Given: Oxygen and sulfur belong to the same group in the periodic table, but their melting and boiling points differ significantly. The primary reason for this difference is atomicity, i.e., the number of atoms in a molecule of the element.
- The atomic size of sulfur is larger than oxygen. However, atomic size alone does not explain the large difference in their melting and boiling points. Therefore, this factor is not the main reason.
- Oxygen exists as \( O_2 \), a diatomic molecule, whereas sulfur exists as \( S_8 \), an eight-atom molecule. - The larger atomicity of sulfur leads to stronger intermolecular forces due to more atoms interacting with each other. - The stronger intermolecular forces result in higher boiling and melting points for sulfur compared to oxygen. - **Conclusion**: The primary reason for the difference in melting and boiling points is atomicity.
- Oxygen is more electronegative than sulfur. While electronegativity does play a role in bonding and molecular interactions, it does not significantly affect the large difference in the melting and boiling points of oxygen and sulfur.
- Oxygen has a higher electron gain enthalpy than sulfur. While this difference is important for understanding ionization and electron affinity, it does not explain the large discrepancy in melting and boiling points.
The correct answer is **(2) Atomicity**. The molecular structure of sulfur with \( S_8 \) leads to stronger intermolecular forces, resulting in a higher melting and boiling point compared to oxygen.
The correct orders among the following are:
[A.] Atomic radius : \(B<Al<Ga<In<Tl\)
[B.] Electronegativity : \(Al<Ga<In<Tl<B\)
[C.] Density : \(Tl<In<Ga<Al<B\)
[D.] 1st Ionisation Energy :
In\(<Al<Ga<Tl<B\)
Choose the correct answer from the options given below :
Given below are two statements:
Statement I: $ H_2Se $ is more acidic than $ H_2Te $
Statement II: $ H_2Se $ has higher bond enthalpy for dissociation than $ H_2Te $
In the light of the above statements, choose the correct answer from the options given below.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: