VSEPR Theory Questions in English

(B) The structure of $NO_2^+$ is linear.
In $NO_2^+$,the nitrogen atom is $sp$ hybridized,resulting in a bond angle of $180^{\circ}$ and a linear geometry: $[O=N=O]^+$.
$SO_2$,$NO_2^-$,and $SCl_2$ are bent due to the presence of lone pairs on the central atom.

(B) In $NH_4^+$,the nitrogen atom undergoes $sp^3$ hybridization and has no lone pair of electrons,resulting in a tetrahedral bond angle of $109^\circ 28'$.
In $NH_3$,the presence of one lone pair reduces the bond angle to $107^\circ$.
In $PCl_3$,the bond angle is approximately $100^\circ$ due to the larger size of the central atom and lone pair repulsion.
In $SCl_2$,the bond angle is approximately $103^\circ$.
Therefore,$NH_4^+$ has the maximum bond angle.

(D) The $N$ atom in the $NO_3^-$ ion is bonded to three oxygen atoms.
One double bond and two single bonds are formed to satisfy the octet rule and formal charges,resulting in $4$ bond pairs around the nitrogen atom.
The valence shell of nitrogen has $5$ electrons. In $NO_3^-$,nitrogen forms $4$ bonds (one double and two single),utilizing all $5$ valence electrons (one electron is used for the negative charge).
Therefore,the number of lone pairs on the $N$ atom is $0$.

(C) In $PF_5$,the central atom $P$ undergoes $sp^3d$ hybridization,resulting in a trigonal bipyramidal geometry.
It contains three equatorial $P-F$ bonds and two axial $P-F$ bonds.
The axial bonds are longer than the equatorial bonds due to greater repulsion from the equatorial bond pairs,leading to unequal bond lengths.

(B) In the $NH_3$ molecule,the nitrogen atom is $sp^3$ hybridized. Due to the presence of one lone pair of electrons on the nitrogen atom,the geometry is distorted from tetrahedral to pyramidal.

(C) $NF_3$ and $H_3O^+$ both have $sp^3$ hybridization and a pyramidal geometry.
$BF_3$ and $NO_3^-$ both have $sp^2$ hybridization and a trigonal planar geometry.

(D) In the hydrides of Group $16$ elements,the bond angle decreases as we move down the group.
This is because the electronegativity of the central atom decreases down the group,which leads to an increase in the bond length and a decrease in the bond angle.

(B) The Lewis structure of $OF_2$ is represented as $F-O-F$ with lone pairs on each atom.
Each $F-O$ bond represents one bonding electron pair,so there are $2$ bonding pairs.
The oxygen atom has $2$ lone pairs,and each fluorine atom has $3$ lone pairs.
Total number of lone pairs $= 2 + 3 + 3 = 8$.

(A) The hybridization of $S$ in $SCl_4$ is $sp^3d$,which results in a see-saw geometry due to the presence of one lone pair on the sulfur atom.
$SO_4^{2-}$ has $sp^3$ hybridization and a tetrahedral geometry.
$Ni(CO)_4$ has $sp^3$ hybridization and a tetrahedral geometry.
$NiCl_4^{2-}$ has $sp^3$ hybridization and a tetrahedral geometry.
Therefore,$SCl_4$ is the only species that is not tetrahedral.

(B) In $BeCl_2$,the central atom $Be$ is $sp$ hybridized,resulting in a linear geometry.
In $CO_2$,the central atom $C$ is $sp$ hybridized,resulting in a linear geometry.
In $CH \equiv CH$,both carbon atoms are $sp$ hybridized,resulting in a linear geometry.
In $SO_2$,the central atom $S$ is $sp^2$ hybridized and has one lone pair of electrons,which gives it a bent or angular shape.

(A) The bond angles for the given molecules are as follows:
$OF_2$: $103^o$
$H_2O$: $104.5^o$
$OCl_2$: $111^o$
$ClO_2$: $118^o$
Thus,the correct increasing order of bond angles is $OF_2 < H_2O < OCl_2 < ClO_2$.

