Calculate the screening constant of `Li` for valency electrons.

Explanation

To calculate the screening constant (σ) for the valence electrons of lithium (Li), we will follow these steps:

### Step 1: Understand the Electronic Configuration
Lithium has an atomic number of 3, which means it has 3 electrons. The electronic configuration of lithium is:
- 1s² 2s¹

This indicates that there are two electrons in the 1s subshell and one electron in the 2s subshell.

### Step 2: Identify the Shells
For the valence electron in the 2s subshell, we need to consider the electrons in the inner shells:
- The 1s shell contains 2 electrons (n=1).

### Step 3: Apply the Screening Constant Formula
The screening constant (σ) can be calculated using the formula:
\[ σ = 0.35 \times (n_{shell}) + 0.85 \times (n_{-1 shell}) + 1 \times (n_{inertial}) \]
Where:
- \( n_{shell} \) = number of electrons in the same shell (2s)
- \( n_{-1 shell} \) = number of electrons in the previous shell (1s)
- \( n_{inertial} \) = number of inertial electrons (not applicable here)

### Step 4: Calculate Each Component
1. **For the 2s shell (n_shell)**: There are no other electrons in the 2s shell, so \( n_{shell} = 0 \).
2. **For the 1s shell (n_-1 shell)**: There are 2 electrons in the 1s shell, so \( n_{-1 shell} = 2 \).
3. **Inertial electrons**: There are no inertial electrons in this case, so \( n_{inertial} = 0 \).

### Step 5: Substitute Values into the Formula
Now, substituting the values into the formula:
\[ σ = 0.35 \times 0 + 0.85 \times 2 + 1 \times 0 \]
\[ σ = 0 + 1.7 + 0 \]
\[ σ = 1.7 \]

### Final Answer
The screening constant (σ) for the valence electrons of lithium (Li) is **1.7**.

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