Question

Which of the following statements is false about oxygen and sulphur?

• A. Atoms of oxygen and sulphur consist two unpaired electrons in valence shell.
• B. Oxygen and sulphur show -2, +4 and +6 oxidation states.
• C. Oxygen is gas while sulphur is solid at room temperature.
• D. Hydride of oxygen is more stable than hydride of sulphur.

Hint:

The absence of vacant d-orbitals is the primary reason why second-period elements (N, O, F) exhibit anomalous behavior compared to their heavier group members! Oxygen's maximum covalency is restricted strictly to 4.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to evaluate several physical and chemical properties of Group 16 elements (Oxygen and Sulphur) to identify the factually incorrect statement.

Step 2: Detailed Explanation:
Let's evaluate each statement logically based on p-block chemistry:
(a) Both have two unpaired electrons: Oxygen is $1s^2 2s^2 2p^4$. Sulphur is $[\text{Ne}] 3s^2 3p^4$. The $np^4$ configuration fills the three p-orbitals as (pair, single, single), leaving exactly two unpaired electrons. This statement is True.
(c) Physical state: Oxygen exists as a small, non-polar diatomic molecule ($\text{O}_2$) with very weak van der Waals forces, making it a gas. Sulphur exists as a massive, puckered $\text{S}_8$ ring with much stronger dispersion forces, making it a solid. This statement is True.
(d) Hydride stability: Thermal stability of hydrides depends heavily on bond dissociation enthalpy. Since the O atom is much smaller than S, the O-H bond is shorter and significantly stronger than the S-H bond. Therefore, $\text{H}_2\text{O}$ is thermally much more stable than $\text{H}_2\text{S}$. This statement is True.
(b) Oxidation states: Sulphur possesses empty 3d orbitals, allowing it to unpair and promote its valence electrons to exhibit higher positive oxidation states like +4 (in $\text{SO}_2$) and +6 (in $\text{SO}_3$). Oxygen, however, lacks d-orbitals in its principal shell ($n=2$). Because of its high electronegativity and lack of expandable octet, Oxygen primarily shows a -2 oxidation state. It almost never shows positive oxidation states (except for +2 in $\text{OF}_2$ and +1 in $\text{O}_2\text{F}_2$). It absolutely cannot show +4 or +6. Therefore, this statement is False.

Step 3: Final Answer:
The false statement is that both show -2, +4, and +6 oxidation states, matching option (b).