Which of the following have greater `Z_(eff)` than `Zn:`

Options

A. `Cu^(+)`
B. `Cu^(2+)`
C. `Fe^(3+)`
D. `Zn^(2+)`

Answer

To determine which of the given ions have a greater effective nuclear charge (Z_eff) than zinc (Zn), we will analyze the electronic configurations of each ion and compare them to that of Zn.

### Step-by-Step Solution:

1. **Identify the Electronic Configuration of Zn:**
- Zinc (Zn) has the atomic number 30 and its electronic configuration is:
\[ \text{Zn: } [\text{Ar}] 4s^2 3d^{10} \]

2. **Determine the Electronic Configurations of the Given Ions:**
- Copper (Cu) has two oxidation states to consider:
- Copper(I) ion (Cu^+):
\[ \text{Cu}^+: [\text{Ar}] 4s^0 3d^{10} \]
- Copper(II) ion (Cu^{2+}):
\[ \text{Cu}^{2+}: [\text{Ar}] 4s^0 3d^9 \]
- Iron(III) ion (Fe^{3+}):
\[ \text{Fe}^{3+}: [\text{Ar}] 4s^0 3d^5 \]
- Zinc ion (Zn^{2+}):
\[ \text{Zn}^{2+}: [\text{Ar}] 4s^0 3d^{10} \]

3. **Analyze the Screening Effect:**
- The effective nuclear charge (Z_eff) is influenced by the number of electrons in the outer shell and the inner shell electrons that shield the nuclear charge.
- Zn has 30 protons and 30 electrons, leading to a certain amount of screening.
- The ions we are comparing have fewer electrons than Zn, which means they experience less electron-electron repulsion and more effective nuclear charge.

4. **Comparison of Z_eff:**
- For Cu^+, Cu^{2+}, Fe^{3+}, and Zn^{2+}, since they all have fewer electrons than Zn, they will have a greater Z_eff due to reduced shielding.
- Specifically:
- Cu^+: 29 electrons (greater Z_eff than Zn)
- Cu^{2+}: 28 electrons (greater Z_eff than Zn)
- Fe^{3+}: 26 electrons (greater Z_eff than Zn)
- Zn^{2+}: 28 electrons (greater Z_eff than Zn)

5. **Conclusion:**
- All the given ions (Cu^+, Cu^{2+}, Fe^{3+}, and Zn^{2+}) have a greater effective nuclear charge (Z_eff) than Zn.

### Final Answer:
All of the given options (A, B, C, and D) have a greater Z_eff than Zn.

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