JEE Chemistry 2025 Question with Solution

Given: Group $14$ elements A and B have first ionisation enthalpy values $708 \, \text{kJ mol}^{-1}$ and $715 \, \text{kJ mol}^{-1}$ respectively, and these are the lowest in the group.

Find: The nature of ions $A^{2+}$ and $B^{4+}$ as reducing or oxidising agents.

Ionisation enthalpy is the energy required to remove an electron from a neutral gaseous atom. Lower ionisation enthalpy means the element loses electrons more easily.

From the given values and the group trend, the two elements correspond to the heavier group $14$ members, namely Sn and Pb.

Thus,

$$A = \text{Sn}, \qquad B = \text{Pb}$$

So the ions are

$$A^{2+} = \text{Sn}^{2+}, \qquad B^{4+} = \text{Pb}^{4+}$$

Now,

• $\text{Sn}^{2+}$ can lose electrons further to form $\text{Sn}^{4+}$, so it behaves as a reducing agent.
• $\text{Pb}^{4+}$ tends to gain electrons to form $\text{Pb}^{2+}$, so it behaves as an oxidising agent.

Therefore, the nature of $A^{2+}$ and $B^{4+}$ respectively is reducing and oxidising.

The correct option is C.

Given: The first ionisation enthalpies are $708 \, \text{kJ mol}^{-1}$ and $715 \, \text{kJ mol}^{-1}$, and these are the lowest among the group $14$ elements.

Find: Whether $A^{2+}$ and $B^{4+}$ are reducing or oxidising.

Group $14$ contains carbon, silicon, germanium, tin and lead. The lowest first ionisation enthalpy values in this group belong to the heavier elements.

Using the given values,

$$708 \, \text{kJ mol}^{-1} \approx \text{Sn}, \qquad 715 \, \text{kJ mol}^{-1} \approx \text{Pb}$$

Hence,

$$A = \text{Sn}, \qquad B = \text{Pb}$$

Now examine their ions:

1. $\text{Sn}^{2+}$ can be oxidised to $\text{Sn}^{4+}$.
2. A species that gets oxidised itself acts as a reducing agent.
3. $\text{Pb}^{4+}$ can be reduced to $\text{Pb}^{2+}$.
4. A species that gets reduced itself acts as an oxidising agent.

So,

$$A^{2+} : \text{reducing}, \qquad B^{4+} : \text{oxidising}$$

Therefore, the correct answer is reducing and oxidising, that is, option C.