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:

  • A
    $A < B < C < D$
  • B
    $B < A < D < C$
  • C
    $B < A < C < D$
  • D
    $B < D < A < C$

Explore More

Similar Questions

The elements of Group $13$ with highest and lowest first ionisation enthalpies are respectively$:$

Ionization energy increases in the order

The electronic configuration of elements $A, B$ and $C$ are $[He] 2s^1, [Ne] 3s^1$ and $[Ar] 4s^1$ respectively. Which one of the following orders is correct for the first ionization potentials (in $kJ \ mol^{-1}$) of $A, B$ and $C$?

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

The ionization potentials of $Li$ and $K$ are $5.4 \ eV$ and $4.3 \ eV$ respectively. The ionization potential of $Na$ will be .............. $eV$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo