VSEPR Theory Questions in English

(C) To determine the geometry,we calculate the hybridization of the central atom:
$1$. For $SF_6$,the central atom is $S$ (Sulfur). It has $6$ valence electrons and forms $6$ bonds with $F$ atoms. The steric number is $6 + 0 = 6$,which corresponds to $sp^3d^2$ hybridization and an octahedral geometry.
$2$. $TeF_4$ has a steric number of $5$ ($4$ bond pairs + $1$ lone pair),resulting in a see-saw shape.
$3$. $SeCl_2$ has a steric number of $4$ ($2$ bond pairs + $2$ lone pairs),resulting in a bent shape.
$4$. $SeF_4$ has a steric number of $5$ ($4$ bond pairs + $1$ lone pair),resulting in a see-saw shape.
Therefore,$SF_6$ is the only octahedral molecule among the given options.

(A) To determine the shape of the molecules,we look at their hybridization and the presence of lone pairs:
$1$. $SO_2$: The sulfur atom has $sp^2$ hybridization with one lone pair. Due to the lone pair-bond pair repulsion,the molecule adopts a bent (or angular) shape.
$2$. $BeBr_2$: The beryllium atom has $sp$ hybridization with no lone pairs,resulting in a linear shape.
$3$. $CO_2$: The carbon atom has $sp$ hybridization with no lone pairs,resulting in a linear shape.
$4$. $BF_3$: The boron atom has $sp^2$ hybridization with no lone pairs,resulting in a trigonal planar shape.
Therefore,$SO_2$ is the molecule with a bent shape.

(B) To determine the shape of the molecules,we look at their hybridization and the presence of lone pairs:
$1$. $HBr$: It is a diatomic molecule,so it is linear.
$2$. $H_2S$: The central sulfur atom has $6$ valence electrons. It forms $2$ bonds with hydrogen atoms and has $2$ lone pairs. According to $VSEPR$ theory,the steric number is $4$ ($2$ bond pairs + $2$ lone pairs),resulting in $sp^3$ hybridization and a bent (angular) shape.
$3$. $BeBr_2$: The central beryllium atom has $2$ valence electrons and forms $2$ bonds with bromine atoms. It has no lone pairs,resulting in $sp$ hybridization and a linear shape.
$4$. $CO_2$: The central carbon atom forms $2$ double bonds with oxygen atoms. It has no lone pairs,resulting in $sp$ hybridization and a linear shape.
Therefore,$H_2S$ is the only molecule that is not linear.

(B) In the $H_2O$ molecule,the oxygen atom undergoes $sp^3$ hybridization.
Due to the presence of two lone pairs on the oxygen atom,the bond angle is compressed from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $104^{\circ} 28^{\prime}$ according to the $VSEPR$ theory.

(D) The $SO_2$ molecule has a bent geometry with $sp^2$ hybridization at the sulfur atom.
It contains one lone pair of electrons on the sulfur atom.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion.
This causes the bond angle to decrease from the ideal trigonal planar angle of $120^{\circ}$ to approximately $119.5^{\circ}$.

(D) The central atom $Sb$ has $5$ valence electrons and forms $5$ bonds with $F$ atoms,resulting in $5$ bond pairs and $0$ lone pairs.
According to $VSEPR$ theory,a molecule with $5$ bond pairs and $0$ lone pairs has $sp^3d$ hybridization.
The geometry of $SbF_5$ is Trigonal bipyramidal.

(B) To determine the geometry,we calculate the hybridization of the central atom using the formula: $H = \frac{1}{2} (V + M - C + A)$,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.
$1$. For $SeCl_2$: $H = \frac{1}{2} (6 + 2) = 4$ ($sp^3$ hybridization). With $2$ lone pairs,it has a bent shape.
$2$. For $SeF_4$: $H = \frac{1}{2} (6 + 4) = 5$ ($sp^3d$ hybridization). With $1$ lone pair,it has a seesaw shape,which is derived from a trigonal bipyramidal electron geometry.
$3$. For $SF_6$: $H = \frac{1}{2} (6 + 6) = 6$ ($sp^3d^2$ hybridization). It has an octahedral geometry.
$4$. For $TeF_6$: $H = \frac{1}{2} (6 + 6) = 6$ ($sp^3d^2$ hybridization). It has an octahedral geometry.
Among the given options,$SeF_4$ is the only one associated with the trigonal bipyramidal electron geometry.

