VSEPR Theory Questions in English

(C) The central atom in $IF_3$ is $I$ (iodine),which has $7$ valence electrons.
It forms $3$ bond pairs with $F$ atoms and has $2$ lone pairs of electrons.
The total number of electron pairs is $3 + 2 = 5$,which corresponds to $sp^3d$ hybridization.
According to $VSEPR$ theory,the presence of $2$ lone pairs in the equatorial positions of a trigonal bipyramidal geometry results in a $T$-shaped molecular geometry.

(A) The geometries of the given Xenon compounds are as follows:
$(A)$ $XeF_2$: The molecule has $sp^3d$ hybridization with $3$ lone pairs on the central $Xe$ atom,resulting in a linear geometry $(A-3)$.
$(B)$ $XeO_3$: The molecule has $sp^3$ hybridization with $1$ lone pair on the central $Xe$ atom,resulting in a trigonal pyramidal geometry $(B-1)$.
$(C)$ $XeF_6$: The molecule has $sp^3d^3$ hybridization with $1$ lone pair on the central $Xe$ atom,resulting in a distorted octahedral geometry $(C-2)$.
$(D)$ $XeOF_4$: The molecule has $sp^3d^2$ hybridization with $1$ lone pair on the central $Xe$ atom,resulting in a square pyramidal geometry $(D-4)$.
Thus,the correct matching is $A-3, B-1, C-2, D-4$.

(B) The central atom $Xe$ has $8$ valence electrons.
In $XeF_4$,$Xe$ forms $4$ bonds with $F$ atoms and has $2$ lone pairs of electrons.
The total number of electron pairs is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridisation.
According to $VSEPR$ theory,the presence of $2$ lone pairs in an octahedral geometry results in a square planar shape.

(C) To determine the $AB_nE_m$ type for a molecule,we identify the central atom $(A)$,the number of atoms bonded to it $(B)$,and the number of lone pairs on the central atom $(E)$:
$1$. For $SO_2$: The central atom $S$ has $2$ bond pairs and $1$ lone pair,so it is $AB_2E$.
$2$. For $H_2O_2$: The structure is $H-O-O-H$. Each oxygen atom is bonded to $2$ atoms and has $2$ lone pairs,but it is not a simple $AB_2E_2$ system due to the $O-O$ bond.
$3$. For $H_2O$: The central oxygen atom is bonded to $2$ hydrogen atoms $(B=2)$ and has $2$ lone pairs $(E=2)$. Thus,it is $AB_2E_2$.
$4$. For $XeF_2$: The central xenon atom is bonded to $2$ fluorine atoms $(B=2)$ and has $3$ lone pairs $(E=3)$. Thus,it is $AB_2E_3$.
Therefore,the correct molecule is $H_2O$.

(A) The geometry of the phosphorus pentachloride $(PCl_5)$ molecule is trigonal bipyramidal,where the central $P$ atom is $sp^3d$ hybridized.
In this structure,three $Cl$ atoms are located at the equatorial positions,forming a trigonal plane.
The bond angle between any two equatorial $Cl-P-Cl$ bonds is $120^{\circ}$.
Since there are three such equatorial $Cl$ atoms,the number of $120^{\circ}$ angles is $3$ (specifically,the angles between equatorial-equatorial bonds).

(B) The central iodine atom in the $I_3^{-}$ ion is surrounded by $3$ lone pairs and $2$ bond pairs,resulting in a total of $5$ electron pairs.
According to $VSEPR$ theory,this corresponds to $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
Since the $3$ lone pairs occupy the equatorial positions to minimize repulsion,the remaining $2$ iodine atoms occupy the axial positions,resulting in a linear molecular geometry.
Hence,the correct option is $B$.

(D) The bond angles for the given molecules are as follows:
$BF_3$: $120^{\circ}$ (trigonal planar geometry)
$NH_3$: $107^{\circ}$ (pyramidal geometry)
$NF_3$: $102^{\circ}$ (pyramidal geometry,lower than $NH_3$ due to higher electronegativity of $F$)
$PCl_3$: $100^{\circ}$ (pyramidal geometry,lower than $NH_3$ due to larger size of $P$ atom)
Thus,the correct order of bond angles is $BF_3 > NH_3 > NF_3 > PCl_3$.
Therefore,the correct option is $(D)$.

