Draw the Lewis structure for the SO32−. Draw the molecule by placing atoms on the grid and connec…

Concepts and reasonHybridization: During the bond formation the valence atomic orbital of different energy level mix together and result in the formation of new hybrid orbital of same energy level.VSEPR theory: The Valence Shell Electron Pair Repulsion (VSEPR) theory determines the shape (geometry) of the molecule based on the number and kind of atomic orbitals that hybridize to give the hybridized orbitals. The geometries are such that the hybrid orbitals have minimal repulsive interactions.Electron Geometry: It describes the shape of the molecule, which includes bonded electrons as well as lone pair of electrons Formal charge: It’s the method of identifying the number of electrons associated with the atom in a molecule. Formal charge is much helpful for predicting the reactivity of molecules.
FundamentalsThe formal charge (FC) of the atom can be calculated by using the following formula.

Step 1
(A)Lewis Structure of the given molecule is :
Explanation: The sulfur has and oxygen has 6 valence electrons. The Lewis structure is drawn based on the number of electrons present in each atom. The Lewis structure of is shown above. The formal charge of each atom was calculated based on the Lewis structure. Thus it was calculated as
Common mistakes: Ensure that the Lewis structure was drawn properly.Correct:Incorrect:
Hint for next step: Identify the wrong choices for the hybridization of .

Step 2
(B)The wrong choices are
Explanation: The explanation for the wrong choices were given here, if the given molecules posses hybridization, then it geometry will be linear, then there should be only two atoms around the central atom, but the molecules contains three oxygen atom surrounding the central sulfur atom, so this option is not correct. The geometry of hybridized orbital was trigonal planar, so it’s a wrong option for the given molecule. In the given molecule orbital of sulfur doesn’t involve in hybridization, therefore the options are not applicable for the given structure.
Common mistakes: Make sure that the wrong choices were properly identified.Correct:Incorrect:
Hint for next step: Identify the correct hybridization of

Step 3
From the Lewis structure of , its electron geometry was identified as tetrahedral and its corresponding hybridization will be .
Explanation: Lewis structure of was given in step-1, from which it was identified that the central atom is sulfur and it was surrounded by three oxygen atom and a lone pair of electrons. Three sigma bond and lone pair of electrons arose from , one -bond arose from un un-hybridized d-orbital.
Common mistakes: Make sure that the hybridization was properly identified Incorrect:The hybridization of was identified as .Correct:The hybridization of was identified as .
Hint for next step: Draw the Lewis structure and indicate the formal charge for the given.

Step 4
(C)Lewis Structure of the given molecule is:
Explanation: The phosphorous has 5 valence electrons and fluorine has 7 valence electrons. The Lewis structure is drawn based on the number of electrons present in each atom. The Lewis structure of is shown above. The formal charge of each atom is indicated based on the Lewis structure, Thus it was calculated as
Common mistakes: Incorrect:Correct:
Hint for next step: Identify the wrong choices for the hybridization of .

Step 5
(D)Wrong hybridization for the molecule is
Explanation: In the given molecule six fluorine atoms were surrounding the central phosphorous atoms so the above given option are not correct, in the above given hybridization the central atom can posses only maximum of five atoms around them, therefore the above given option are incorrect.
Common mistakes: Make sure that the wrong choices were properly identified.Correct:Incorrect:

Step 6
In the given molecule , the central atom is phosphorus , it was surrounded by six fluorine atom, so the electron pair geometry of is octahedral and its corresponding hybridization is .
Explanation: The electron geometry of a central atom includes both the number of bonding groups and the lone pairs of electrons. For , the central phosphorous atom contains 6 bonding groups and zero lone pair of electrons. Hence, the electron geometry is octahedral.
Common mistakes: Incorrect:The hybridization of electron geometry:.Correct:The hybridization of electron geometry:.

Answer
Part AThe Lewis structure of was given below;
‎ Formal charge for is

Answer
Part BThe correct hybridization for the central atom in based on the electron geometry is:

Answer
Part CThe Lewis structure for is given below;Formal charge in is

Answer
Part DThe correct hybridization for the central atom in based on the electron geometry is