(D) In $NH_{3}$ molecule,the nitrogen atom is $sp^{3}$ hybridized.
It has one lone pair and three bond pairs of electrons.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion.
Due to this,the bond angle decreases from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $107^{\circ} 18^{\prime}$.

(B) The $PCl_{5}$ molecule has a trigonal bipyramidal geometry.
In this structure,three $Cl$ atoms are present in the equatorial plane,forming a trigonal arrangement around the central $P$ atom.
The bond angle between these equatorial $Cl-P-Cl$ bonds is $120^{\circ}$.
The other two $Cl$ atoms are in the axial positions,making a $90^{\circ}$ angle with the equatorial plane and a $180^{\circ}$ angle with each other.

(A) The $C-O-H$ bond angle in $CH_3OH$ (methanol) is $108.9^{\circ}$.
In methanol,the oxygen atom is $sp^3$ hybridized and has two lone pairs of electrons.
According to $VSEPR$ theory,the lone pair-lone pair repulsion is greater than the lone pair-bond pair repulsion,which in turn is greater than the bond pair-bond pair repulsion.
This repulsion causes the bond angle to deviate from the ideal tetrahedral angle of $109.5^{\circ}$ to $108.9^{\circ}$.

(A) According to the $VSEPR$ theory,the repulsive interaction between electron pairs follows the order: $lone \ pair-lone \ pair > lone \ pair-bond \ pair > bond \ pair-bond \ pair$.
This is because a $lone \ pair$ is under the influence of only one nucleus (the central atom),whereas a $bond \ pair$ is shared between two nuclei.
Consequently,the electron cloud of a $lone \ pair$ occupies more space around the central atom,leading to greater repulsion compared to $bond \ pair$ interactions.

(C) $1$. $ClF_{3}$ has $sp^{3}d$ hybridization with two lone pairs,resulting in a $T$-shaped geometry.
$2$. $XeF_{2}$ has $sp^{3}d$ hybridization with three lone pairs,resulting in a linear geometry.
$3$. $BeF_{2}$ has $sp$ hybridization with no lone pairs,resulting in a linear geometry.
$4$. $SO_{2}$ has $sp^{2}$ hybridization with one lone pair,resulting in a bent geometry.
$5$. Both $XeF_{2}$ and $BeF_{2}$ have linear geometry. However,in standard multiple-choice questions of this type,$BeF_{2}$ is the most fundamental example of linear geometry due to $sp$ hybridization.

(B) The central atom in $BrF_5$ is Bromine $(Br)$,which has $7$ valence electrons.
It forms $5$ bonds with $5$ Fluorine $(F)$ atoms and has $1$ lone pair of electrons.
The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization.
According to $VSEPR$ theory,a molecule with $5$ bond pairs and $1$ lone pair has a square pyramidal geometry.

(B) The central iodine atom in $I_{3}^{-}$ undergoes $sp^{3}d$ hybridization.
It has $3$ lone pairs of electrons in the equatorial positions and $2$ bonding pairs in the axial positions.
According to $VSEPR$ theory,the presence of $3$ lone pairs results in a linear geometry with a bond angle of $180^{\circ}$.

(D) In $SnCl_{2}$,the central atom $Sn$ has a valence shell configuration of $5s^{2} 5p^{2}$. It forms two covalent bonds with two $Cl$ atoms,utilizing two $5p$ electrons. This leaves one lone pair in the $5s$ orbital. The total number of electron domains around $Sn$ is $3$ ($2$ bond pairs + $1$ lone pair),which corresponds to $sp^{2}$ hybridization. Due to the presence of one lone pair,the geometry is bent or $V$-shaped.