(A) In $Cl_2O_7$,each chlorine atom is bonded to four oxygen atoms in a tetrahedral arrangement. Two such $ClO_4$ tetrahedra share a common oxygen atom at the vertex,resulting in a structure where each $Cl$ atom is at the center of a tetrahedron.

(C) The trihalide cation $I_3^+$ has a bent geometry due to the presence of two lone pairs on the central iodine atom,whereas $IBrF^-$,$ICl_2^-$,and $I_3^-$ are linear due to $sp^3d$ hybridization with three lone pairs on the central atom.

(B) The central chlorine atom in $ClO_3^-$ has $7$ valence electrons.
It forms $3$ double bonds with oxygen atoms and has $1$ lone pair of electrons.
The total number of electron pairs around the central atom is $3 + 1 = 4$,which corresponds to $sp^3$ hybridization.
Due to the presence of one lone pair,the geometry is distorted from tetrahedral to pyramidal.

(A) The central atom $Cl$ in $ClO_3^-$ has $7$ valence electrons.
It forms $3$ single bonds with $O$ atoms and has $1$ lone pair.
According to $VSEPR$ theory,the steric number is $3 + 1 = 4$,which corresponds to $sp^3$ hybridization.
Due to the presence of one lone pair,the geometry is distorted from tetrahedral to trigonal pyramidal.

(D) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \text{Steric number} - \text{Number of bonded atoms}$.

Compound	Lone pairs on central atom
$A$. $[ClO_3]^-$	$1$
$B$. $XeF_4$	$2$
$C$. $SF_4$	$1$
$D$. $[I_3]^-$	$3$

In $[I_3]^-$,the central iodine atom has $3$ lone pairs. Therefore,it has the maximum number of lone pairs.

(A) Both $XeF_2$ and $IF_2^-$ have $sp^3d$ hybridization with $3$ lone pairs on the central atom,resulting in a linear geometry. Therefore,they are isostructural.

(A) In $XeF_2$,the central atom $Xe$ has $8$ valence electrons. It forms $2$ bond pairs with $F$ atoms and has $3$ lone pairs of electrons. According to $VSEPR$ theory,the arrangement of $3$ lone pairs and $2$ bond pairs results in a trigonal bipyramidal electron geometry with a linear molecular shape ($sp^3d$ hybridization).

(A) The central atom $Xe$ in $XeF_2$ undergoes $sp^3d$ hybridization.
According to $VSEPR$ theory,the molecule has a linear geometry.
$Xe$ has $8$ valence electrons,$2$ are used in bonding with $F$ atoms,leaving $6$ electrons which form $3$ lone pairs.
Therefore,the number of lone pairs on $Xe$ is $3$.

(B) $(B)$ In $XeF_4$, the Xenon atom undergoes $sp^3d^2$ hybridisation.
It has $4$ bonding pairs and $2$ lone pairs.
The lone pairs occupy the axial positions to minimize repulsion, resulting in a square planar geometry.
Thus, it is a planar molecule.

(A) The central atom $Xe$ in $XeF_4$ undergoes $sp^3d^2$ hybridization.
It has $4$ bond pairs and $2$ lone pairs of electrons.
According to $VSEPR$ theory,the presence of $2$ lone pairs results in a square planar geometry.

(A) The number of lone pairs on the central atom $Xe$ can be determined using the formula: $\text{Lone pairs} = \frac{1}{2} (V - B)$,where $V$ is the number of valence electrons of the central atom ($8$ for $Xe$) and $B$ is the number of bonding pairs (number of $F$ atoms attached).

$\text{Compound}$	$\text{Lone pairs on } Xe$
$XeF_2$	$\frac{1}{2}(8-2) = 3$
$XeF_4$	$\frac{1}{2}(8-4) = 2$
$XeF_6$	$\frac{1}{2}(8-6) = 1$

Thus,the number of lone pairs are $3, 2$,and $1$ respectively.

