Question

$S{{F}_{4}}$ has ____________ shape.
(A) T-Shape
(B) Bend
(C) Octahedral
(D) see-saw

Answer 1

Use VSEPR Theory postulates such as pair of electrons in valence shell repel one another since their cloud is negatively charged and if there is lone pair that means more repulsion since $S{{F}_{4}}$ also have one lone pair.

Complete answer:
As says the VEPER Theory
A lone pair of electrons takes up more space round the central atom than a bond pair, since the lone pair is attracted to one nucleus while the bond pair is shared by two nuclei. Thus, the presence of lone pairs causes changes such as:

$bond\text{ }pair\left( bp \right)\text{ }\text{ }bond\text{ }pair\left( bp \right)~<\text{ }lone\text{ }pair\left( lp \right)\text{ }\text{ }bond\text{ }pair\left( bp \right)~~~<\text{ }Lone\text{ }pair\left( lp \right)~\text{ }lone\text{ }pair\left( lp \right)$

As we know $S{{F}_{4}}$ has 4-bond pairs and 1-lone pair so it’s assumed geometry should be Trigonal bi-pyramidal but due to its lone pair its shape more like a see-saw because lone pair-bond pair repulsion is much more than the bond pair-bond pair repulsion so the angle is reduced to ${{102}^{\circ }}$ from ${{120}^{\circ }}$.

Besides this we can also see that in (a) the lone pair is present at axial position so there are three lone pair-bond repulsions at ${{90}^{\circ }}$. In (b) the lone pair is in equatorial position and there are two lone pairs –bond pair repulsion. Hence (b) is more stable which is also called as a distorted tetrahedron, a folded square and a see-saw.

Hence, among the options if we check option (D) i.e. See Saw is the correct answer for this question.
Note: VSEPR theory has certain postulates which one must follow to deduce the geometrical structure of a compound.
Here $S{{F}_{4}}$ has one lone pair of electrons. The presence of a lone pair changes the actual angle. Always remember that the more stable structure is preferable, so even if we have different possibilities only the most stable one among them is considered.