(A) In $SiO_{4}^{4-}$,the central silicon atom $(Si)$ has $4$ valence electrons.
Each of the $4$ oxygen atoms contributes $1$ electron to form $4$ single bonds with $Si$,and carries a negative charge.
The total number of electron pairs around the central $Si$ atom is $4$ (all bond pairs).
According to $VSEPR$ theory,$4$ bond pairs correspond to $sp^3$ hybridization.
Therefore,the geometry of the $SiO_{4}^{4-}$ anion is tetrahedral.

(A) In $NH_3$,the nitrogen atom undergoes $sp^3$ hybridization.
It has $3$ bond pairs and $1$ lone pair of electrons.
According to $VSEPR$ theory,the presence of a lone pair causes repulsion,which distorts the expected tetrahedral geometry into a pyramidal shape.

(B) In $H_2S$,the $H-S-H$ bond angle is found to be $92.1^{\circ}$.
This is significantly smaller than the $104.5^{\circ}$ bond angle observed in $H_2O$.
As we move down the group in the hydrides of the oxygen family $(H_2O, H_2S, H_2Se, H_2Te)$,the electronegativity of the central atom decreases.
This leads to less hybridization and the bond angles approach $90^{\circ}$ as the bonds are formed primarily by pure $p$-orbitals.

(A) In the $H_2O$ molecule,the oxygen atom undergoes $sp^3$ hybridization.
It has two bond pairs and two lone pairs of electrons.
According to $VSEPR$ theory,the presence of two lone pairs causes repulsion,which distorts the bond angle from the ideal tetrahedral angle of $109.5^{\circ}$ to $104.5^{\circ}$.
Therefore,the molecular geometry is described as bent or $V$-shaped,which is derived from a distorted tetrahedral electron geometry.

(D) The structure of the ozone $(O_3)$ molecule consists of a central oxygen atom double-bonded to one terminal oxygen atom and single-bonded to another terminal oxygen atom.
The central oxygen atom has $6$ valence electrons.
In the resonance structure,the central oxygen atom forms $3$ bonds (total $6$ bonding electrons) and has $1$ lone pair ($2$ electrons).
Using the formula: $\text{Formal charge} = V - L - \frac{1}{2}B$,where $V$ is the number of valence electrons,$L$ is the number of lone pair electrons,and $B$ is the number of bonding electrons.
For the central oxygen atom: $V = 6$,$L = 2$,$B = 6$.
$\text{Formal charge} = 6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1$.

(A) The $SO_2$ molecule has a bent or $V$-shaped geometry due to the presence of one lone pair on the sulfur atom.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion,which causes the bond angle to decrease from the ideal $120^{\circ}$ (for $sp^2$ hybridization) to approximately $119.3^{\circ}$.
Among the given options,$119.5^{\circ}$ is the closest value to the experimental bond angle.

(B) The chemical formula for chlorous acid is $HClO_2$.
In the structure of $HClO_2$,the chlorine atom is bonded to one hydroxyl group $(-OH)$ and one oxygen atom via a double bond.
The valence shell configuration of chlorine is $3s^2 3p^5$,meaning it has $7$ valence electrons.
In $HClO_2$,chlorine forms $3$ bonds ($1$ with $O$ of $-OH$ group and $2$ with the other $O$ atom).
Out of $7$ valence electrons,$3$ are involved in bonding,leaving $4$ electrons as lone pairs.
Since $2$ electrons make $1$ lone pair,$4$ electrons correspond to $2$ lone pairs on the chlorine atom.