(D) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (monovalent atoms).
$1$. For $SF_4$: Central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (6 - 4) = 1$.
$2$. For $CF_4$: Central atom $C$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (4 - 4) = 0$.
$3$. For $XeF_4$: Central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (8 - 4) = 2$.
Thus,the number of lone pairs on the central atoms of $SF_4$,$CF_4$,and $XeF_4$ are $1, 0$,and $2$ respectively.

(D) $1$. $SiF_4$ has a tetrahedral geometry ($sp^3$ hybridization) where all $Si-F$ bond lengths are equal.
$2$. $XeF_4$ has a square planar geometry ($sp^3d^2$ hybridization) with two lone pairs at axial positions,making all $Xe-F$ bond lengths equal.
$3$. $BF_4^-$ has a tetrahedral geometry ($sp^3$ hybridization) where all $B-F$ bond lengths are equal.
$4$. $SF_4$ has a see-saw geometry ($sp^3d$ hybridization) with one lone pair at an equatorial position. Due to the presence of the lone pair,the axial $S-F$ bonds and equatorial $S-F$ bonds have different bond lengths.

(D) The central atom in $IF_7$ is Iodine $(I)$,which has $7$ valence electrons. All $7$ electrons are involved in bonding with $7$ Fluorine atoms. The steric number is $7 + 0 = 7$. $A$ steric number of $7$ corresponds to $sp^3d^3$ hybridization,which results in a pentagonal bipyramidal geometry.

(A) The bond angles for the given species are as follows:
$Cl_2O$: $111^\circ$
$ClO_2^-$: $110^\circ$ (approximate,due to two lone pairs on $Cl$)
$ClO_2$: $118^\circ$
Comparing these values,the increasing order of bond angles is $ClO_2^- < Cl_2O < ClO_2$.

(D) The most preferred Lewis structure for a molecule is the one that minimizes formal charges on all atoms.
For $SO_3$,the structure where the sulfur atom is bonded to three oxygen atoms with double bonds $(S=O)$ results in a formal charge of $0$ on all atoms (Sulfur has $6$ valence electrons and forms $6$ bonds; Oxygen has $6$ valence electrons and forms $2$ bonds).
This structure minimizes the formal charge to $0$ for every atom,making it the most stable and lowest energy structure.

(B) In $(SiH_3)_3N$,the lone pair on $N$ is involved in $p\pi-d\pi$ back-bonding with the empty $d$-orbitals of $Si$. This makes the $N$ atom $sp^2$ hybridized,resulting in a trigonal planar geometry.
In $(SiH_3)_3P$,the $P$ atom has a lone pair,but the energy gap between the $P$ $3p$ orbital and $Si$ $3d$ orbital is large,making $p\pi-d\pi$ back-bonding ineffective. Thus,$P$ remains $sp^3$ hybridized with a lone pair,resulting in a pyramidal (tetrahedral arrangement of electron pairs) geometry.

(C) $1$. For $OSX_2$ series: As the electronegativity of the halogen $X$ decreases $(F > Cl > Br)$,the bond pair electrons move closer to the central atom $S$. This increases the repulsion between bond pairs,but the bond angle is primarily determined by the electronegativity of the substituent. Lower electronegativity of $X$ leads to a smaller bond angle. Thus,$OSF_2$ has the smallest bond angle.
$2$. For $SbX_3$ series: As the size of the halogen increases $(Cl < Br < I)$,the bond length increases and the repulsion between the bonding pairs decreases,leading to a decrease in the bond angle. Thus,$SbI_3$ has the smallest bond angle.
$3$. For $AI_3$ series: As the electronegativity of the central atom $A$ decreases $(P > As > Sb)$,the bond pair electrons move further away from the central atom,reducing the repulsion between them. Thus,$SbI_3$ has the smallest bond angle.
Correct option is $C$.

