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 will evaluate each statement one by one and determine whether they are true or false.

### Step 1: Evaluate Statement (a)
**Statement (a):** In `(BeH_2)_n`, all bonds are 3c - 2e.

- **Analysis:** The structure of beryllium hydride, `(BeH_2)_n`, involves three-center two-electron (3c-2e) bonds, also known as banana bonds. In this structure, each beryllium atom is bonded to two hydrogen atoms, and the bonding involves three atomic centers (two hydrogen atoms and one beryllium atom) sharing two electrons.
- **Conclusion:** This statement is **True**.

### Step 2: Evaluate Statement (b)
**Statement (b):** In `(B_2H_6)`, all H-atoms are present in a plane.

- **Analysis:** The structure of diborane, `(B_2H_6)`, consists of four terminal hydrogen atoms that lie in the same plane, while two bridging hydrogen atoms are positioned above and below this plane. Therefore, not all hydrogen atoms are in the same plane.
- **Conclusion:** This statement is **False**.

### Step 3: Evaluate Statement (c)
**Statement (c):** In `I_2Cl_6`, I-atom is `sp^3d^2` hybridized.

- **Analysis:** Iodine in `I_2Cl_6` has seven valence electrons. In this molecule, iodine forms bonds with chlorine atoms and has lone pairs. The hybridization can be determined by counting the bond pairs and lone pairs around the iodine atom. There are four bond pairs and two lone pairs, which means iodine uses six orbitals for hybridization, leading to `sp^3d^2` hybridization.
- **Conclusion:** This statement is **True**.

### Step 4: Evaluate Statement (d)
**Statement (d):** In `C_2H_6`, C-atom used vacant orbital in hybridization.

- **Analysis:** In ethane (`C_2H_6`), each carbon atom has the electronic configuration of `1s^2 2s^2 2p^2`. The carbon atoms hybridize their `2s` and `2p` orbitals to form four equivalent `sp^3` hybrid orbitals. There are no vacant orbitals involved in this hybridization process.
- **Conclusion:** This statement is **False**.

### Final Evaluation of Statements
- Statement (a): True
- Statement (b): False
- Statement (c): True
- Statement (d): False

### Correct Code
The correct code based on the evaluations is: **TFTF** (True, False, True, False).