On treatment with cold water, an element (A) reacts readily liberating a colourless, odourless gas (B) and a solution (C). Lithium is reacted with (B) yeildng a solid product (D) which effervesce with water to give a strongely basic solution (E). When `CO_(2)` gas is bubbled through solution (C), a white ppt. (F) is formed but this redissolved forming solution (G) when more `CO_(2)` is passed. Precipitate (F) effervesced when moistened with conc. HCl and give deep red colouration to Bunsen burner flame. (F) on heating with excess of carbon at `2000^(@)`C give (H).
Answer the following question on the basis of above passage.
Q. Metal (A) may be :

Options

A. `Be`
B. `Ca`
C. `Mg`
D. `Ba`

Answer

To determine the identity of metal (A) based on the provided information, we can break down the problem step by step.

### Step 1: Identify the Reaction of Metal (A) with Cold Water
- When metal (A) reacts with cold water, it liberates a colorless, odorless gas (B) and forms a solution (C).
- The most common metals that react with cold water to produce hydrogen gas (B) are alkali and alkaline earth metals.

**Hint:** Think about which metals from the s-block can react with water to produce hydrogen gas.

### Step 2: Determine the Nature of Gas (B) and Solution (C)
- The colorless, odorless gas (B) produced is likely hydrogen (H₂).
- The solution (C) formed is likely a hydroxide of the metal (A). For example, if (A) is calcium, then (C) would be calcium hydroxide (Ca(OH)₂).

**Hint:** Recall the products formed when alkaline earth metals react with water.

### Step 3: Reaction of Lithium with Gas (B)
- When lithium reacts with gas (B), it yields a solid product (D).
- Lithium reacts with hydrogen gas to form lithium hydride (LiH), which is the solid product (D).

**Hint:** Consider the reaction of lithium with hydrogen and what solid it might form.

### Step 4: Solid Product (D) Reacting with Water
- The solid product (D), lithium hydride (LiH), effervesces with water to produce a strongly basic solution (E).
- The reaction is: LiH + H₂O → LiOH + H₂, where (E) is lithium hydroxide (LiOH).

**Hint:** Think about how metal hydrides behave when they come into contact with water.

### Step 5: Bubbling CO₂ through Solution (C)
- When CO₂ is bubbled through solution (C), a white precipitate (F) is formed.
- If (C) is calcium hydroxide, then the reaction with CO₂ forms calcium carbonate (CaCO₃), which is the white precipitate (F).

**Hint:** Consider the reaction between a metal hydroxide and carbon dioxide.

### Step 6: Behavior of Precipitate (F) with Additional CO₂
- The white precipitate (F) dissolves upon the addition of more CO₂, forming solution (G).
- This indicates the formation of calcium bicarbonate (Ca(HCO₃)₂), which is soluble in water.

**Hint:** Think about how carbonates can react with carbon dioxide to form bicarbonates.

### Step 7: Precipitate (F) with Concentrated HCl
- Precipitate (F) effervesces when moistened with concentrated hydrochloric acid (HCl).
- The reaction is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, confirming that (F) is calcium carbonate.

**Hint:** Consider how carbonates react with acids.

### Step 8: Flame Test and Heating of Precipitate (F)
- The white precipitate (F) gives a deep red coloration to a Bunsen burner flame, indicating the presence of calcium.
- When (F) is heated with excess carbon at 2000°C, it produces calcium carbide (CaC₂) and other carbon oxides.

**Hint:** Recall the characteristic flame colors of different metal ions.

### Conclusion
Based on the reactions and products formed, metal (A) is identified as **Calcium (Ca)**.

### Final Answer
Metal (A) may be: **Calcium (Option 2)**.