Question

The value of $I{P}_1$, $I{P}_2$, $I{P}_3$ and $I{P}_4$ of an atom are respectively 7.5 eV, 25.6 eV, 48.6 eV and 170.6 eV. The electronic configuration of the atom will be

• A. 1s², 2s², 2p⁶, 3s¹
• B. 1s², 2s², 2p⁶, 3s², 3p¹
• C. 1s², 2s², 2p⁶, 3s², 3p³
• D. 1s², 2s², 2p⁶, 3s²

Answer 1

The correct option is B

1s², 2s², 2p⁶, 3s², 3p¹

First, let's get the data from equation

So, from the data, we can say that

Ip, was small, then difference between $I{P}_1$ and $I{P}_2$ and the $I{P}_2$and $I{P}_3$ was also small (little bit mover than $I{P}_1$) , but the difference between $I{P}_3$ and $I{P}_4$ was large (122eV).

We can deduce the following:

After losing three electrons, i.e., after $I{P}_3$ the atom became very stable, that's why we had to apply very large amount of energy to remove the next electron.

So, we can say that after losing three electrons, it achieved noble gas configuration.

(i) 1$s^2$ 2$s^2$ 2$p^6$ 3$s^1$ $\to$ (after IP3) 1$s^2$ 2$s^2$ 2$p^4$

(ii) 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$ 3$p^1$ $\to$ 1$s^2$ 2$s^2$ 2$p^6$

(iii) 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$ 3$p^3$ $\to$ 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$

(iv) 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$$\to$ 1$s^2$ 2$s^2$ 2$p^5$

Only (ii) configuration achieved noble gas configuration, and you can also see after $I{P}_1$ also, there is an increase of double of the $I{P}_1$ when it is going to $I{P}_2$.

It is because after losing 1 electron also, it became more stable because then valence electrons were in s-orbital, which is closer to the nucleus because of penetration effect. Closer the sub-shells are, more is ionization energy

Ionization energy: s > p > d > f for any given shell.

Hence the ionization energy of s electron is greater than that of the electron which in turn is greater than that of a d electron, and so on.