, never seen in a stable molecule. $\endgroup$ Nov 7 at 10:44

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