Question

In analogy to $O_2^{+}[Pt{F}_6{]}^{-}$ a compound $N_2^{+}[Pt{F}_6{]}^{-}$ will not be formed because:

• A. the ionization enthalpy of N₂ gas is higher than that of O₂ gas
• B. the ionization enthalpy of N₂ gas is lower than that of O₂ gas
• C. the ionization enthalpy of N₂ gas is higher than that of N atom.
• D. None of these.

Answer 1

The correct option is A the ionization enthalpy of N₂ gas is higher than that of O₂ gas

Ionization enthalpy of $N_2$ gas $\left(1503kJmo{l}^{-1}\right)$ is higher than that of $O_2$ gas $\left(1175kJmo{l}^{-1}\right)$ and it cannot lose its electron so easily as $O_2$ does in forming $O_2^{+}[Pt{F}_6{]}^{-}$ compound.