(B) The central $P$ atom in $PF_3$ has $5$ valence electrons. It forms $3$ $P-F$ bonds and has $1$ lone pair of electrons.
Total electron pairs = $3$ (bonding) + $1$ (lone pair) = $4$.
According to $VSEPR$ theory,$4$ electron pairs result in a tetrahedral electron geometry.
Due to the presence of one lone pair,the molecular geometry is trigonal pyramidal.
Thus,the correct option is $B$.

(A) The central $B$ atom in $BH_4^{-}$ undergoes $sp^3$ hybridisation.
This results in a tetrahedral geometry with $4$ bond pairs and $0$ lone pairs of electrons,as shown in the structure.

(A) The central oxygen atom in $H_3O^{+}$ is $sp^3$ hybridized.
It has $3$ bond pairs and $1$ lone pair of electrons,resulting in a total of $4$ electron domains.
According to the $\text{VSEPR}$ theory,the electron geometry is tetrahedral,but the presence of one lone pair distorts the shape.
Therefore,the molecular geometry of $H_3O^{+}$ is trigonal pyramidal.
Hence,the correct option is $(A)$.

(C) To determine isostructural molecules,we check the hybridization and geometry of the central atom:
$CO_2$: The central $C$ atom is $sp$-hybridized,resulting in a linear geometry.
$[NO_2]^{+}$: The central $N$ atom is $sp$-hybridized,resulting in a linear geometry.
$SiO_2$: Exists as a giant covalent network (quartz) with $sp^3$ hybridization.
$SO_2$ and $TeO_2$: Both have $sp^2$ hybridization with one lone pair,resulting in a bent (angular) geometry.
Since both $CO_2$ and $[NO_2]^{+}$ are linear,they are isostructural.
Thus,the correct option is $C$.

(C) The arrangement of electron pairs around the central atom is octahedral when the steric number is $6$ (i.e.,$sp^3d^2$ hybridization). $A$ molecule has a non-octahedral shape if it contains one or more lone pairs of electrons.
$(i)$ $SF_6$: Steric number = $6$ ($6$ bond pairs,$0$ lone pairs). Shape is octahedral.
$(ii)$ $XeF_6$: Steric number = $7$ ($6$ bond pairs,$1$ lone pair). The electron pair arrangement is pentagonal bipyramidal,not octahedral.
$(iii)$ $BrF_5$: Steric number = $6$ ($5$ bond pairs,$1$ lone pair). The electron pair arrangement is octahedral,but the shape is square pyramidal (not octahedral).
$(iv)$ $XeO_2F_4$: Steric number = $6$ ($6$ bond pairs,$0$ lone pairs). Shape is octahedral.
Therefore,$BrF_5$ is the molecule where the electron pair arrangement is octahedral but the shape is not.

(C) To determine the geometry,we calculate the hybridization and the number of lone pairs for each molecule:
$1$. $PCl_5$: $sp^3d$ hybridization,trigonal bipyramidal geometry.
$2$. $BrF_5$: $sp^3d^2$ hybridization with $1$ lone pair,square pyramidal geometry.
$3$. $ClF_5$: $sp^3d^2$ hybridization with $1$ lone pair,square pyramidal geometry.
$4$. $PF_5$: $sp^3d$ hybridization,trigonal bipyramidal geometry.
$5$. $ClF_3$: $sp^3d$ hybridization with $2$ lone pairs,$T$-shaped geometry.
$6$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,square planar geometry.
$7$. $XeF_2$: $sp^3d$ hybridization with $3$ lone pairs,linear geometry.
$8$. $IF_5$: $sp^3d^2$ hybridization with $1$ lone pair,square pyramidal geometry.
Thus,the molecules with square pyramidal geometry are $BrF_5, ClF_5$,and $IF_5$.
The total count is $3$.

