The first,second and third ionisation energie | vedclass

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

Question

(D) The stability of an oxidation state is determined by the jump in ionization energy.Given: $E_1 = 7 \ eV$,$E_2 = 12.5 \ eV$,and $E_3 = 42.5 \ eV$.T

Vedclass · Accepted Answer

$+2$

Vedclass · Answer

(D) The stability of an oxidation state is determined by the jump in ionization energy.Given: $E_1 = 7 \ eV$,$E_2 = 12.5 \ eV$,and $E_3 = 42.5 \ eV$.The jump from $E_2$ to $E_3$ is $42.5 - 12.5 = 30 \ eV$,which is significantly large compared to the jump from $E_1$ to $E_2$ $(12.5 - 7 = 5.5 \ eV)$.$A$ large jump in ionization energy indicates that the electron is being removed from a stable,noble gas-like configuration after the removal of two electrons.Therefore,the element most easily loses two electrons to achieve a stable configuration,making the $+2$ oxidation state the most stable.Hence,$D$ is the correct answer.

The first,second and third ionisation energies ($E_1, E_2$ and $E_3$) for an element are $7 \ eV$,$12.5 \ eV$ and $42.5 \ eV$ respectively. The most stable oxidation state of the element will be

Similar Questions

The first four ionization energy values of an element are $191$,$578$,$872$ and $5962 \ kcal$. The number of valence electrons in the element is

Which of the following gaseous atoms has the highest value of $IE$?

Which among the following elements has the lowest value of $IE_1$?

The correct order of first ionisation enthalpy of group-$13$ elements is

Among $Be$,$B$,$Mg$ and $Al$,the second ionisation potential is maximum for

Vedclass Test Series

Exam Paper Generator

Online Exam Module