Consider the following statements and arrange in the order of true/false as given in the codes.
`S_1`:The hydrogen bond in salt, `KHF_2` is stronger than that in gasous HF.
`S_2` :The dipole moment of `CH_3F` is greater than that of `CH_3Cl`.
`S_3`:`sp^2` hybrid orbitals have equals s and p character.
`S_4`:Two types of bond present in `B_2H_6` are covalent and three centre-two electrons bonds.

Options

A. TTFF
B. TFFT
C. FFFT
D. F F T T

Explanation

To solve the question, we need to evaluate each statement and determine whether it is true or false. Let's analyze each statement step by step.

### Step 1: Analyze Statement S1
**Statement S1**: The hydrogen bond in salt, KHF2, is stronger than that in gaseous HF.

- **Analysis**: KHF2 exists in an ionic form, and the hydrogen bond formed between HF and F⁻ in KHF2 is indeed stronger due to the presence of ionic interactions and the delocalization of electrons. The enthalpy of hydrogen bonding in KHF2 is significantly higher than that in gaseous HF.
- **Conclusion**: **True**

### Step 2: Analyze Statement S2
**Statement S2**: The dipole moment of CH3F is greater than that of CH3Cl.

- **Analysis**: The dipole moment depends on the electronegativity difference and the molecular geometry. Chlorine is more electronegative than fluorine, leading to a greater charge separation in CH3Cl compared to CH3F. Therefore, CH3Cl has a greater dipole moment.
- **Conclusion**: **False**

### Step 3: Analyze Statement S3
**Statement S3**: sp² hybrid orbitals have equal s and p character.

- **Analysis**: In sp² hybridization, one s orbital and two p orbitals combine to form three sp² hybrid orbitals. The percentage of s character is calculated as (1 s orbital / 3 total orbitals) × 100 = 33.33% s character and (2 p orbitals / 3 total orbitals) × 100 = 66.67% p character. They do not have equal s and p character.
- **Conclusion**: **False**

### Step 4: Analyze Statement S4
**Statement S4**: Two types of bonds present in B2H6 are covalent and three-center two-electron bonds.

- **Analysis**: B2H6 (diborane) contains terminal B-H bonds (covalent bonds) and also features three-center two-electron bonds where two boron atoms share a hydrogen atom. This unique bonding is characteristic of diborane.
- **Conclusion**: **True**

### Final Arrangement of Statements
Based on the analysis:
- S1: True
- S2: False
- S3: False
- S4: True

### Final Answer
The order of true/false statements is: **True, False, False, True**.