Question

First, second and third I.P. values are 100 eV, 150 eV and 1500 eV. Element can be

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

This solution is in the data and electronic configurations of Be, B, F and Na.

Let's start with electronic configurations.

Be $\to$ 1$s^2$ 2$s^2$

B $\to$ 1$s^2$ 2$s^2$ 2$p^1$

F $\to$ 1$s^2$ 2$s^2$ 2$p^5$

Na $\to$ 1$s^2$ 2$s^2$ 2$p^6$ 3$s^1$

$I{P}_1$ = 100 eV $I{P}_2$ = 150 eV $I{P}_3$ = 1500 eV

Now, you can see after $I{P}_2$, very large potential was needed to remove an electron.

So, we are looking for an atom which will attain very high stability or noble gas configuration after losing two electrons.

By looking at electronic configurations, we can very easily say that it is very much possible in case of Be.

After giving away two electrons, it will achieve noble gas configuration (He).

So most oblivious answer is Be.