Identify the elements from the given first ionisation enthalpies

In group 14, the elements are: $\text{C, Si, Ge, Sn, Pb}$

The given first ionisation enthalpies are approximately:

• $708\ \text{kJ mol}^{-1}$
• $715\ \text{kJ mol}^{-1}$

These are the lowest among group 14 members, so they must correspond to the heavier elements.

Known approximate values are:

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

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

Find the nature of $A^{2+}$ and $B^{4+}$

So we need the nature of: $\text{Sn}^{2+} \quad \text{and} \quad \text{Pb}^{4+}$

Use inert pair effect in group 14

In heavier group 14 elements, the oxidation state $+2$ becomes more stable down the group due to the inert pair effect.

• For tin, both $+2$ and $+4$ states are possible, but $\text{Sn}^{2+}$ tends to get oxidised to $\text{Sn}^{4+}$.
• Hence $\text{Sn}^{2+}$ acts as a reducing agent.

So, $\text{Sn}^{2+} : \text{reducing}$

Nature of $\text{Pb}^{4+}$

For lead, due to stronger inert pair effect, the $+2$ state is more stable than $+4$.

Therefore $\text{Pb}^{4+}$ tends to get reduced to $\text{Pb}^{2+}$.

Hence $\text{Pb}^{4+}$ acts as an oxidising agent.

So, $\text{Pb}^{4+} : \text{oxidising}$

Match with the options

• $A^{2+}$ i.e. $\text{Sn}^{2+}$ is reducing
• $B^{4+}$ i.e. $\text{Pb}^{4+}$ is oxidising

Therefore the correct option is: $\boxed{\text{D: reducing and oxidising}}$

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, they agree.