(B) The chemical formula for hypochlorous acid is $HOCl$.
In this molecule,the chlorine atom is bonded to one oxygen atom and one hydrogen atom.
The valence shell configuration of chlorine is $3s^2 3p^5$.
Chlorine forms one single bond with oxygen $(Cl-O)$ and one single bond with hydrogen is not present; rather,the structure is $H-O-Cl$.
In $H-O-Cl$,the oxygen atom is bonded to both $H$ and $Cl$.
Oxygen has $2$ lone pairs,and chlorine,having $7$ valence electrons,uses $1$ electron for bonding with oxygen,leaving $6$ electrons as $3$ lone pairs on the chlorine atom.

(A) To determine the shape,we use the $VSEPR$ theory based on hybridization and lone pairs:
$1$. $BeCl_2$ ($sp$ hybridization,$0$ lone pairs) is linear.
$2$. $CO_2$ ($sp$ hybridization,$0$ lone pairs) is linear.
$3$. $XeF_2$ ($sp^3d$ hybridization,$3$ lone pairs) is linear.
$4$. $ICl_2^-$ ($sp^3d$ hybridization,$3$ lone pairs) is linear.
Both pairs $(BeCl_2, CO_2)$ and $(XeF_2, ICl_2^-)$ have identical linear shapes. Given the standard options provided in typical chemistry problems of this type,$BeCl_2$ and $CO_2$ are the most common examples of linear molecules.

(A) In $PH_{3}$,the phosphorus atom has one lone pair of electrons. According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion,which causes the bond angle to decrease from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to approximately $93.6^{\circ}$,resulting in a pyramidal geometry.
In $PH_{4}^{+}$,the phosphorus atom has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsions and the presence of four identical bond pair-bond pair interactions,$PH_{4}^{+}$ adopts a perfect tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.

(C) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - B)$,where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons (or atoms bonded to it).
$1$. $I_{3}^{+}$: The central $I$ atom has $7$ valence electrons. It forms $2$ bonds with other $I$ atoms and has a positive charge,so $V = 7 - 1 = 6$. Bonding electrons $B = 2$. Lone pairs = $\frac{1}{2} (6 - 2) = 2$.
$2$. $H_{2}O$: The central $O$ atom has $6$ valence electrons. It forms $2$ bonds with $H$ atoms. Lone pairs = $\frac{1}{2} (6 - 2) = 2$.
Both $I_{3}^{+}$ and $H_{2}O$ have $2$ lone pairs on their central atoms. Therefore,option $C$ is correct.

(D) The pyramidal structure of the covalent molecule $AB_{3}$ is shown below:
In this structure,the central atom $A$ is bonded to three $B$ atoms,forming $3$ bond pairs.
There is $1$ lone pair of electrons present on the central atom $A$.
Therefore,the number of lone pairs is $1$ and the number of bond pairs is $3$.

(C) According to $VSEPR$ theory:
For tetrahedral geometry: No. of electron pairs $= 4$,Bond pairs $= 4$,Lone pairs $= 0$.
For pyramidal geometry (e.g.,$NH_3$): No. of electron pairs $= 4$,Bond pairs $= 3$,Lone pairs $= 1$.
For trigonal planar geometry: No. of electron pairs $= 3$,Bond pairs $= 3$,Lone pairs $= 0$.
For octahedral geometry: No. of electron pairs $= 6$,Bond pairs $= 6$,Lone pairs $= 0$.
Therefore,the structure with three bond pairs and one lone pair is pyramidal.

(D) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} \times (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (monovalent atoms attached).
$I) \ ClF_3$: Central atom $Cl$ ($7$ valence electrons). $3$ $F$ atoms attached. $\text{Lone pairs} = \frac{1}{2} \times (7 - 3) = 2$.
$II) \ XeF_2$: Central atom $Xe$ ($8$ valence electrons). $2$ $F$ atoms attached. $\text{Lone pairs} = \frac{1}{2} \times (8 - 2) = 3$.
$III) \ SF_4$: Central atom $S$ ($6$ valence electrons). $4$ $F$ atoms attached. $\text{Lone pairs} = \frac{1}{2} \times (6 - 4) = 1$.
$IV) \ SiH_4$: Central atom $Si$ ($4$ valence electrons). $4$ $H$ atoms attached. $\text{Lone pairs} = \frac{1}{2} \times (4 - 4) = 0$.
Comparing the values: $IV (0) < III (1) < I (2) < II (3)$.
Therefore,the increasing order is $IV < III < I < II$.

