Consider the following elements In, Tl, Al, Pb, and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are:
Show Hint
- +2 and +3
- +4 and +1
- +4 and +3
- +1 and +4
The Correct Option is B
Solution and Explanation
Given elements are: Indium (In), Thallium (Tl), Aluminum (Al), Lead (Pb), Tin (Sn), and Germanium (Ge).
We need to determine the most stable oxidation states of the elements with the highest and lowest first ionization enthalpies from this group.
Position in the periodic table:
- Aluminum (Al): Group 13, Period 3
- Germanium (Ge): Group 14, Period 4
- Tin (Sn): Group 14, Period 5
- Indium (In): Group 13, Period 5
- Lead (Pb): Group 14, Period 6
- Thallium (Tl): Group 13, Period 6
Ionization enthalpy trends:
- Ionization enthalpy generally increases across a period and decreases down a group.
- Among these elements, **Aluminum (Al)** has the highest first ionization enthalpy because it is the lightest and topmost element in its group.
- Thallium (Tl) has the lowest first ionization enthalpy due to its position in Period 6, where the inert pair effect becomes significant.
Oxidation states:
Aluminum (Al), with the highest ionization enthalpy, typically shows a +3 oxidation state. This is the most stable oxidation state for Al, as it readily loses all three valence electrons.
Thallium (Tl), with the lowest ionization enthalpy, shows both +3 and +1 oxidation states. However, due to the **inert pair effect**, the +1 oxidation state is more stable than +3.
This trend is also seen in other heavier p-block elements like Pb and Sn, where the +2 oxidation state becomes more stable compared to the +4 state because of the inert pair effect.
Conclusion:
- The element with the highest first ionization enthalpy is **Aluminum (Al)**, which has a most stable oxidation state of **+3**.
- The element with the lowest first ionization enthalpy is **Thallium (Tl)**, which has a most stable oxidation state of **+1**.
Final Answer:
The final answer is $ (2)\ +4\ \text{and}\ +1 $.
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