(D) To determine if all bond lengths are the same,we examine the geometry and hybridization of each molecule:
$1$. $SF_4$: It has $sp^3d$ hybridization with a see-saw shape. Due to the presence of a lone pair,the axial and equatorial $S-F$ bonds have different lengths.
$2$. $C_2H_6$: It is ethane. The $C-C$ bond is single,but the $C-H$ bonds are not equivalent to the $C-C$ bond length.
$3$. $PCl_5$: It has $sp^3d$ hybridization with a trigonal bipyramidal geometry. It has two types of $P-Cl$ bonds: axial and equatorial,which have different lengths.
$4$. $SiF_4$: It has $sp^3$ hybridization with a perfectly tetrahedral geometry. All four $Si-F$ bonds are equivalent and have the same bond length.

(C) The central atom $Cl$ in $ClF_3$ has $7$ valence electrons. It forms $3$ $Cl-F$ bonds and has $2$ lone pairs,resulting in $sp^3d$ hybridization.
According to $VSEPR$ theory,lone pairs occupy equatorial positions to minimize $lp-lp$ and $lp-bp$ repulsions.
If the lone pairs were to occupy axial positions,the $lp-lp$ repulsion would be higher,and the molecule would not be stable in that configuration.
In the actual $T$-shaped structure,the two axial $Cl-F$ bonds are longer than the equatorial $Cl-F$ bond due to the repulsion from the lone pairs.
Therefore,the statement that 'All bond lengths are equal' is incorrect.

(C) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (or atoms attached).
$1$. In $SiF_4$: $Si$ has $4$ valence electrons and forms $4$ bonds. Lone pairs = $\frac{1}{2} (4 - 4) = 0$.
$2$. In $BrF_3$: $Br$ has $7$ valence electrons and forms $3$ bonds. Lone pairs = $\frac{1}{2} (7 - 3) = 2$.
$3$. In $XeF_2$: $Xe$ has $8$ valence electrons and forms $2$ bonds. Lone pairs = $\frac{1}{2} (8 - 2) = 3$.
$4$. In $SF_4$: $S$ has $6$ valence electrons and forms $4$ bonds. Lone pairs = $\frac{1}{2} (6 - 4) = 1$.
Comparing these,$XeF_2$ has the maximum number of lone pairs $(3)$.

(C) The molecule $O_2F_2$ has a non-planar,open-book structure similar to $H_2O_2$. In both molecules,the dihedral angle is approximately $90^{\circ}$ to $100^{\circ}$ due to the repulsion between lone pairs on the oxygen atoms.

(A) The geometries of the given xenon compounds are as follows:
$A$. $XeF_4$: It has $4$ bond pairs and $2$ lone pairs,resulting in a square planar geometry $(4)$.
$B$. $XeF_6$: It has $6$ bond pairs and $1$ lone pair,resulting in a distorted octahedral geometry $(3)$.
$C$. $XeO_3$: It has $3$ bond pairs and $1$ lone pair,resulting in a pyramidal geometry $(1)$.
$D$. $XeOF_2$: It has $3$ bond pairs and $2$ lone pairs,resulting in a $T$-shape geometry $(2)$.
Therefore,the correct matching is $A-4, B-3, C-1, D-2$.

(C) The molecule $SF_4$ has a see-saw geometry due to $sp^3d$ hybridization with one lone pair on the sulfur atom.
In this structure,there are two types of $F$ atoms: equatorial and axial.
Due to the presence of the lone pair,the bond angles deviate from the ideal trigonal bipyramidal angles ($90^{\circ}$ and $120^{\circ}$).
The different types of $F-S-F$ bond angles present are:
$1$. The angle between the two equatorial $F$ atoms is approximately $102^{\circ}$.
$2$. The angle between the two axial $F$ atoms is approximately $173^{\circ}$.
$3$. The angle between an axial $F$ atom and an equatorial $F$ atom is approximately $87^{\circ}$.
Thus,there are $3$ distinct types of $F-S-F$ bond angles in $SF_4$.

