The valence atomic orbital on `C` in silver acetylide is ……………hybridised.

Explanation

To determine the hybridization of the valence atomic orbitals on carbon in silver acetylide (Ag2C2), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Silver Acetylide**:
- The formula for silver acetylide is Ag2C2. This indicates that there are two carbon atoms (C) involved in the structure.

2. **Understand the Bonding in Acetylide**:
- In silver acetylide, the carbon atoms are connected by a triple bond. The structure can be represented as Ag^+ - C≡C - Ag^+. The triple bond consists of one sigma bond and two pi bonds.

3. **Determine the Steric Number**:
- The steric number (S) is calculated using the formula:
\[ S = \text{Number of sigma bonds} + \text{Number of lone pairs} \]
- For each carbon atom in the triple bond:
- There is **1 sigma bond** (from the triple bond).
- Each carbon also has a **lone pair** due to the negative charge associated with the acetylide ion.
- Therefore, for each carbon:
\[ S = 1 \text{ (sigma bond)} + 1 \text{ (lone pair)} = 2 \]

4. **Determine the Hybridization**:
- The hybridization can be determined from the steric number:
- A steric number of 2 corresponds to **sp hybridization**.
- Thus, both carbon atoms in silver acetylide are **sp hybridized**.

5. **Conclusion**:
- The valence atomic orbital on carbon in silver acetylide is **sp hybridized**.

### Final Answer:
The valence atomic orbital on C in silver acetylide is **sp hybridised**.