In which of the following electronic configur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The ionization energy increases significantly when an electron is removed from a stable,noble gas-like configuration or a completely filled shell.

Vedclass · Accepted Answer

$1s^2\, 2s^2\, 2p^6\, 3s^2$

Vedclass · Answer

(D) The ionization energy increases significantly when an electron is removed from a stable,noble gas-like configuration or a completely filled shell.For option $D$,the configuration is $1s^2\, 2s^2\, 2p^6\, 3s^2$ (Magnesium,$Z=12$).The first ionization energy removes one $3s$ electron.The second ionization energy removes the second $3s$ electron.After removing two electrons,the configuration becomes $1s^2\, 2s^2\, 2p^6$,which is the stable neon noble gas configuration.Therefore,removing the third electron requires a very high amount of energy,leading to a large difference between the second and third ionization energies.

In which of the following electronic configurations will there be a large difference between the second and third ionization energy?

Similar Questions

Outermost electronic configurations of four elements $A, B, C, D$ are given below:
$A: 3s^{2}$
$B: 3s^{2} 3p^{1}$
$C: 3s^{2} 3p^{3}$
$D: 3s^{2} 3p^{4}$
The correct order of first ionization enthalpy for them is:

Consider the elements $Mg$,$Al$,$S$,$P$ and $Si$. The correct increasing order of their first ionization enthalpy is:

The order of ionisation potential between $He^{+}$ ion and $H$ atom (both species are in gaseous state) is

The correct order of $IE_2$ is

Starting from the $1^{st}$,the successive ionization potentials of an element are respectively $5.98, 18.8, 28.4, 120.1, 154 \ eV$. The element is

Vedclass Test Series

Exam Paper Generator

Online Exam Module