Question

Geometry as sulphur is bent in
I. ${{\text{S}}_{\text{8}}}$, II. ${\text{S}}{{\text{F}}_2}$, III.${\text{S}}{{\text{O}}_2}$
A. I, II, III
B.I, II
C. II, III
D.I, III

Answer 1

Hint:We can draw the Lewis structure of all the molecules to determine the electron pair of the central atom. Then by using valence shell electron pair repulsion theory the geometry can be determined.

Complete answer:
Write the Lewis structure as follows
-Write the basic structure. Write the central atom around which writes all atoms of the molecule. The least electronegative atom is the central atom.
-Count total valence electrons.
-Two electrons are used in the formation of a bond.
-Count the total electron used in bond formation.
-Subtracts the electrons used in bond formation from the total valence electrons.
-Arrange the remaining electrons around each atom to complete the octet.
-The valence shell electron pair repulsion theory is as follows:
-Electron pair is the number of electron pairs present around the central atom in a molecule.
According to VSEPR the electron pairs present around the central atom repel each other. So, all the pairs get arranged to minimize the repulsion. Based on the number of electron pair the geometry can be determined as follows:

Number of electron pair	Geometry
$2$	Linear
$3$	Trigonal planar
$4$	Tetrahedral
$5$	Trigonal bipyramid
$6$	Octahedral

Lewis structure of the pair ${{\text{S}}_8}$ is as follows:
Total valence electrons in ${{\text{S}}_8}$ are as follows:
$ = \,\left( {6 \times 8} \right)$
$ = \,48$

The S atom has six valence electrons. Each S is connected to the other two S atoms by single covalent bond. Now, S is bigger in size, to gain stability and to complete octet,it attains the shape of a crown containing 8 S atoms.
Lewis structure of the pair is ${{\text{S}}_8}$ as follows:
Total valence electrons in ${\text{S}}{{\text{F}}_{\text{2}}}$ are as follows:
$ = \,\left( {6 \times 1} \right) + \left( {7 \times 2} \right)$
$ = \,20$

The total electron pair around sulphur in ${\text{S}}{{\text{F}}_{\text{2}}}$ is four so, the geometry will be tetrahedral but all four electron pairs are not same. Out of four two are bond pairs and two are lone pairs, so the geometry of ${\text{S}}{{\text{F}}_{\text{2}}}$ is bent.
The geometry of ${\text{S}}{{\text{F}}_{\text{2}}}$ is shown as follows:

Lewis structure of the pair ${\text{S}}{{\text{O}}_{\text{2}}}$ is as follows:
Total valence electrons in ${\text{S}}{{\text{O}}_{\text{2}}}$ are as follows:
$ = \,\left( {6 \times 1} \right) + \left( {6 \times 2} \right)$
$ = \,18$

The total electron pair around sulphur in is three, so the geometry will be trigonal planar but all three electron pairs are not the same .Out of three, two are bond pairs and one is a lone pair, so the geometry of ${\text{S}}{{\text{O}}_{\text{2}}}$ is bent.
The geometry of ${\text{S}}{{\text{O}}_{\text{2}}}$ is shown as follows:

Geometry as sulphur is bent in all ${{\text{S}}_{\text{8}}}$, ${\text{S}}{{\text{F}}_2}$, and${\text{S}}{{\text{O}}_2}$.

Therefore, option (A) I, II, III, is the answer correct.

Note:To determine the total valence electrons of a molecule, sum all the valence electrons of the atoms present in the molecule. Subtract one for every positive charge and add one for every negative charge. Geometry around the central atom is decided only on the basis of sigma bond pair and lone pair only. Pi bonds are not counted to determine the geometry. In tetrahedral geometry if the number of bond pairs will be three and lone pairs will be one then the geometry will be pyramidal.