Question

Give reasons for the following
I. Sulfur has a greater tendency for catenation than oxygen.
II. ${\text{O}}{{\text{F}}_{\text{6}}}$ compound is not known.
III. ${\text{S}}{{\text{F}}_{\text{4}}}$ is easily hydrolyzed whereas, ${\text{S}}{{\text{F}}_{\text{6}}}$is not easily hydrolyze

Answer 1

i. Catenation is the ability of atoms to make bond chains with other atoms of the same element. This property is prominent in carbon e.g., Diamond, Graphite, and fullerene.
ii. Look at the position of oxygen in the periodic table and some physical trends that follow as we move along the period(from going right to left)
iii. Hydrolysis is the reaction of a compound with water molecules, where the ${{\text{H}}_2}{\text{O}}$ attacks the central atom.

Complete step by step solution:
i). Let’s look at the electronic configuration of sulphur and oxygen
${\text{S}} \to \left[ {{\text{Ne}}} \right]3{{\text{s}}^2}3{{\text{p}}^4} \\\ {\text{O}} \to 1{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^4} \\\$
Oxygen and sulfur belong to the 6th group of the modern periodic table, oxygen is the first element in the group, 6th group is also known as the oxygen family. And we know that the first member of a group shows anomalous behavior from the rest of the group members.
Oxygen is smaller in size and forms a double bond to another oxygen atom, the lone present in the ${{\text{O}}_2}$ repel each other to a greater extent, than the lone pair present in the sulfur atoms, due to its larger size.
ii. By looking at the Electronic configuration of an oxygen atom, ${\text{O}} \to 1{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^4}$
We can say that oxygen needs only 2 electrons to get the stable inert gas electronic configuration of Ne (Neon). Therefore, it cannot accommodate 6 fluorine atoms in its shells, besides 6 fluorine atoms means 6 e, which will create high repulsion, and steric hindrance (crowdedness) which results in instability. All because of the small size of the oxygen atom.
iiii. ${\text{S}}{{\text{F}}_{\text{4}}}$ has 4 fluorine atoms, having enough space/gap for ${{\text{H}}_2}{\text{O}}$ molecules to attack the Sulfur, and form, ${\text{S}}{{\text{F}}_4} + {\text{ 2}}{{\text{H}}_2}{\text{O}} \to {\text{ S}}{{\text{O}}_2}{\text{ + 4HF}}$
But in the case of ${\text{S}}{{\text{F}}_{\text{6}}}$, the 6 fluorine atoms completely cover the Sulfur atom, leaving no space for ${{\text{H}}_2}{\text{O}}$molecules to attack the sulfur atom. Hence, due to the highly packed structure of ${\text{S}}{{\text{F}}_{\text{6}}}$, it cannot be hydrolyzed

Note: while answering, these types of conceptual questions, having a clear knowledge of the basic trends in the periodic table helps a lot, trends like the change in size, electronegativity, electropositivity, metallic character, and electronic configuration have a major influence on an atom's behavior.