An element has successive ionization enthalpies as $940 \, kJ \, mol^{-1}$ (first),$2080 \, kJ \, mol^{-1}$,$3090 \, kJ \, mol^{-1}$,$4140 \, kJ \, mol^{-1}$,$7030 \, kJ \, mol^{-1}$,$7870 \, kJ \, mol^{-1}$,$16000 \, kJ \, mol^{-1}$ and $19500 \, kJ \, mol^{-1}$. To which group of the periodic table does this element belong?

  • A
    $14$
  • B
    $15$
  • C
    $16$
  • D
    $17$

Explore More

Similar Questions

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

Assertion : Element has a tendency to lose the electron$(s)$ to attain the stable configuration.
Reason : Ionization enthalpy is the energy released to remove an electron from an isolated gaseous atom in its ground state.

The electronic configuration with the highest ionization enthalpy is

The ionisation energy of nitrogen is more than oxygen because of $:-$

Mercury is the only metal which is liquid at $0 \, ^{\circ}C$. This is due to its

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