If the successive ionisation energies of an e | vedclass

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

Question

(D) The successive ionisation energies are given as: $IE_1 = 165 \ kcal$,$IE_2 = 190 \ kcal$,$IE_3 = 550 \ kcal$,and $IE_4 = 595 \ kcal$. There is a l

Vedclass · Accepted Answer

$[Ne] 3s^2$

Vedclass · Answer

(D) The successive ionisation energies are given as: $IE_1 = 165 \ kcal$,$IE_2 = 190 \ kcal$,$IE_3 = 550 \ kcal$,and $IE_4 = 595 \ kcal$. There is a large jump in ionisation energy between the second $(IE_2)$ and third $(IE_3)$ ionisation energies ($190 \ kcal$ to $550 \ kcal$). This significant increase indicates that the first two electrons are removed from the valence shell,and the third electron is removed from a more stable,inner shell. Therefore,the element $A$ has $2$ valence electrons. Among the given options,the configuration $[Ne] 3s^2$ corresponds to an element with $2$ valence electrons.

If the successive ionisation energies of an element $A$ are $165$,$190$,$550$ and $595 \ kcal$,respectively,then the ground state electronic configuration of element $A$ is

Similar Questions

Which of the following metals is expected to have the highest third ionization enthalpy?

Assertion $(A)$: Boron has a smaller first ionisation enthalpy than beryllium.
Reason $(R)$: The penetration of a $2s$-electron to the nucleus is more than the $2p$-electron; hence,the $2p$-electron is more shielded by the inner core of electrons than $2s$-electrons.

The electronic configuration of an element with the largest difference between the $1^{st}$ and $2^{nd}$ ionisation energies is

What is second ionization enthalpy?

The successive ionisation enthalpies of an element $X$ in $(kJ \ mol^{-1})$ are $1012$,$1907$,$2955$,$4955$,$6275$ and $21,260$ respectively. The element $X$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module