Question

Thermodynamically, the most stable form of carbon is:

• A. Diamond
• B. Graphite
• C. Fullerenes
• D. Coal

Answer 1

The correct option is B Graphite

As might be expected based on the periodic trend for the effective nuclear charge, ionization energy generally increases to the right across a period and decreases down a given group. The IE again increases progressing across period two from $LitoBe$, but then decreases as we move from $BetoB$. Then, as predicted, the IE increases as we move to $CandN$, but decreases when we move to $O$. The ionization of $Be$ requires removal of an electron from the $2s$ subshell. Ionization of $B$, however, requires removal of an electron from a $2p$ subshell. The $2p$ subshell is higher in energy than the $2s$ subshell resulting in a lower than expected first ionization energy for boron. On comparing $NandO$, we see that oxygen has a fourth electron in its $2p$ subshell. Since this subshell contains only three orbitals, this fourth electron must occupy an orbital that already contains an electron. Electrons, being negatively charged particles, repel each other and it is this increased electron-electron repulsion that causes oxygen to have a lower than expected first ionization energy.

Therefore the order of IE is $B<Be<C<O<N$.