QJEEAdv02Chebf7: JEE Advanced 2002 Chemistry - Class 11 JEEAdv02Chebf7 Solution: Identify the least stable ion amongst the following

To determine the least stable ion among the given options, we need to consider the stability of negative ions (anions) formed by each element in their respective periods. Stability of anions generally decreases as we move across a period from left to right due to the increasing nuclear charge and decreasing atomic size, which causes higher electron-electron repulsion when an extra electron is added.

The options given are:

• Option A: Li^-
• Option B: Be^-
• Option C: B^-
• Option D: C^-

Let's analyze them in detail:

Li^-: Lithium has 3 electrons (1s²2s¹). Adding an extra electron will fill the 2s orbital, resulting in a relatively stable configuration.

Be^-: Beryllium has 4 electrons (1s²2s²). Adding an extra electron will result in a configuration of 1s²2s²2p¹. This electron now enters a higher energy 2p orbital, making the ion less stable due to the destabilizing effect of the electron repulsion in the 2p orbital.

B^-: Boron has 5 electrons (1s²2s²2p¹). Adding an extra electron results in the configuration 1s²2s²2p², which still holds a relatively stable configuration as the extra electron is accommodated in the 2p orbital.

C^-: Carbon has 6 electrons (1s²2s²2p²). Adding an extra electron results in the configuration 1s²2s²2p³, creating a half-filled p-orbital which can be considered relatively stable due to the Hund's Rule of maximum multiplicity.

Among these, Be^- is the least stable ion as the added electron goes into a new, higher-energy 2p orbital, significantly increasing electron-electron repulsion and destabilizing the ion.

Thus, the least stable ion is:

Option B: Be^-