Which of the following have higher `Z_(eff)` than Fluorine.

Options

A. `CI`
B. `O`
C. `F^(-)`
D. none of these

Answer

To determine which elements have a higher effective nuclear charge (Z_eff) than fluorine, we need to understand the concept of Z_eff and how to calculate it.

**Step 1: Understanding Z_eff**
Z_eff is the effective nuclear charge experienced by an electron in an atom. It can be calculated using the formula:
\[ Z_{eff} = Z - S \]
where:
- \( Z \) is the atomic number (the number of protons in the nucleus).
- \( S \) is the shielding constant, which represents the extent to which inner electrons shield the outer electrons from the nuclear charge.

**Step 2: Determine Z and S for Fluorine**
Fluorine has an atomic number \( Z = 9 \). The electron configuration of fluorine is \( 1s^2 2s^2 2p^5 \).

- The inner shell (1s) has 2 electrons, which shield the 7 valence electrons (2s^2 2p^5).
- The shielding constant \( S \) for fluorine can be estimated using Slater's rules, which gives \( S \approx 2 \).

Thus, for fluorine:
\[ Z_{eff} = 9 - 2 = 7 \]

**Step 3: Compare with Other Elements**
Now, we need to compare the Z_eff of fluorine with other elements. We will check elements in the same period (Period 2) and those in Period 3 or higher.

1. **Oxygen (O)**:
- \( Z = 8 \), \( S \approx 1.8 \)
- \( Z_{eff} = 8 - 1.8 = 6.2 \) (lower than fluorine)

2. **Nitrogen (N)**:
- \( Z = 7 \), \( S \approx 1.4 \)
- \( Z_{eff} = 7 - 1.4 = 5.6 \) (lower than fluorine)

3. **Neon (Ne)**:
- \( Z = 10 \), \( S \approx 2.0 \)
- \( Z_{eff} = 10 - 2.0 = 8 \) (higher than fluorine)

4. **Sodium (Na)**:
- \( Z = 11 \), \( S \approx 2.3 \)
- \( Z_{eff} = 11 - 2.3 = 8.7 \) (higher than fluorine)

5. **Chlorine (Cl)**:
- \( Z = 17 \), \( S \approx 10 \)
- \( Z_{eff} = 17 - 10 = 7 \) (equal to fluorine)

**Step 4: Conclusion**
From the above calculations, the elements that have a higher Z_eff than fluorine are Neon (Ne) and Sodium (Na).

**Final Answer:**
Neon (Ne) and Sodium (Na) have higher Z_eff than Fluorine.

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