(C) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} \times (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (shared with surrounding atoms).
$1$. $SF_6$: $S$ has $6$ valence electrons,$6$ bonds. Lone pairs = $\frac{1}{2} \times (6 - 6) = 0$.
$2$. $BF_3$: $B$ has $3$ valence electrons,$3$ bonds. Lone pairs = $\frac{1}{2} \times (3 - 3) = 0$.
$3$. $ClF_3$: $Cl$ has $7$ valence electrons,$3$ bonds. Lone pairs = $\frac{1}{2} \times (7 - 3) = 2$.
$4$. $PCl_5$: $P$ has $5$ valence electrons,$5$ bonds. Lone pairs = $\frac{1}{2} \times (5 - 5) = 0$.
$5$. $BrF_5$: $Br$ has $7$ valence electrons,$5$ bonds. Lone pairs = $\frac{1}{2} \times (7 - 5) = 1$.
$6$. $XeF_4$: $Xe$ has $8$ valence electrons,$4$ bonds. Lone pairs = $\frac{1}{2} \times (8 - 4) = 2$.
$7$. $H_2O$: $O$ has $6$ valence electrons,$2$ bonds. Lone pairs = $\frac{1}{2} \times (6 - 2) = 2$.
$8$. $SF_4$: $S$ has $6$ valence electrons,$4$ bonds. Lone pairs = $\frac{1}{2} \times (6 - 4) = 1$.
The molecules with two lone pairs are $ClF_3$,$XeF_4$,and $H_2O$. Total count is $3$.

(D) To determine the number of molecules with a lone pair on the central atom,we analyze the hybridization and structure of each molecule:
$1$. $BF_3$: Central atom $B$ has $3$ valence electrons,all involved in bonding. Lone pairs = $0$.
$2$. $SF_4$: Central atom $S$ has $6$ valence electrons,$4$ used for bonding,$1$ lone pair remains.
$3$. $SiCl_4$: Central atom $Si$ has $4$ valence electrons,all used for bonding. Lone pairs = $0$.
$4$. $XeF_4$: Central atom $Xe$ has $8$ valence electrons,$4$ used for bonding,$2$ lone pairs remain.
$5$. $NCl_3$: Central atom $N$ has $5$ valence electrons,$3$ used for bonding,$1$ lone pair remains.
$6$. $XeF_6$: Central atom $Xe$ has $8$ valence electrons,$6$ used for bonding,$1$ lone pair remains.
$7$. $PCl_5$: Central atom $P$ has $5$ valence electrons,all used for bonding. Lone pairs = $0$.
$8$. $HgCl_2$: Central atom $Hg$ has $2$ valence electrons,all used for bonding. Lone pairs = $0$.
$9$. $SnCl_2$: Central atom $Sn$ has $4$ valence electrons,$2$ used for bonding,$1$ lone pair remains.
The molecules with at least one lone pair are $SF_4, XeF_4, NCl_3, XeF_6, SnCl_2$. The total count is $5$.

(D) In $BrF_5$,the central atom $Br$ has $7$ valence electrons. $5$ electrons form bonds with $5$ $F$ atoms,leaving $2$ electrons ($1$ lone pair).
In $XeO_3$,$Xe$ has $8$ valence electrons. $6$ electrons form double bonds with $3$ $O$ atoms,leaving $2$ electrons ($1$ lone pair).
In $SO_2$,$S$ has $6$ valence electrons. $4$ electrons form double bonds with $2$ $O$ atoms,leaving $2$ electrons ($1$ lone pair).
Thus,the number of lone pairs are $1, 1, 1$ respectively.

