J.C. Slater proposed an empirical constant that represents the cumulative extent to which the other electrons of an atom shield (or screen) any particular electron from the nuclear charge. Thus, slater's screening contant `sigma` is as : `Z^(**)=Z-sigma`
Here, Z is the atomic number of the atom, and hence is equal to the actual number of protons in the atom. the parameter `Z^(**)` is the effective nuclear charge, which according to is smaller than Z, since the electron in question is screened (shielded) from Z by an amount `sigma`. Conversely, an electron that is well shielded from the nuclear charge Z experiences a small effective nuclear charge `Z^(**)`.
The value of `sigma` for any one electron in a given electron configuration (i.e., in the presence of the other electrons of the atom in question) is calculated using a set of empirical rules developed by slater. according to these rules, the value of `sigma` for the electron in question is the cumulative total provided by the various other electrons of the atom.
Q. Which of the following statement is correct?

Options

A. A 4s-orbital is filled earlier than a 3d-orbital because, `Z^(**)` for `3d gt Z^(**)` for 4s.
B. A 4s-orbital is filled earlier than a 3d-orbital because, `Z^(**)` for `4s gt Z^(**)` for 3d
C. The effective nuclear charge for 3d-and 4s-orbitals are same, but energy of 3d-orbital becomes higher.
D. The effective nuclear charge for 3d and 4s-orbitals are same but, energy of 4s-orbital becomes higher.

Answer

To solve the question regarding the effective nuclear charge (Z^**) and the screening constant (σ) as proposed by J.C. Slater, we will analyze the given statements based on the calculations of Z effective for the 4s and 3d orbitals of scandium (atomic number 21).

### Step-by-step Solution:

1. **Understanding Z effective**:
- The effective nuclear charge (Z^**) is calculated using the formula:
\[ Z^{**} = Z - \sigma \]
- Here, Z is the atomic number (number of protons), and σ is the screening constant that accounts for the shielding effect of other electrons.

2. **Electronic Configuration of Scandium**:
- The electronic configuration of Scandium (Sc) is:
\[ 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^1 \]
- This means Sc has a total of 21 electrons.

3. **Calculating σ for 4s Electrons**:
- For the 4s electrons, we apply Slater's rules:
- For electrons in the same group (n=4), σ = 0.35 per electron.
- For electrons in the n-1 shell (n=3), σ = 0.85 per electron.
- For inner electrons (n=2 and below), σ = 1 per electron.
- Calculation:
- There are 2 electrons in 4s: \( 2 \times 0.35 = 0.70 \)
- There are 9 electrons in 3s and 3p: \( 9 \times 0.85 = 7.65 \)
- There are 10 inner electrons: \( 10 \times 1 = 10 \)
- Total σ for 4s:
\[ \sigma = 0.70 + 7.65 + 10 = 18.35 \approx 18 \]

4. **Calculating Z effective for 4s**:
- Using the values calculated:
\[ Z^{**}_{4s} = 21 - 18 = 3 \]

5. **Calculating σ for 3d Electron**:
- For the 3d electron:
- There is 1 electron in 3d: \( 0 \times 0.35 = 0 \)
- There are 18 inner electrons: \( 18 \times 1 = 18 \)
- Total σ for 3d:
\[ \sigma = 0 + 18 = 18 \]

6. **Calculating Z effective for 3d**:
- Using the values calculated:
\[ Z^{**}_{3d} = 21 - 18 = 3 \]

7. **Comparing Z effective**:
- We find that:
\[ Z^{**}_{4s} = Z^{**}_{3d} = 3 \]
- Therefore, the effective nuclear charge for both orbitals is the same.

8. **Energy Consideration**:
- According to the n + l rule:
- For 3d: n = 3, l = 2 → n + l = 5
- For 4s: n = 4, l = 0 → n + l = 4
- Since 3d has a higher n + l value, it has higher energy than 4s.

9. **Evaluating the Statements**:
- **Option 1**: Incorrect, as Z effective for 3d and 4s is the same.
- **Option 2**: Incorrect, as Z effective for both is the same.
- **Option 3**: Correct, as Z effective is the same but the energy of 3d is higher.
- **Option 4**: Incorrect, as the energy of 3d is higher, not 4s.

### Conclusion:
The correct statement is **Option 3**: The effective nuclear charge for 3d and 4s orbitals are the same, but the energy of the 3d orbital becomes higher.