Question

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

• A. both reducing
• B. both oxidising
• C. reducing and oxidising
• D. oxidising and reducing

Hint:

In general, elements with low ionization enthalpies are reducing agents because they can easily lose electrons, while elements with high ionization enthalpies tend to attract electrons and thus act as oxidizing agents.

Answer 1

The Correct Option is C

Step 1: Understanding Ionization Enthalpy
Ionization enthalpy refers to the energy required to remove an electron from a neutral atom in its gaseous state. Lower ionization enthalpy indicates that the element can lose electrons more easily. The elements A and B have the lowest ionization enthalpy values in their group. This suggests that they are relatively easier to ionize compared to other members of their group. 
Step 2: Analyzing the Nature of Their Ions
Element A: Since A has a low ionization enthalpy, it will lose electrons easily and thus, the A\(^{2+}\) ion will readily accept electrons (reduced). Hence, element A behaves as a 
reducing agent
Element B: On the other hand, element B has a slightly higher ionization enthalpy compared to A. The B\(^{4+}\) ion tends to attract electrons, which makes it a strong 
oxidizing agent
Thus, the nature of their ions is reducing (for A\(^{2+}\)) and oxidizing (for B\(^{4+}\)).

Answer 2

The question asks about the nature of ions A²⁺ and B⁴⁺ from group 14 elements based on their ionization enthalpy. To answer this, we need to understand the relationship between ionization enthalpy, group properties, and the behavior of ions as reducing or oxidizing agents.

Step 1: Understand the Context

Group 14 elements are carbon (C), silicon (Si), germanium (Ge), tin (Sn), and lead (Pb). The ionization enthalpy values provided (708 and 715 kJ mol^-1) are the lowest among their group, indicating that A and B are the heavier elements at the bottom of the group. Typically, among these elements, tin (Sn) and lead (Pb) have the lowest ionization enthalpy.

Step 2: Identifying Elements and Their Ions

Given the ionization enthalpies, it is reasonable to assume:

• A is Sn with ionization enthalpy approximately 708 kJ mol^-1.
• B is Pb with ionization enthalpy approximately 715 kJ mol^-1.

We need to evaluate the nature of Sn²⁺ and Pb⁴⁺ ions.

Step 3: Evaluating the Ions as Reducing or Oxidizing Agents

The reducing or oxidizing nature of an ion depends on its ability to donate or accept electrons:

• Sn²⁺ as a Reducing Agent: Sn²⁺ can lose two more electrons to form Sn⁴⁺, making it a reducing agent as it can donate electrons.
• Pb⁴⁺ as an Oxidizing Agent: On the other hand, Pb⁴⁺ can accept electrons to form Pb²⁺, thereby acting as an oxidizing agent.

Step 4: Conclusion

The ions Sn²⁺ and Pb⁴⁺ align as reducing and oxidizing agents, respectively. Thus, the correct answer is:

Reducing and Oxidizing

This choice is justified by the tendency of Sn²⁺ to oxidize itself (act as a reducing agent) and Pb⁴⁺ to reduce itself (act as an oxidizing agent). Such behavior is typical for these ions based on their position in the periodic table and the stability of their oxidation states.