To solve the question, we need to evaluate the correctness of the three statements provided:
### Statement Analysis:
**Statement 1 (S₁):** The first ionization energy of magnesium is less than that of aluminium.
1. **Electronic Configuration:**
- Magnesium (Mg) has an atomic number of 12, so its electronic configuration is \(1s^2 2s^2 2p^6 3s^2\).
- Aluminium (Al) has an atomic number of 13, so its electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^1\).
2. **Ionization Energy Explanation:**
- The first ionization energy is the energy required to remove the outermost electron.
- In magnesium, the outermost electrons are in the 3s subshell, while in aluminium, the outermost electron is in the 3p subshell.
- The 3p electron is farther from the nucleus and experiences a lower effective nuclear charge compared to the 3s electrons.
- Therefore, it is easier to remove the 3p electron from aluminium than the 3s electrons from magnesium.
**Conclusion for S₁:** This statement is **false**. The first ionization energy of aluminium is less than that of magnesium.
---
**Statement 2 (S₂):** ClO₃⁻, ClO₄⁻, MnO₄²⁻, SO₄²⁻, ZnO₂²⁻, BO₃³⁻ all have the suffix 'ate'.
1. **Analyzing the Anions:**
- ClO₃⁻ is called chlorate.
- ClO₄⁻ is called perchlorate.
- MnO₄²⁻ is called manganate.
- SO₄²⁻ is called sulfate.
- ZnO₂²⁻ is called zincate.
- BO₃³⁻ is called borate.
**Conclusion for S₂:** This statement is **true**. All these anions have the suffix 'ate'.
---
**Statement 3 (S₃):** The increasing order of negative electron gain enthalpy of the 17th group elements is I < Br < F < Cl.
1. **Understanding Electron Gain Enthalpy:**
- Electron gain enthalpy is the energy change when an electron is added to an atom.
- Generally, as we move up the group in the periodic table, the electron gain enthalpy becomes more negative because the atomic size decreases, leading to a stronger attraction for the added electron.
- However, fluorine has a smaller atomic size compared to chlorine, which leads to greater electron-electron repulsion in the compact 2p orbital of fluorine.
**Conclusion for S₃:** This statement is **false**. The correct order should be I < Br < Cl < F.
---
### Final Evaluation:
- S₁: False
- S₂: True
- S₃: False
Thus, the correct option regarding the statements is **False, True, False**.