The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called as first ionization energy `(IE_(1))`. Similarly the amount of energies required to knock out second, third etc. electrons from the isolated and `IE_(3)gt IE_(2)gt IE_(1)`.
(i) Nuclear charge (ii) Atomic size (iii) penetration effect of the electrons (iv) shielding effect of the inner electrons and (b) electronic configurations (exactly half filled and completely filled configurations are extra stable) are the important factors which affect the ionisation energies. Similarly, the amount of energy released when a neutral isolated gaseous atom accepts an extra electron to from gaseous anion is called electron affinity.
`(X(g)+e^(-)(g)rarr X^(-)(g)+` energy
A positive elecrton affinity idicates that the ion `X^(-)` has a lower more negative energy than the neutral atom X. The second electron affinity for the addition of a second electron to an initially neutral atom is negative because the electron replusion outweights the nuclear attraction, e.g.,
`O(g)+e^(-)overset("Exothermic")rarr O^(-)(g),E_(a)=+141 kJ mol^(-)` ....(i)
`O^(-)(g)+e^(-)overset("Excothermic")rarr, E_(a)=-780 kJ mol^(-)` ...(ii)
The electron affinity of an element depends upon (i) atomic size (ii) nuclear charge and (iii) electronic configuration. In general, in a group, ionisation energy and electron affinity decrease as the atomic size increases. The members of third period have some higher (e.g., S and Cl) electron affinity values than the members of second period (e.g., O and F) because second period elements have very small atomic size. Hence, there is tendency of electron-electron repulsion, which resultss in less evolution of energy in the formation of correcsponding anion.
Identify the least stable ion amongst the following:

Options

A. `Li^(-)`
B. `Be^(-)`
C. `B^(-)`
D. `C^(-)`

Answer

B