Question

Which of the following statements regarding sulphur is incorrect?
[A] ${{S}_{2}}$ molecule is paramagnetic
[B] The vapour at ${{200}^{\circ }}C$ consists mostly of ${{S}_{8}}$ rings
[C] At ${{600}^{\circ }}C$ the gas mostly consists of ${{S}_{2}}$ molecules
[D] The oxidation state of sulphur is never less than +4 in its compounds.

Answer 1

To get the correct answer here you have to proceed option wise and find out the incorrect statement about sulphur. You should remember that it'll be paramagnetic if it has unpaired electrons. Also remember that elemental sulphur breaks at higher temperature.

Complete step by step solution:
To find the correct answer here let us discuss the statements mentioned in the options and find out if they are correct or incorrect.
Firstly, we have that ${{S}_{2}}$ molecule is paramagnetic. We know that a substance which is attracted by a magnetic field due to the presence of unpaired electrons are known as paramagnetic substances. When electrons are unpaired in an orbital, they have a net spin which is not zero and thus they attract magnetic fields. For an element / atom / substance to own paramagnetic property, it should have unpaired electrons. We can write electronic configuration of disulphide according to MOT as- ${{\left( {{\sigma }_{3s}} \right)}^{2}}{{\left( {{\sigma }^{*}}_{3s} \right)}^{2}}{{\left( {{\sigma }_{3{{p}_{z}}}} \right)}^{2}}{{\left( {{\pi }_{3{{p}_{yz}}}} \right)}^{4}}{{\left( {{\pi }^{*}}_{3{{p}_{yz}}} \right)}^{2}}$. We can see it has 2 unpaired electrons in an antibonding orbital so it is paramagnetic, so this option is incorrect. Then we have that the vapour at ${{200}^{\circ }}C$ consists mostly of ${{S}_{8}}$ rings. This is correct as at ${{200}^{\circ }}C$ the vapour contains rings of ${{S}_{8}}$.
Then we have that at ${{600}^{\circ }}C$ the gas mostly consists of ${{S}_{2}}$ molecules. This is also correct because at a temperature as high as ${{600}^{\circ }}C$, the ${{S}_{8}}$ breaks into ${{S}_{2}}$ molecules. Therefore, this option is correct.
And lastly we have the oxidation state of sulphur is never less than +4 in its compounds. This option is incorrect as we know sulphur belongs to group 16 and its oxidation state varies from -2 to +6. In sulphuric acid it is +6 and in hydrogen sulphide it is -2 and that in ${{S}_{8}}$ is 0. Therefore, this statement is incorrect about sulphur.

Therefore, the correct answer is option [D] the oxidation state of sulphur is never less than +4 in its compounds.

Note: We need to remember that sulphur exists as ${{S}_{2}}$ in vapour state but elemental sulphur is ${{S}_{8}}$. Elemental sulphur is diamagnetic unlike sulphur in vapour which is paramagnetic. A substance which is not attracted by a magnetic field due to the absence of unpaired electrons is known as diamagnetic substances. When two electrons are paired with each other in an orbital, their total spin is zero and they repel magnetic fields.
Elemental sulphur does not have any unpaired electrons so it is diamagnetic.