Search any question & find its solution
Question:
Answered & Verified by Expert
If the successive ionisation energies of an element $A$ are 165, 190, 550 and $595 \mathrm{kcal}$, respectively, then the ground state electronic configuration of element $A$ is
Options:
Solution:
1153 Upvotes
Verified Answer
The correct answer is:
$[\mathrm{Ne}] 3 \mathrm{~s}^2$
$$
\begin{aligned}
& \mathrm{IE}_1=165 \mathrm{kcal}, \mathrm{IE}_2=190 \mathrm{kcal} . \\
& \mathrm{IE}_3=550 \mathrm{kcal}, \mathrm{IE}_4=595 \mathrm{kcal}
\end{aligned}
$$
As ionisation energy value (first and second) of element $A$ is lower, than third and fourth ionisation energy.
So, it indicates that valence shell posess 2 electrons.
$\therefore$ Configuration of element $A$ is $[\mathrm{Ne}] 3 \mathrm{~s}^2$.
\begin{aligned}
& \mathrm{IE}_1=165 \mathrm{kcal}, \mathrm{IE}_2=190 \mathrm{kcal} . \\
& \mathrm{IE}_3=550 \mathrm{kcal}, \mathrm{IE}_4=595 \mathrm{kcal}
\end{aligned}
$$
As ionisation energy value (first and second) of element $A$ is lower, than third and fourth ionisation energy.
So, it indicates that valence shell posess 2 electrons.
$\therefore$ Configuration of element $A$ is $[\mathrm{Ne}] 3 \mathrm{~s}^2$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.