(C) The molecular geometries are determined by $VSEPR$ theory:

Molecule	Type	Shape
$BrF_5$	$AB_5E$	Square pyramidal
$SF_4$	$AB_4E$	See-saw
$XeF_4$	$AB_4E_2$	Square planar
$ClF_3$	$AB_3E_2$	$T$-shape

Matching the items:
$A$ $(BrF_5)$ matches $III$ ($AB_5E$,square pyramidal).
$B$ $(SF_4)$ matches $I$ ($AB_4E$,see-saw).
$C$ $(XeF_4)$ matches $II$ ($AB_4E_2$,square planar).
$D$ $(ClF_3)$ matches $IV$ ($AB_3E_2$,$T$-shape).
Therefore,the correct match is $A-III, B-I, C-II, D-IV$.

(B) Statement $(i)$ is correct: $ClF_3$ has $3$ bond pairs and $2$ lone pairs $(AB_3E_2)$,and $SO_2$ has $2$ bond pairs and $1$ lone pair $(AB_2E)$.
Statement $(ii)$ is correct: $SF_4$ has $4$ bond pairs and $1$ lone pair,resulting in a "See-saw" geometry.
Statement $(iii)$ is incorrect: $HgCl_2$ has $2$ bond pairs and $0$ lone pairs,resulting in a linear shape. $PbCl_2$ has $2$ bond pairs and $1$ lone pair,resulting in a bent shape.
Therefore,only statement $(iii)$ is incorrect.

(A) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pair} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons.
$(i)$ $SO_2$: $S$ has $6$ valence electrons. It forms $2$ double bonds with $O$ atoms,so $N = 4$. $\text{Lone pair} = \frac{1}{2} (6 - 4) = 1$.
$(ii)$ $XeF_4$: $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms,so $N = 4$. $\text{Lone pair} = \frac{1}{2} (8 - 4) = 2$.
$(iii)$ $PbCl_2$: $Pb$ (Group $14$) has $4$ valence electrons. It forms $2$ single bonds with $Cl$ atoms,so $N = 2$. $\text{Lone pair} = \frac{1}{2} (4 - 2) = 1$.
$(iv)$ $SF_4$: $S$ has $6$ valence electrons. It forms $4$ single bonds with $F$ atoms,so $N = 4$. $\text{Lone pair} = \frac{1}{2} (6 - 4) = 1$.
$(v)$ $ClF_3$: $Cl$ has $7$ valence electrons. It forms $3$ single bonds with $F$ atoms,so $N = 3$. $\text{Lone pair} = \frac{1}{2} (7 - 3) = 2$.
Thus,molecules $(i)$,$(iii)$,and $(iv)$ have only one lone pair on their central atoms.

(B) The bonds that are projecting out of the plane are $W$ and $Z$.
Here,normal lines $(-)$ show the bonds lying in the plane of the paper.
Solid-wedge shows the bond projecting out of the plane of paper towards the observer $(W)$.
Dashed-wedge shows the bond projecting out of the plane of paper away from the observer $(Z)$.

(A) In the $CH_4$ molecule,the carbon atom is $sp^3$ hybridized with a tetrahedral geometry.
In the provided structure:
$1$. The bonds $X$ and $Y$ are represented by simple lines,indicating they lie in the plane of the paper.
$2$. The bond $Z$ is represented by a dashed wedge,indicating it projects away from the observer (behind the plane).
$3$. The bond $W$ is represented by a solid wedge,indicating it projects towards the observer (in front of the plane).
Therefore,the correct mapping is: In-plane $(A)$ = $X, Y$; Away from observer $(B)$ = $Z$; Towards observer $(C)$ = $W$.

(B) $XeF_2$ and $PF_5$ both have $sp^3d$ hybridization.
According to the $\text{VSEPR}$ theory,their shapes are different.
$XeF_2$ has $3$ lone pairs on the central $Xe$ atom,resulting in a linear geometry.
$PF_5$ has $0$ lone pairs on the central $P$ atom,resulting in a trigonal bipyramidal geometry.

(B) To determine the geometry,we use $VSEPR$ theory:
$a$. $SF_4$: Steric number is $5$ ($4$ bond pairs + $1$ lone pair),resulting in a See-Saw shape $(iv)$.
$b$. $BrF_5$: Steric number is $6$ ($5$ bond pairs + $1$ lone pair),resulting in a Square Pyramidal shape $(v)$.
$c$. $ClF_3$: Steric number is $5$ ($3$ bond pairs + $2$ lone pairs),resulting in a $T$-shape $(ii)$.
$d$. $PbCl_2$: This is an ionic compound with a bent geometry in the gas phase due to the lone pair on $Pb^{2+}$ $(iii)$.
Therefore,the correct matching is $a-iv, b-v, c-ii, d-iii$.

