In analogy to \(\mathrm{O}_{2}^{+}\left[\mathrm{PtF}_{6}\right]^{-}\) a compound \(\mathrm{N}_{2}+\left[\mathrm{PtF}_{6}\right]^{-}\) will not be formed because: (a) The I.E. of \(\mathrm{N}_{2}\) gas is higher than that of \(\mathrm{N}\) atom (b) The I.E. of \(\mathrm{N}_{2}\) gas is lower than that of \(\mathrm{O}_{2}\) gas (c) The I.E. of \(\mathrm{N}_{2}\) gas is higher than that of \(\mathrm{O}_{2}\) gas (d) None of these

Ionization energy (I.E.) is a fundamental concept in chemistry that relates to how strongly an atom or molecule holds onto its electrons. The higher the I.E., the more energy is required to remove an electron from the atom to form a positively charged ion, or cation. This is an important consideration when analyzing the reactivity of elements and the formation of chemical compounds.

Understanding I.E. helps explain why some atoms are less likely to form cations and thus engage in ionic bonding. For instance, elements with low I.E.s such as alkali metals easily lose their outermost electron and typically form ionic compounds. Conversely, noble gases have very high I.E.s, making them relatively unreactive.

The formation of chemical compounds involves the interaction of atoms to achieve a more stable electron arrangement. Atoms will transfer or share electrons in a process driven by their intrinsic properties, including their ionization energies.

In ionic bonding, an atom with a relatively low I.E. will lose an electron to an atom with a higher electron affinity, resulting in the formation of cations and anions that attract each other to form a compound. In covalent bonding, atoms with more similar I.E.s tend to share electrons. The balance between these energetic considerations explains why certain compounds form and others do not.

When considering ionic compounds similar to the exercise, the I.E. of the atom or molecule that is expected to form a cation is critical. If its I.E. is too high, it is less likely to lose an electron and hence, less likely to form the chemical compound.

Comparing ionization energies between different atoms and molecules is crucial when predicting chemical reactions and the formation of compounds. Generally, the element with the lower I.E. is more likely to lose an electron and form a cation.

This can be illustrated by comparing the atomic I.E.s of nitrogen and oxygen. As we've seen in the exercise, if we're not able to create a compound with nitrogen similar to one with oxygen, it can often be attributed to their respective I.E.s. Since oxygen can form an ionic compound with \(\mathrm{PtF}_{6}^{-}\), but nitrogen cannot, we can infer that the I.E. of nitrogen gas must be higher than that of oxygen gas, preventing the formation of \(\mathrm{N}_{2}^{+}\big[\mathrm{PtF}_{6}\big]^{-}\).

Therefore, comparing I.E.s helps determine which elements or molecules can participate in reactions leading to compound formation and which cannot, serving as a predictive tool for chemists.