Lowest `I^(st)` IP is of : -

Options

A. Pb
B. Si
C. Sn
D. C

Answer

To determine which element has the lowest first ionization potential (IP) among the group 14 elements (carbon, silicon, germanium, tin, and lead), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Group**:
- The elements in question belong to group 14 of the periodic table: Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), and Lead (Pb).

2. **Understand Ionization Potential**:
- Ionization potential (or ionization energy) is the energy required to remove the most loosely bound electron from an isolated gaseous atom.

3. **Trend in Ionization Potential**:
- Generally, ionization potential decreases as we move down a group in the periodic table. This is because the atomic size increases, leading to a weaker attraction between the nucleus and the outermost electrons.

4. **Consider Atomic Size**:
- As we move from carbon to lead, the atomic size increases. Therefore, we would expect the ionization potential to decrease.

5. **Examine Specific Elements**:
- While lead (Pb) is at the bottom of the group and typically would have the lowest ionization potential due to its larger size, we must consider the effect of **lanthanide contraction**.

6. **Lanthanide Contraction**:
- The presence of lanthanides causes a contraction in the size of the atoms. This means that although lead is larger than tin (Sn), the effective nuclear charge experienced by the outer electrons in lead is higher due to the increased charge density.

7. **Compare Tin and Lead**:
- Tin (Sn) has a larger atomic radius than lead (Pb) due to lanthanide contraction, which results in a lower ionization potential for tin compared to lead.

8. **Conclusion**:
- Therefore, the element with the lowest first ionization potential in group 14 is tin (Sn).

### Final Answer:
The element with the lowest first ionization potential is **Tin (Sn)**.

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