Q17 / 75 · JEE Main 2025 (24 Jan Shift 2) · Chemistry

The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824 \mathrm{~kJ} / \mathrm{mol}$, respectively. By using the above values predict the group in which the above element is present :

  • Correct answer
    A
    Group 13
  • B
    Group 14
  • C
    Group 2
  • D
    Group 4

Solution

The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824 \mathrm{~kJ} / \mathrm{mol}$, respectively. $\begin{aligned} & \frac{\mathrm{IE}_2}{\mathrm{IE}_1}=\frac{2427}{800}=3.03 \\ & \frac{\mathrm{IE}}{\mathrm{IE}_2}=\frac{3658}{2427}=1.51 \\ & \frac{\mathrm{IE}}{\mathrm{IE}_3}=\frac{25024}{3658}=6.84 \end{aligned}$ $\frac{\mathrm{IE}_5}{\mathrm{IE}_4}=\frac{32824}{25024}=1.31$ Since ( $\mathrm{IE}_4 / \mathrm{IE}_3$ ) value is maximum, the element belongs to group 13.

Hence, the correct answer is option: 'A'.

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