(D) The central atom $Br$ has $7$ valence electrons. In $BrF_5$,it forms $5$ single bonds with $F$ atoms,utilizing $5$ electrons. This leaves $2$ electrons,which form $1$ lone pair. The total number of electron pairs 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 adopts a square pyramidal geometry.

(D) In $SF_4$,the central atom is sulfur $(S)$.
Sulfur has $6$ valence electrons.
It forms $4$ single bonds with $4$ fluorine atoms,using $4$ valence electrons.
This leaves $6 - 4 = 2$ electrons,which form $1$ lone pair on the sulfur atom.
Thus,there are $4$ bond pairs and $1$ lone pair.
The steric number is $4 + 1 = 5$,which corresponds to $sp^3d$ hybridization.
Due to the presence of $1$ lone pair in the equatorial position of a trigonal bipyramidal geometry,the shape of the $SF_4$ molecule is see-saw.

(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 electrons involved in bonding (monovalent atoms).
$1$. For $NH_3$: Central atom $N$ has $5$ valence electrons. It forms $3$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (5 - 3) = 1$.
$2$. For $H_2O$: Central atom $O$ has $6$ valence electrons. It forms $2$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (6 - 2) = 2$.
$3$. For $ClF_3$: Central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (7 - 3) = 2$.
$4$. For $XeF_2$: Central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (8 - 2) = 3$.
Thus,$XeF_2$ has the maximum number of lone pairs $(3)$ on the central atom.

(B) In $PCl_5$ (phosphorus pentachloride),the geometry is trigonal bipyramidal.
Due to the presence of three equatorial bonds and two axial bonds,the axial $P-Cl$ bonds are longer than the equatorial $P-Cl$ bonds because of greater repulsion from the equatorial bond pairs.
In $SF_6$,$BCl_3$,and $CCl_4$,all bond lengths are equivalent due to the high symmetry of the molecules.

(A) According to $VSEPR$ theory,the bond angle depends on the hybridization and the number of lone pairs on the central atom.1. $SO_2$ $(IV)$: The central atom $S$ is $sp^2$ hybridized with one lone pair. The bond angle is approximately $119.5^\circ$.2. $CH_4$ $(III)$: The central atom $C$ is $sp^3$ hybridized with no lone pairs. It has a perfect tetrahedral bond angle of $109.5^\circ$.3. $NH_3$ $(II)$: The central atom $N$ is $sp^3$ hybridized with one lone pair. Due to lone pair-bond pair repulsion,the bond angle reduces to approximately $107^\circ$.4. $H_2O$ $(I)$: The central atom $O$ is $sp^3$ hybridized with two lone pairs. Due to greater lone pair-lone pair repulsion,the bond angle reduces further to approximately $104.5^\circ$.Therefore,the correct order is: $SO_2$ $(IV)$ $> CH_4$ $(III)$ $> NH_3$ $(II)$ $> H_2O$ $(I)$.

(D) The structure of ozone $(O_3)$ consists of a central oxygen atom double-bonded to one terminal oxygen atom and single-bonded to another terminal oxygen atom.
Using the formal charge formula: $FC = V - L - \frac{1}{2}B$,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: $FC = 6 - 2 - \frac{1}{2}(6) = +1$.
For the terminal oxygen atom with a double bond: $FC = 6 - 4 - \frac{1}{2}(4) = 0$.
For the terminal oxygen atom with a single bond: $FC = 6 - 6 - \frac{1}{2}(2) = -1$.
Thus,the formal charges on the terminal oxygen atoms are $0$ and $-1$.

(A) The structure of a molecule is determined by its hybridization and the number of lone pairs on the central atom.
$BF_3$ has $sp^2$ hybridization with no lone pairs on the boron atom,resulting in a trigonal planar geometry.
$NH_3$ and $NF_3$ have $sp^3$ hybridization with one lone pair,resulting in a trigonal pyramidal geometry.
$H_2S$ has $sp^3$ hybridization with two lone pairs,resulting in a bent ($V$-shaped) geometry.
Since $BF_3$ is the only molecule with a trigonal planar structure,the correct option is $A$.

