The bonding energy of the two Ti―O bonds at TiO2 in the gas phase, sharing one Ti atom (i.e. O―Ti―O) can be seen as jointly corresponding to that oxide enthalpy of atomization , accordingly: TiO2 (g) → Ti (g) + 2O (g). However, such reaction, presumably taken to somehow occur at some sufficiently high temperature, it is not the reverse reaction that would correspond to that embedded at the standard enthalpy of formation of TiO2 (g), since last figure correspond to the oxide formation from molecular oxygen, not from atomic oxygen. And of course, we would still have to bear in mind the usual enthalpy corrections associated with Hess' law application to the proposed problem, due to the inherent phase changes and temperatures to be considered, to properly estimate the requested bonding energy based on the standard enthalpy of formation of solid TiO2 (anatase).

Given the reaction Ti(g) + 2O(g) --> TiO2(g), the formation enthalpy deltaH = E(TiO2(g)) - E(Ti(g)) - 2*(O(g))

Thank you Lian. Actually I did it. My question was rather about how to calculate atomic bonding energy, for example Ti-O bonding energy in Anatase, from the formation enthalpy. I could find this parameter from anatase formation enthalpy.

However, it seems that it is not correct method. I mean for other materials like BN and B4C for example you cannot find B-N and B-C bonding energy from the formation enthalpy. I don't know why !