At what temperature can molecules no longer form?
The bond will break if you can put more than its dissociation energy into it; $\ce{CO}$ has a huge dissociation energy of $\pu{1080 kJ/mol}$, $\ce{I2}$ less than most and is only $\pu{151 kJ/mol}$. This energy can be easily obtained using a laser of the correct wavelength. (You will need to convert units to electron volts to get wavelength needed. $\pu{1 kJ/mol} = \pu{1.036E-2 eV}$, $\pu{1eV} = \pu{8065 cm^-1}$).

However, when we excite using a laser we only use only a small range of energy, using heat at a given temperature is different as temperature is an average measure of a distribution of energies given by the Boltzmann expression $n_i \approx \exp(-E_i/k_\mathrm{B}T)$ where $n_i$ is the number of molecules with energy $E_i$ at temperature $T$, $k_\mathrm{B}$ is the Boltzmann constant. If  dealing with molar units replace $k_\mathrm{B}$ by $R$, the gas constant.

Thus, even at room temperature ($\pu{300 K}$),  there are, in principle, some molecules of $\ce{CO}$ that will be decomposed  even though the dissociation energy is so large; the chance of being decomposed is $\exp(-1080000/(300\times8.314) \approx 10^{-190}$. It is utterly minute even for $6\cdot 10^{23}$ molecules in a mole, but for iodine $\approx 10^{-27}$, still minute so these molecules are stable at room temperature. At $\pu{1000 K}$, 1 in $\approx 10^8$ $\ce{I2}$ molecule decompose and $10^{-4}$ at $\pu{200 K}$ so this is appreciable over a period of time. 

(Strictly we should integrate from $T$ to infinity to get the exact answer but the exponential decays so quickly just one value gives us a good idea)