Which of the following are true ?

Options

A. `He_(2) and BiCl_(5)` do not exist
B. There are two `p pi-d pi` bonds in `SO_(3)`
C. `SeF_(4) and CH_(4)` are tetrahedral
D. `I_(3)^(-)` is linear molecule with `sp^(3)d` hybridisation

Explanation

To determine which statements are true regarding the given compounds, we will analyze each option step by step.

### Step 1: Analyze Option A - He2 and BiCl5
1. **He2 (Helium Dimer)**:
- Helium (He) has a complete outer shell with 2 electrons and does not form bonds.
- According to molecular orbital theory, He2 has a bond order of 0 (calculated as: (bonding electrons - antibonding electrons)/2 = (2 - 2)/2 = 0).
- A bond order of 0 indicates that He2 cannot exist.

2. **BiCl5 (Bismuth Pentachloride)**:
- Bismuth (Bi) typically exhibits a +3 oxidation state due to the inert pair effect, which makes the +5 oxidation state less stable.
- BiCl5 cannot accommodate five chlorine atoms because of steric hindrance and the instability of Bi in the +5 state.
- Therefore, BiCl5 does not exist.

**Conclusion for Option A**: Both He2 and BiCl5 do not exist. Thus, Option A is true.

### Step 2: Analyze Option B - SO3 and ppi, dpi bonds
1. **SO3 (Sulfur Trioxide)**:
- The structure of SO3 involves sulfur as the central atom bonded to three oxygen atoms.
- Sulfur has 6 valence electrons and forms 6 bonds (3 sigma and 3 pi bonds).
- The pi bonds are formed using the 3p and 3d orbitals of sulfur.
- There are indeed 2 ppi (from sulfur's 3p) and 2 dpi (from sulfur's 3d) bonds in SO3.

**Conclusion for Option B**: There are 2 ppi and 2 dpi bonds in SO3. Thus, Option B is true.

### Step 3: Analyze Option C - SeF4 and CH4
1. **SeF4 (Selenium Tetrafluoride)**:
- Selenium has 6 valence electrons and forms 4 bonds with fluorine, leaving one lone pair.
- The molecular geometry is not tetrahedral; it has a seesaw shape (C-SO geometry) due to the lone pair.

2. **CH4 (Methane)**:
- Carbon has 4 valence electrons and forms 4 bonds with hydrogen, resulting in a tetrahedral geometry.

**Conclusion for Option C**: CH4 is tetrahedral, but SeF4 is not tetrahedral. Thus, Option C is false.

### Step 4: Analyze Option D - I3- (Triiodide ion)
1. **I3- (Triiodide Ion)**:
- The I3- ion has a linear shape due to the three iodine atoms and one extra electron (from the -1 charge).
- The central iodine atom has 5 electron pairs (2 bonding pairs and 3 lone pairs), leading to sp3d hybridization.

**Conclusion for Option D**: I3- is linear and has sp3d hybridization. Thus, Option D is true.

### Final Conclusion:
- **True Statements**: A, B, D
- **False Statement**: C