Question

Which of the following is correct for $I.E{.}_2$? (Here, $I.E{.}_2=2^{nd}$ ionization enthalpy)

• A. $Li<Be<C<O$
• B. $Li<Be<C<O$
• C. $Be<C<O<Li$
• D. $Li<Be<C<O$

Answer 1

The correct option is C $Be<C<O<Li$

Second ionisation enthalpy is the energy required to remove the second most loosely bound electron from a neutral gaseous isolated atom.
Electronic configuration of ionic species are:
$L{i}^{+}:1s^2$
$B{e}^{+}:1s^{2}2s^1$
$C^{+}:1s^{2}2s^{2}2p^1$
$O^{+}:1s^{2}2s^{2}2p^3$
The correct answer is option (c).
The second ionization involves the removal of an electron from the $1s^2$ electronic configuration of $L{i}^{+}$. Hence, the highest eneegy is required in this case . Oxygen is placed second because we need to remove an electron from the stable $2p^3$ orbital. In carbon, after the removal of the second electron, it will have an electronic configuration $1s^{2}2s^2$. Thus, less energy is required to remove the second electron of carbon as compared to oxygen. The lowest $2^{nd}$ ionisation energy is for beryllium because after removing the second electron, it will get a stable electronic configuration $1s^2$ and effective nuclear charge for beryllium is also low.