Periodic Classification And Periodic Properties Ques 14
14. 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) $\mathrm{K}$
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Answer:
Correct Answer: 14.(d)
Solution:
- The electronic configuration of given elements are as follows : $\mathrm{K}(19)=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}$
$\operatorname{Mg}(12)=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$
$\operatorname{Sr}(38)=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2}$
$\operatorname{Sc}(21)=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{1}$ First ionisation enthalpy (I.E.) of $\mathrm{K}$ 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. $E _{2}-I . E _{1}$ ) value is maximum for $\mathrm{K}$ (potassium).
BETA