(C) $SF_4$ does not have equal bond angles because of its structure,which is see-saw (derived from trigonal bipyramidal).
Sulfur in $SF_4$ has a total of six valence electrons,four of which form bonds with fluorine atoms and two form a lone pair.
Due to the presence of a lone pair,the bond angles deviate from the ideal trigonal bipyramidal angles of $90^\circ$ and $120^\circ$.
The equatorial $F-S-F$ bond angle is approximately $102^\circ$ and the axial $F-S-F$ bond angle is approximately $173^\circ$ due to lone pair-bond pair repulsion.
In contrast,$SiF_4$ ($sp^3$,tetrahedral),$ICl_4^-$ ($sp^3d^2$,square planar),and $PCl_4^+$ ($sp^3$,tetrahedral) have equivalent bond angles due to their symmetric geometries.

(A) The cyanate ion $(OCN^-)$ consists of one oxygen atom,one carbon atom,and one nitrogen atom with a total of $6 + 4 + 5 + 1 = 16$ valence electrons.
In the correct Lewis structure,the oxygen atom is double-bonded to the carbon atom,which is in turn double-bonded to the nitrogen atom.
This arrangement satisfies the octet rule for all atoms: $[\ddot{O}=C=\ddot{N}:]^{-}$.

(A) The bond angle depends on the electronegativity of the central atom and the presence of lone pairs.
$1$. For $NH_3$ $(107^\circ)$ and $H_2O$ $(104.5^\circ)$,the central atoms are $N$ and $O$ (period $2$).
$2$. For $PH_3$ $(93.5^\circ)$ and $H_2S$ $(92^\circ)$,the central atoms are $P$ and $S$ (period $3$).
$3$. Elements in period $2$ have higher electronegativity and smaller size,leading to greater repulsion between bond pairs compared to period $3$ elements.
$4$. Thus,the bond angles of $NH_3$ and $H_2O$ are significantly larger than those of $PH_3$ and $H_2S$.
$5$. Comparing $NH_3$ and $H_2O$: $NH_3$ has one lone pair,while $H_2O$ has two lone pairs,resulting in more repulsion in $H_2O$ and a smaller bond angle.
$6$. Comparing $PH_3$ and $H_2S$: $PH_3$ has one lone pair and $H_2S$ has two,leading to a similar trend.
$7$. The correct order is $NH_3 > H_2O > PH_3 > H_2S$.

(C) To determine the bond angle,we look at the hybridization and the number of lone pairs on the central atom:
$1$. $BeF_2$: $sp$ hybridization,linear geometry,bond angle = $180^{\circ}$.
$2$. $CH_4$: $sp^3$ hybridization,tetrahedral geometry,bond angle = $109.5^{\circ}$.
$3$. $NH_3$: $sp^3$ hybridization,trigonal pyramidal geometry,one lone pair,bond angle = $107^{\circ}$.
$4$. $H_2O$: $sp^3$ hybridization,bent geometry,two lone pairs. The lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion,which compresses the bond angle to $104.5^{\circ}$.
Comparing these,$H_2O$ has the smallest bond angle.

(D) The bond angle depends on the electronegativity of the central atom and the size of the atoms attached.
In $NH_3$ and $PH_3$,the central atoms are $N$ and $P$ respectively. $PH_3$ has a bond angle of approximately $93.5^\circ$ due to the lower electronegativity of $P$ compared to $N$ $(NH_3 \approx 107^\circ)$.
In $H_2O$ and $H_2S$,the central atoms are $O$ and $S$ respectively. $H_2S$ has a bond angle of approximately $92^\circ$ because $S$ is less electronegative than $O$ and has a larger atomic size,leading to less repulsion between bond pairs.
Comparing $PH_3$ and $H_2S$,$H_2S$ has the smallest bond angle among the given options.