(A) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pair} = \frac{V - (M + C - A)}{2}$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the charge of the cation,and $A$ is the charge of the anion.

Molecule	Lone Pairs
$BrF_5$	$1$
$SF_4$	$1$
$SnCl_2$	$1$
$XeF_4$	$2$
$NH_3$	$1$
$ClF_3$	$2$

Based on the calculation,the molecules $BrF_5$,$SF_4$,and $SnCl_2$ each contain exactly $1$ lone pair on the central atom.

(D) In the $XeOF_4$ molecule,the central atom $Xe$ is bonded to four $F$ atoms via single bonds and one $O$ atom via a double bond.
Thus,the total number of bond pairs is $4 + 2 = 6$.
Regarding lone pairs:
- Each of the four $F$ atoms has $3$ lone pairs $(4 \times 3 = 12)$.
- The $O$ atom has $2$ lone pairs.
- The $Xe$ atom has $1$ lone pair.
Total lone pairs $= 12 + 2 + 1 = 15$.

(D) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{V - N}{2}$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of monovalent atoms bonded to it.
$1$. For $IF_7$: Central atom $I$ has $7$ valence electrons. It forms $7$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{7 - 7}{2} = 0$.
$2$. For $IF_5$: Central atom $I$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{7 - 5}{2} = 1$.
$3$. For $ClF_3$: Central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{7 - 3}{2} = 2$.
$4$. For $XeF_2$: Central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{8 - 2}{2} = 3$.
The number of lone pairs is: $IF_7 (0) < IF_5 (1) < ClF_3 (2) < XeF_2 (3)$.
Thus,the correct order is $IF_7 < IF_5 < ClF_3 < XeF_2$.

(B) To determine the geometry,we look at the hybridization and lone pairs of each species:
$1$. $BO_3^{3-}$: Central atom $B$ has $3$ bonding pairs and $0$ lone pairs ($sp^2$ hybridization). Geometry: Trigonal planar.
$2$. $NH_3$: Central atom $N$ has $3$ bonding pairs and $1$ lone pair ($sp^3$ hybridization). Geometry: Trigonal pyramidal.
$3$. $PCl_3$: Central atom $P$ has $3$ bonding pairs and $1$ lone pair ($sp^3$ hybridization). Geometry: Trigonal pyramidal.
$4$. $BCl_3$: Central atom $B$ has $3$ bonding pairs and $0$ lone pairs ($sp^2$ hybridization). Geometry: Trigonal planar.
$5$. $ClF_3$: Central atom $Cl$ has $3$ bonding pairs and $2$ lone pairs ($sp^3d$ hybridization). Geometry: $T$-shaped.
$6$. $XeO_3$: Central atom $Xe$ has $3$ bonding pairs and $1$ lone pair ($sp^3$ hybridization). Geometry: Trigonal pyramidal.
Thus,only $BO_3^{3-}$ and $BCl_3$ have a trigonal planar structure. The total count is $2$.

(B) According to the $VSEPR$ theory,a molecule with a $T$-shaped geometry arises from an $AX_3E_2$ type molecule,where $A$ is the central atom,$X$ are the bonding atoms,and $E$ are the lone pairs of electrons.
In this configuration,there are $3$ bond pairs and $2$ lone pairs.
The total number of electron pairs = $(\text{Number of bond pairs} + \text{Number of lone pairs}) = 3 + 2 = 5$.
Therefore,the total number of electron pairs in the valence shell of the central atom is $5$.

(A) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pair} = \frac{1}{2} (V - M - C + A)$,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. Alternatively,we can use the structure:

Molecule	Lone pairs on central atom
$SO_3$	$0$
$SnCl_2$	$1$
$ClF_3$	$2$
$XeF_2$	$3$

The increasing order of lone pairs is $SO_3 (0) < SnCl_2 (1) < ClF_3 (2) < XeF_2 (3)$.

