All Zn(+II) compounds are white because:

Options

A. `Zn^(2+)` has a `d^(10)` configuration and the d-subshell is full
B. `Zn^(2+)` shows d-d transition
C. `Zn^(2+)` has no electron in the 4s-subshell
D. Zn is not a transition element

Answer

To solve the question "All Zn(+II) compounds are white because:", we will analyze the options provided and understand the underlying concepts related to the electronic configuration of zinc.

### Step-by-Step Solution:

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

2. **Determine the Configuration of Zn(+II):**
- When zinc loses two electrons to form Zn(+II), the electrons are removed from the 4s orbital first, followed by the 3d orbital. Thus, the electronic configuration of Zn(+II) is:
\[ \text{Zn}^{2+}: [\text{Ar}] 3d^{10} 4s^0 \]

3. **Analyze the d-Orbital Configuration:**
- In Zn(+II), the 3d subshell is completely filled with 10 electrons (3d^10). This means all the electrons in the d-orbitals are paired.

4. **Relate Electron Configuration to Color:**
- The color of a compound is often due to the presence of unpaired electrons in the d-orbitals, which can undergo d-d transitions when light is absorbed. Since Zn(+II) has a completely filled d-orbital (d^10), there are no unpaired electrons available for such transitions.

5. **Conclusion:**
- Because there are no unpaired electrons in Zn(+II), it does not absorb visible light and thus appears colorless (or white in the case of solid compounds). Therefore, the correct answer to the question is that all Zn(+II) compounds are white because Zn(+II) has a d10 configuration.

### Final Answer:
All Zn(+II) compounds are white because Zn(+II) has a d10 configuration.

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