(D) To determine the $O-N-O$ bond angle,we look at the hybridization and geometry of the central nitrogen atom in each species:
$1$. $NO^{-}_3$: Nitrogen is $sp^2$ hybridized with a trigonal planar geometry. The bond angle is approximately $120^{\circ}$.
$2$. $NO^{-}_2$: Nitrogen is $sp^2$ hybridized with one lone pair. Due to lone pair-bond pair repulsion,the bond angle is slightly less than $120^{\circ}$ (approx. $115^{\circ}$).
$3$. $NO_2$: Nitrogen is $sp^2$ hybridized with one unpaired electron. The bond angle is approximately $134^{\circ}$.
$4$. $NO^{+}_2$: Nitrogen is $sp$ hybridized with a linear geometry. The bond angle is $180^{\circ}$.
Comparing these,$NO^{+}_2$ has the maximum bond angle of $180^{\circ}$.

(C) To determine the bond angle,we look at the hybridization and the number of lone pairs on the central nitrogen atom:
$1$. $NO^{+}_2$: The nitrogen atom is $sp$ hybridized with $0$ lone pairs. The geometry is linear,so the bond angle is $180^{\circ}$.
$2$. $NO_2$: The nitrogen atom is $sp^2$ hybridized with $1$ unpaired electron. The geometry is bent,and the bond angle is approximately $134^{\circ}$.
$3$. $NO^{-}_2$: The nitrogen atom is $sp^2$ hybridized with $1$ lone pair. The geometry is bent,and the bond angle is approximately $115^{\circ}$.
Therefore,the correct decreasing order of bond angles is $NO^{+}_2 > NO_2 > NO^{-}_2$.

(A) The structure of $CO_2$ is linear with $sp$ hybridization at the central carbon atom.
$HgCl_2$ also has a linear structure with $sp$ hybridization at the central mercury atom.
$H_2O$ has a bent shape ($sp^3$ hybridization).
$SnCl_2$ has a bent shape ($sp^2$ hybridization).
$NO_2^-$ has a bent shape ($sp^2$ hybridization).
Therefore,$HgCl_2$ is isostructural with $CO_2$.

(D) To determine if species are isostructural,we calculate the hybridization and the number of lone pairs on the central atom.
$1$. $NO^{-}_3$: Central atom $N$ has $5$ valence electrons. Hybridization = $\frac{1}{2}(5 + 1 + 0) = 3$ $(sp^2)$. It has $0$ lone pairs. Geometry: Trigonal planar.
$2$. $CO^{2-}_3$: Central atom $C$ has $4$ valence electrons. Hybridization = $\frac{1}{2}(4 + 2 + 0) = 3$ $(sp^2)$. It has $0$ lone pairs. Geometry: Trigonal planar.
$3$. $SO_3$: Central atom $S$ has $6$ valence electrons. Hybridization = $\frac{1}{2}(6 + 0 + 0) = 3$ $(sp^2)$. It has $0$ lone pairs. Geometry: Trigonal planar.
$4$. $ClO^{-}_3$: Central atom $Cl$ has $7$ valence electrons. Hybridization = $\frac{1}{2}(7 + 1 + 0) = 4$ $(sp^3)$. It has $1$ lone pair. Geometry: Pyramidal.
Since $NO^{-}_3, CO^{2-}_3,$ and $SO_3$ all have $sp^2$ hybridization and $0$ lone pairs,they are isostructural (trigonal planar). Therefore,both $b$ and $c$ are correct pairs.

(A) The central iodine atom in $I_3^-$ has $7$ valence electrons.
It forms one bond with each of the other two iodine atoms,using $2$ electrons.
This leaves $5$ electrons,plus the $1$ negative charge,totaling $6$ electrons,which form $3$ lone pairs.
Thus,the steric number is $2$ (bond pairs) $+ 3$ (lone pairs) $= 5$.
$A$ steric number of $5$ corresponds to trigonal bipyramidal geometry.
According to $VSEPR$ theory,to minimize repulsion,the $3$ lone pairs occupy the equatorial positions,and the $2$ bonding iodine atoms occupy the axial positions,resulting in a linear molecular shape.

