wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Thermodynamically, the most stable form of carbon is:

A
Diamond
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Graphite
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Fullerenes
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Coal
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

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 Li to Be, but then decreases as we move from Be to B. Then, as predicted, the IE increases as we move to C and N, 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 N and O, 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.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon