The group 14 elements A and B have the first ionisation enthalpy values of $708$ and $715 \mathrm{~kJ}…

The group 14 elements A and B have the first ionisation enthalpy values of $708$ and $715 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions $\mathrm{A}^{2+}$ $\mathrm{B}^{4+}$ respectively is
  1. both reducing
  2. both oxidising
  3. reducing and oxidising
  4. oxidising and reducing

Solution

As per given information of ionisation energy
$\mathrm{A}=\mathrm{Sn} \& \mathrm{~B}=\mathrm{Pb}$
$\begin{aligned} & \mathrm{A}^{+2}=\mathrm{Sn}^{2+}=\text { Reducing agent } \\ & \mathrm{B}^{+4}=\mathrm{Pb}^{+4}=\text { Oxidising agent }\end{aligned}$

Asked in: JEE Main 2025 (07 Apr Shift 1)

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