First ionization energy is the lowest with:

Options

A. Lead
B. Carbon
C. Silicon
D. Tin

Answer

To determine which element has the lowest first ionization energy among lead (Pb), carbon (C), silicon (Si), and tin (Sn), we can follow these steps:

### Step 1: Understand Ionization Energy
Ionization energy is the energy required to remove an outermost electron from an atom. Generally, ionization energy decreases down a group in the periodic table and increases across a period.

### Step 2: Identify the Elements and Their Electronic Configurations
- **Lead (Pb)**: Atomic number 82, electronic configuration: [Xe] 4f14 5d10 6s2 6p2
- **Carbon (C)**: Atomic number 6, electronic configuration: [He] 2s2 2p2
- **Silicon (Si)**: Atomic number 14, electronic configuration: [Ne] 3s2 3p2
- **Tin (Sn)**: Atomic number 50, electronic configuration: [Kr] 4d10 5s2 5p2

### Step 3: Analyze the Position of Each Element in the Periodic Table
- **Carbon** and **Silicon** are in the second and third periods, respectively, and have relatively high ionization energies due to their smaller atomic radii and higher effective nuclear charge.
- **Tin** is in the fifth period, and its ionization energy is lower than that of carbon and silicon but higher than lead.
- **Lead**, being in the sixth period, has more electron shells, which means that the outermost electrons are further from the nucleus and experience more shielding. This generally leads to lower ionization energy.

### Step 4: Compare the Ionization Energies
- As we move down the group from carbon to lead, the ionization energy decreases.
- Among the given elements, lead (Pb) has the lowest ionization energy due to its larger atomic size and the shielding effect of the inner electrons.

### Conclusion
Thus, the element with the lowest first ionization energy is **Lead (Pb)**.