Periodic Classification And Periodic Properties Result Question 2
2. The element having greatest difference between its first and second ionisation energy, is
(2019 Main, 9 April I)
(a) $\mathrm{Ca}$
(b) $\mathrm{Sc}$
(c) $\mathrm{Ba}$
(d) $\mathrm{K}$
Show Answer
Answer:
Correct Answer: 2. ( d )
Solution:
- The electronic configuration of given elements are as follows :
$ \begin{aligned} & \mathrm{K}(19)=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1 \\ & \mathrm{Mg}(12)=1 s^2 2 s^2 2 p^6 3 s^2 \\ & \operatorname{Sr}(38) \quad 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^{10} 4 p^6 5 s^2 \\ & \operatorname{Sc}(21) \quad 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^1 \end{aligned} $
First ionisation enthalpy (I.E.) of $K$ is lowest among the given options. Here, the energy required to remove an electron from $4 s^1$ is least as only one electron is present in the outermost shell. I.E. (I) is comparatively high for $\mathrm{Mg}$ and $\mathrm{Sr}$ and two electrons (fully-filled) are placed in $s$-orbital. Second ionisation enthalpy of $\mathrm{K}$ is highest among the given options.
Now, removal of an electron occur from $p^6$ (fully-filled). So, high energy is required to remove the electron. From the above discussion, it can be concluded that (I. $\mathrm{E} _2 \quad \mathrm{I} . \mathrm{E} _1$ ) value is maximum for $\mathrm{K}$ (potassium).
BETA
