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\(^{4+}\)).