Question

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:

• A. +2 and +3
• B. +4 and +1
• C. +4 and +3
• D. +1 and +4

Hint:

Ionisation enthalpy trends in the periodic table influence the common oxidation states of elements, particularly for p-block elements where the inert pair effect is significant.

Answer 1

The Correct Option is B

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 $.

Answer 2

Step 1: Understand the given question.
We are given the elements: In (Indium), Tl (Thallium), Al (Aluminium), Pb (Lead), and Ge (Germanium). We have to determine the most stable oxidation states of the elements having the highest and lowest first ionisation enthalpies among these elements.

Step 2: Recall the concept of ionisation enthalpy.
Ionisation enthalpy is the energy required to remove the most loosely bound electron from a gaseous atom. In general, within a group, the ionisation enthalpy decreases down the group due to the increase in atomic size and shielding effect. Across a period, it increases due to increased nuclear charge.

Step 3: Identify positions of the given elements.
- Al (Aluminium) → Group 13, Period 3
- In (Indium) → Group 13, Period 5
- Tl (Thallium) → Group 13, Period 6
- Pb (Lead) → Group 14, Period 6
- Ge (Germanium) → Group 14, Period 4

Step 4: Compare their first ionisation enthalpies.
Among these elements:
- Aluminium and Germanium have relatively high ionisation enthalpies because they are higher up in their respective groups.
- Thallium has the lowest ionisation enthalpy because it lies much lower in the periodic table and experiences strong shielding from inner d and f electrons.

Thus:
- Highest ionisation enthalpy → Germanium (Ge)
- Lowest ionisation enthalpy → Thallium (Tl)

Step 5: Identify the most stable oxidation states.
- For Germanium (Ge) (Group 14), the most stable oxidation state is **+4**.
- For Thallium (Tl) (Group 13), the most stable oxidation state is **+1** due to the inert pair effect (the s-electrons remain non-bonding).

Step 6: Conclusion.
- The element with the highest ionisation enthalpy (Ge) has the stable oxidation state **+4**.
- The element with the lowest ionisation enthalpy (Tl) has the stable oxidation state **+1**.

Final Answer:
\[ \boxed{+4 \text{ and } +1} \]