Statement-1: Third ionisation energy of phosphorous is larger than sulphur.
Statement-2: There is a larger amount of stability associated with filled s-and p-sub-shells (a noble gas electron configuration) which corresponds to having eight electrons in the valence shell of an atom or ion.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To analyze the statements given in the question, we will break down the reasoning step by step.

### Step 1: Understanding the Electron Configurations
- **Phosphorus (P)** has the electron configuration:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3 \]
- **Sulfur (S)** has the electron configuration:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4 \]

### Step 2: Analyzing the Ionization Energies
- The **first ionization energy** refers to the energy required to remove the outermost electron.
- Phosphorus has a half-filled p subshell (3p^3), which is a stable configuration. This stability results in a higher ionization energy compared to sulfur, which has a p subshell that is not half-filled (3p^4).

### Step 3: Considering the Third Ionization Energy
- The **third ionization energy** involves removing two electrons first, leading to the following configurations:
- For Phosphorus after removing two electrons:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^1 \]
- For Sulfur after removing two electrons:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^2 \]
- In this case, the stability of the remaining configurations must be considered. Phosphorus has a p subshell with one electron (3p^1), while sulfur has two (3p^2). The half-filled configuration of phosphorus contributes to its stability.

### Step 4: Evaluating Statement 1
- **Statement 1** claims that the third ionization energy of phosphorus is larger than that of sulfur. This is **incorrect** because the stability of the configurations after two electrons are removed does not favor phosphorus over sulfur in this case.

### Step 5: Evaluating Statement 2
- **Statement 2** states that there is a larger amount of stability associated with filled s and p subshells, which is true. The noble gas configuration is indeed stable due to having eight electrons in the valence shell.

### Conclusion
- **Statement 1** is **false**.
- **Statement 2** is **true**.
- Therefore, the correct answer is that Statement 1 is incorrect, and Statement 2 is a valid explanation for the stability associated with noble gas configurations.

### Final Answer
- Statement 1: False
- Statement 2: True