(B) The structure of $CO_2$ is linear $(O=C=O)$ with $sp$ hybridization.
$HgCl_2$ is linear $(Cl-Hg-Cl)$ with $sp$ hybridization.
$C_2H_2$ is linear $(H-C \equiv C-H)$ with $sp$ hybridization.
$ZnCl_2$ is linear $(Cl-Zn-Cl)$ with $sp$ hybridization.
$SnCl_2$ has a bent (angular) structure due to the presence of one lone pair on the $Sn$ atom ($sp^2$ hybridization).
Therefore,$CO_2$ is not isostructural with $SnCl_2$.

(A) To determine the shape,we calculate the number of electron pairs around the central atom using the formula: $\text{Number of electron pairs} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $PCl_3$: $V=5$ (for $P$),$M=3$ (for $Cl$). $\text{Electron pairs} = \frac{1}{2} [5 + 3] = 4$. This corresponds to $sp^3$ hybridization with one lone pair and three bond pairs,resulting in a pyramidal shape.
For $SO_3$: $V=6$ (for $S$),$M=0$. $\text{Electron pairs} = \frac{1}{2} [6] = 3$. This corresponds to $sp^2$ hybridization with zero lone pairs,resulting in a trigonal planar shape.
For $CO_3^{2-}$: $V=4$ (for $C$),$M=0$,$A=2$. $\text{Electron pairs} = \frac{1}{2} [4 + 2] = 3$. This corresponds to $sp^2$ hybridization,resulting in a trigonal planar shape.
For $NO_3^-$: $V=5$ (for $N$),$M=0$,$A=1$. $\text{Electron pairs} = \frac{1}{2} [5 + 1] = 3$. This corresponds to $sp^2$ hybridization,resulting in a trigonal planar shape.
Therefore,only $PCl_3$ has a pyramidal shape.

(B) In $XeF_2$,the central atom $Xe$ has $8$ valence electrons.
It forms $2$ single bonds with $F$ atoms and has $3$ lone pairs of electrons.
According to $VSEPR$ theory,the total number of electron pairs is $2 + 3 = 5$,which corresponds to $sp^3d$ hybridization and trigonal bipyramidal electron geometry.
Due to the presence of $3$ lone pairs in the equatorial positions to minimize repulsion,the $F-Xe-F$ bond angle is $180^{\circ}$,resulting in a linear molecular shape.

(C) The molecule $ClF_3$ has a central $Cl$ atom with $7$ valence electrons. It forms $3$ $Cl-F$ bonds and has $2$ lone pairs,resulting in a total of $5$ electron pairs ($sp^3d$ hybridization). According to $VSEPR$ theory,to minimize repulsion,the lone pairs prefer the equatorial positions in a trigonal bipyramidal geometry. This arrangement minimizes the $90^{\circ}$ lone pair-bond pair repulsions. Therefore,the structure with both lone pairs in the equatorial positions is the most stable.

(B) To determine the shape of $SF_2Cl_2$,we calculate the steric number using the formula: $Steric \ Number = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom $(S = 6)$,$M$ is the number of monovalent atoms ($F = 2, Cl = 2$,so $M = 4$),$C$ is the cationic charge $(0)$,and $A$ is the anionic charge $(0)$.
$Steric \ Number = \frac{1}{2} [6 + 4] = 5$.
$A$ steric number of $5$ corresponds to $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
In $SF_2Cl_2$,the central sulfur atom has $5$ electron pairs ($4$ bonding pairs and $1$ lone pair).
According to $VSEPR$ theory,the lone pair occupies an equatorial position to minimize repulsion.
This arrangement results in a seesaw shape.