Select correct code :-
(a) In `(BeH_2)_n` all bond's are 3c - 2e.
(b) In `(B_2H_6)`, all H-atom are present in a plane
(c) In `I_2Cl_6`, I-Atom `sp^3d^2` hybridised
(d) In `C_2H_6`, C-atom used vacant orbital in a hybridisation

Options

A. T F T T
B. F T T F
C. T F T F
D. T T T T

Answer

To solve the question, we need to evaluate each statement one by one and determine whether they are true or false.

### Step-by-Step Solution:

1. **Evaluate Statement (a):**
- **Statement:** In `(BeH_2)_n`, all bonds are 3c - 2e.
- **Analysis:** The structure of beryllium hydride, `(BeH_2)`, involves 3-center 2-electron (3c-2e) bonds. In `(BeH_2)_n`, beryllium shares its electrons with three centers (two hydrogen atoms and one beryllium atom). Thus, this statement is **true**.

2. **Evaluate Statement (b):**
- **Statement:** In `(B_2H_6)`, all H-atoms are present in a plane.
- **Analysis:** In the structure of diborane `(B_2H_6)`, there are terminal hydrogen atoms that are in the same plane, but the bridging hydrogen atoms are not in the same plane. Therefore, this statement is **false**.

3. **Evaluate Statement (c):**
- **Statement:** In `I_2Cl_6`, I-atom is `sp^3d^2` hybridised.
- **Analysis:** In `I_2Cl_6`, iodine has 4 bond pairs and 2 lone pairs, which corresponds to an `sp^3d^2` hybridization. Therefore, this statement is **true**.

4. **Evaluate Statement (d):**
- **Statement:** In `C_2H_6`, C-atom used vacant orbital in a hybridisation.
- **Analysis:** In ethane `(C_2H_6)`, the carbon atoms use their own `s` and `p` orbitals for hybridization to form `sp^3` hybridization. There are no vacant orbitals used in this process. Therefore, this statement is **false**.

### Summary of Evaluations:
- (a) True
- (b) False
- (c) True
- (d) False

### Final Answer:
The correct code is **3** (True, False, True, False).