Question

The value of $IP_{1},IP_{2},IP_{3}$ and $IP_{4}$ of an atom are, respectively, $7.5eV,25.6eV,48.6eV$ and $170.6eV$ . The electronic configuration of the atom will be:

• A. $1s^{2},2s^{2},2p^{6},3s^{1}$
• B. $1s^{2},2s^{2},2p^{6},3s^{2},3p^{1}$
• C. $1s^{2},2s^{2},2p^{6},3s^{2},3p^{3}$
• D. $1s^{2},2s^{2},2p^{6},3s^{2}$

Answer 1

Answer: (B)

As we can see there is small difference in between $IP_{1},IP_{2},IP_{3}$ which means removal of three electron will be easy, but when we go for removal of fourth electron then we find ionisation enthalpy that is $IP_{4}$ is increased by a huge amount. This increment in energy indicates that electron of the next shell was released or in other words, we can say it has three electron in its valence shell.
So answer is:
$1s^{2},2s^{2},2p^{6},3s^{2},3p^{1}$