(D) The bond angles for the given molecules are as follows:
$1.$ $HgCl_2$: Linear geometry ($sp$ hybridization),bond angle = $180^{\circ}$.
$2.$ $SO_3$: Trigonal planar geometry ($sp^2$ hybridization),bond angle = $120^{\circ}$.
$3.$ $SiCl_4$: Tetrahedral geometry ($sp^3$ hybridization),bond angle = $109.5^{\circ}$.
$4.$ $NH_3$: Trigonal pyramidal geometry ($sp^3$ hybridization with one lone pair),bond angle = $107^{\circ}$.
Therefore,the correct order is $HgCl_2 > SO_3 > SiCl_4 > NH_3$.

(C) To determine the isostructural pairs,we analyze the geometry of each species:
$1$. $XeO_3$: $sp^3$ hybridized with $1$ lone pair,resulting in a pyramidal shape.
$2$. $CO_3^{2-}$: $sp^2$ hybridized with $0$ lone pairs,resulting in a trigonal planar shape.
$3$. $SO_3$: $sp^2$ hybridized with $0$ lone pairs,resulting in a trigonal planar shape.
$4$. $H_3O^{+}$: $sp^3$ hybridized with $1$ lone pair,resulting in a pyramidal shape.
$5$. $ClF_3$: $sp^3d$ hybridized with $2$ lone pairs,resulting in a $T$-shape.
Comparing the shapes:
- $(XeO_3, H_3O^{+})$ are both pyramidal.
- $(SO_3, CO_3^{2-})$ are both trigonal planar.
Therefore,the correct set of isostructural pairs is $(XeO_3, H_3O^{+})$ and $(SO_3, CO_3^{2-})$.

(A) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{V - M - C + A}{2}$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$(A)$ For $XeF_2$: $V=8, M=2, C=0, A=0$. $\text{Lone pairs} = \frac{8-2}{2} = 3$. Thus,$A-III$.
$(B)$ For $XeO_3$: $V=8, M=0$ (Oxygen is divalent),$C=0, A=0$. $\text{Lone pairs} = \frac{8-0}{2} = 4$ valence electrons involved in bonding,but since $O$ is divalent,$3$ double bonds use $6$ electrons,leaving $1$ lone pair. Thus,$B-IV$.
$(C)$ For $XeF_4$: $V=8, M=4, C=0, A=0$. $\text{Lone pairs} = \frac{8-4}{2} = 2$. Thus,$C-I$.
$(D)$ For $PF_6^-$: $V=5, M=6, C=0, A=1$. $\text{Lone pairs} = \frac{5-6+1}{2} = 0$. Thus,$D-II$.
The correct matching is $A-III, B-IV, C-I, D-II$.

$ (A) $ The shapes of the given molecules/ions are determined by $VSEPR$ theory:
$A. I_3^-$: The central iodine atom has $3$ lone pairs and $2$ bond pairs, resulting in a linear geometry ($sp^3d$ hybridization). Thus, $A-4$.
$B. ClF_3$: The central chlorine atom has $2$ lone pairs and $3$ bond pairs, resulting in a $T$-shaped geometry ($sp^3d$ hybridization). Thus, $B-1$.
$C. H_2O$: The central oxygen atom has $2$ lone pairs and $2$ bond pairs, resulting in an angular or bent geometry ($sp^3$ hybridization). Thus, $C-3$.
$D. SF_4$: The central sulfur atom has $1$ lone pair and $4$ bond pairs, resulting in a see-saw geometry ($sp^3d$ hybridization). Thus, $D-2$.
Therefore, the correct matching is $A-4, B-1, C-3, D-2$.

(B) $Sn$ and $Pb$ both belong to group $14$ of the periodic table.
Both $SnCl_2$ and $PbCl_2$ have a central atom with $2$ bond pairs and $1$ lone pair,resulting in a bent or angular molecular geometry.
Therefore,they are isostructural.

