`K_(4)[Fe(CN)_(6)]` can be used to detect one or more out of `Fe^(2+), Fe^(3+). Zn^(2+), Cu^(2+), Ca^(2+)`:

Options

A. only `Fe^(2+),Fe^(3+)`
B. only `Fe^(3+), Zn^(2+), Cu^(2+)`
C. all but not `Ca^(2+)`
D. All of these

Explanation

To determine which ions can be detected by potassium ferrocyanide, \( K_4[Fe(CN)_6] \), we will analyze the reactions of the given ions with the reagent. The ions to be tested are \( Fe^{2+}, Fe^{3+}, Zn^{2+}, Cu^{2+}, \) and \( Ca^{2+} \).

### Step-by-Step Solution:

1. **Testing \( Fe^{2+} \)**:
- When \( Fe^{2+} \) reacts with \( K_4[Fe(CN)_6] \), it forms \( K_2[Fe(CN)_6] \), which is a white precipitate.
- **Conclusion**: \( Fe^{2+} \) can be detected by \( K_4[Fe(CN)_6] \).

2. **Testing \( Fe^{3+} \)**:
- When \( Fe^{3+} \) reacts with \( K_4[Fe(CN)_6] \), it forms \( [Fe_3[Fe(CN)_6]_2] \), which is a blue precipitate known as Prussian blue.
- **Conclusion**: \( Fe^{3+} \) can also be detected by \( K_4[Fe(CN)_6] \).

3. **Testing \( Zn^{2+} \)**:
- When \( Zn^{2+} \) reacts with \( K_4[Fe(CN)_6] \), it forms \( Zn_2[Fe(CN)_6] \), which is a white precipitate.
- **Conclusion**: \( Zn^{2+} \) can be detected by \( K_4[Fe(CN)_6] \).

4. **Testing \( Cu^{2+} \)**:
- When \( Cu^{2+} \) reacts with \( K_4[Fe(CN)_6] \), it forms \( Cu_2[Fe(CN)_6] \), which is a chocolate brown precipitate.
- **Conclusion**: \( Cu^{2+} \) can be detected by \( K_4[Fe(CN)_6] \).

5. **Testing \( Ca^{2+} \)**:
- When \( Ca^{2+} \) reacts with \( K_4[Fe(CN)_6] \), it forms \( Ca_2[Fe(CN)_6] \), which does not form a precipitate.
- **Conclusion**: \( Ca^{2+} \) cannot be detected by \( K_4[Fe(CN)_6] \).

### Final Summary:
- The ions that can be detected by \( K_4[Fe(CN)_6] \) are:
- \( Fe^{2+} \)
- \( Fe^{3+} \)
- \( Zn^{2+} \)
- \( Cu^{2+} \)
- The ion that cannot be detected is:
- \( Ca^{2+} \)