Question

The electronic configuration of the atom having maximum difference between first and second ionisation energies is

• A. $1s^{2}2s^{2}2p^{6}3s^1$
• B. $1s^{2}2s^{2}2p^{6}3s^2$
• C. $1s^{2}2s^{2}2p^1$
• D. $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$

Answer 1

The correct option is A $1s^{2}2s^{2}2p^{6}3s^1$

In case of option 'a', after removal of one electron, we get $1s^{2}2s^{2}2p^6$, which becomes a stable configuration. In addition to that, the atom acquires positive charge. So, the second ionisation enthalpy becomes very high than the first ionisation enthalpy and the difference between first and second ionisation enthalpy becomes more.
In option 'b', the first ionisation enthalpy is comparatively more due to the stable electronic configuration. After removal of one electron, we get $1s^{2}2s^{2}2p^{6}3s^1$ which isn't that stable configuration. So, the difference between first and second ionisation enthalpy is comparatively less.
In case of option 'c', after removal of one electron, we get $1s^{2}2s^2$, which becomes a stable configuration. In addition to that, the atom acquires positive charge. So, the second ionisation enthalpy becomes high than the first ionisation enthalpy. But due to small nuclear charge, the difference between first and second ionisation enthalpy is comparatively lower than the option 'a'.
In option 'd', the first ionisation enthalpy is comparatively more due to the stable electronic configuration. After removal of one electron, we get $1s^{2}2s^{2}2p^{6}3s^{2}3p^2$ which isn't that stable configuration but has a positive charge on atom. So, the difference between first and second ionisation enthalpy is comparatively less.