(D) To determine the shapes,we use the $VSEPR$ theory:
$1$. $XeF_2$: The central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms and has $3$ lone pairs. The steric number is $2 + 3 = 5$ ($sp^3d$ hybridization). The shape is linear.
$2$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$ ($sp^3d^2$ hybridization). The shape is square planar.
$3$. $XeO_3$: The central atom $Xe$ has $8$ valence electrons. It forms $3$ double bonds with $O$ atoms and has $1$ lone pair. The steric number is $3 + 1 = 4$ ($sp^3$ hybridization). The shape is pyramidal.

(C) $TeCl_4$ has a see-saw shape due to $sp^3d$ hybridization with one lone pair. $SeF_4$ also has a see-saw shape due to $sp^3d$ hybridization with one lone pair.
$CH_4$ and $PCl_4^{+}$ both have a tetrahedral shape due to $sp^3$ hybridization with zero lone pairs.
$SF_4$ has a see-saw shape ($sp^3d$,one lone pair),whereas $NH_4^{+}$ has a tetrahedral shape ($sp^3$,zero lone pairs). Thus,they have different shapes.
$SiH_4$ and $CCl_4$ both have a tetrahedral shape due to $sp^3$ hybridization with zero lone pairs.

(D) $XeF_4$: square planar; $PCl_4^{+}$: tetrahedral.
$NH_3$: pyramidal; $ClF_3$: $T$-shaped.
$SF_4$: see-saw; $NH_4^{+}$: tetrahedral.
$XeF_2$: linear; $I_3^{-}$: linear.
Thus,$XeF_2$ and $I_3^{-}$ have the same linear geometry.

(A) To determine the molecule where the number of lone pairs is greater than the number of bond pairs on the central atom,we analyze each option:
$1$. For $XeF_2$: Central atom $Xe$ has $3$ lone pairs and $2$ bond pairs. Here,$3 > 2$.
$2$. For $ClF_3$: Central atom $Cl$ has $2$ lone pairs and $3$ bond pairs. Here,$2 < 3$.
$3$. For $XeF_4$: Central atom $Xe$ has $2$ lone pairs and $4$ bond pairs. Here,$2 < 4$.
$4$. For $SF_4$: Central atom $S$ has $1$ lone pair and $4$ bond pairs. Here,$1 < 4$.
Therefore,the molecule with more lone pairs than bond pairs is $XeF_2$.

(C) The hybridization of $Sn$ in $SnCl_2$ is $sp^2$ with one lone pair and two bond pairs,resulting in a bent geometry.
The hybridization of $Xe$ in $XeF_4$ is $sp^3d^2$ with two lone pairs and four bond pairs,resulting in a square planar geometry.
The hybridization of $Cl$ in $ClF_3$ is $sp^3d$ with two lone pairs and three bond pairs,resulting in a $T$-shape geometry.
The hybridization of $I$ in $IF_5$ is $sp^3d^2$ with one lone pair and five bond pairs,resulting in a square pyramidal geometry.
Therefore,the correct match is $(a-i), (b-v), (c-iv), (d-iii)$.

(A) The geometry of $H_2O$ is bent (angular) due to two lone pairs and two bond pairs on the oxygen atom.
Ammonia $(NH_3)$ has a trigonal pyramidal shape due to one lone pair and three bond pairs on the nitrogen atom.
Since $H_2O$ (bent) and $NH_3$ (trigonal pyramidal) have different geometries,the pair $(H_2O, NH_3)$ is an incorrect match.
$BeCl_2$ and $CO_2$ are both linear.
$SF_4$ and $TeCl_4$ both have a see-saw shape.
$ClF_3$ and $ICl_3$ both have a $T$-shape.
Therefore,the incorrect match is $H_2O, NH_3$.

(C) For $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridisation. Due to $2$ lone pairs,the geometry is square planar.
For $XeOF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom. It has $1$ lone pair. The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridisation. Due to $1$ lone pair,the geometry is square pyramidal.