Hydrogen has the tendency to lose one `e^(-)` and formation of `H^(+)`, In this respect it resembles with :

Options

A. Alkali metal
B. Carbon
C. Alkaline earth metal
D. Halogens

Answer

To solve the question, "Hydrogen has the tendency to lose one electron and formation of H^(+). In this respect, it resembles with:", we will analyze the given options one by one.

### Step-by-Step Solution:

1. **Understanding Hydrogen's Behavior**:
- Hydrogen can lose one electron to form a cation, H^(+). This property is crucial in determining which element it resembles.

2. **Analyzing the Options**:
- **Option A: Alkali Metals**:
- Alkali metals (like Li, Na, K) have one electron in their outermost shell. They tend to lose this one electron easily to form cations (e.g., Na^(+)).
- Since hydrogen also loses one electron to form H^(+), it resembles alkali metals in this aspect.

- **Option B: Carbon**:
- Carbon typically forms covalent bonds and does not lose electrons to form cations like hydrogen does. Instead, it can share electrons or gain electrons to complete its valence shell.
- Therefore, carbon does not resemble hydrogen in this respect.

- **Option C: Alkaline Earth Metals**:
- Alkaline earth metals (like Mg, Ca) have two electrons in their outer shell and tend to lose both electrons to form cations (e.g., Mg^(2+)).
- Since they lose two electrons, they do not resemble hydrogen, which only loses one electron.

- **Option D: Halogens**:
- Halogens (like F, Cl) have seven electrons in their outer shell and tend to gain one electron to achieve a stable octet, forming anions (e.g., Cl^(-)).
- This gaining of electrons is opposite to hydrogen's behavior of losing an electron, so halogens do not resemble hydrogen.

3. **Conclusion**:
- After analyzing all the options, we find that hydrogen most closely resembles alkali metals due to their similar tendency to lose one electron and form cations.
- Therefore, the correct answer is **Option A: Alkali Metals**.

### Final Answer:
**Option A: Alkali Metals**

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