(C) In $SF_6$,the central sulphur atom is bonded to $6$ fluorine atoms.
There are $6$ bond pairs and $0$ lone pairs around the sulphur atom.
According to $VSEPR$ theory,a molecule with $6$ bond pairs and $0$ lone pairs has an octahedral geometry.

(C) The reaction is $BCl_3 + NH_3 \rightarrow BCl_3 \cdot NH_3$ (where $X = BCl_3 \cdot NH_3$).
In $BCl_3$,the boron atom is $sp^2$ hybridized with three bond pairs and no lone pairs,resulting in a trigonal planar geometry.
In the adduct $X$ $(BCl_3 \cdot NH_3)$,the boron atom undergoes a change in hybridization from $sp^2$ to $sp^3$ as it accepts a lone pair from the nitrogen atom of $NH_3$.
Consequently,the boron atom in $X$ adopts a tetrahedral geometry.

(C) $BF_3$ is used as a catalyst in several industrial processes due to its strong Lewis acid nature.
$sp^2$-hybridized boron in $BF_3$ is electron-deficient ($6 \ e^-$ in valence shell,$2 \ e^-$ less than the octet) and acts as a strong Lewis acid.

(D) The structure of $Al_2Cl_6$ (aluminum chloride dimer) consists of two $AlCl_4$ tetrahedra sharing an edge.
In this structure,the terminal $Cl-Al-Cl$ bond angle $(b_1)$ is approximately $118^{\circ}$.
The bridging $Al-Cl-Al$ bond angle $(b_2)$ is approximately $79^{\circ}$.
The $Cl-Al-Cl$ bond angle involving the bridging chlorine atoms $(b_3)$ is approximately $101^{\circ}$.
Therefore,the values for $b_1, b_2, b_3$ are $118^{\circ}, 79^{\circ}, 101^{\circ}$ respectively.

(B) $1$. In $N(SiH_3)_3$,the nitrogen atom has a lone pair which is donated into the vacant $3d$-orbital of the silicon atom,forming a $p\pi-d\pi$ back-bonding. This makes the nitrogen atom $sp^2$ hybridized,resulting in a $planar$ geometry.
$2$. In $Me_3N$ (trimethylamine),the carbon atom does not have vacant $d$-orbitals,so no back-bonding occurs. The nitrogen atom remains $sp^3$ hybridized with a lone pair,resulting in a $pyramidal$ geometry.
$3$. In $(SiH_3)_3P$,the phosphorus atom is larger and has a lower electronegativity compared to nitrogen. The lone pair on phosphorus is stereochemically active and does not participate in significant back-bonding with silicon. Thus,it retains a $pyramidal$ geometry.

(D) The nitrogen atom in the $NO_3^{-}$ ion has a valence shell configuration of $2s^2 2p^3$.
In the nitrate ion,nitrogen forms three sigma bonds (one with each oxygen atom) and one pi bond (delocalized over the three oxygen atoms).
This means nitrogen uses all its $5$ valence electrons to form $4$ bonds ($3$ single bonds and $1$ double bond equivalent in resonance).
Therefore,the number of bond pairs of electrons on the nitrogen atom is $4$.
Since all valence electrons are involved in bonding,there are $0$ lone pairs of electrons on the nitrogen atom.
Thus,the correct answer is $4,0$.

(D) To identify the molecule with a lone pair on the sulphur atom,we examine the oxidation state and bonding of sulphur in each molecule:
$1$. In $H_2SO_5$ (peroxymonosulphuric acid),$H_2S_2O_8$ (peroxydisulphuric acid),and $H_2S_2O_7$ (pyrosulphuric acid),the sulphur atom is in the $+6$ oxidation state. In this state,sulphur uses all its valence electrons for bonding,leaving no lone pair.
$2$. In $H_2SO_3$ (sulphurous acid),the sulphur atom is in the $+4$ oxidation state. The electronic configuration of sulphur is $[Ne] 3s^2 3p^4$. In $H_2SO_3$,sulphur forms two $S-OH$ bonds and one $S=O$ double bond. This utilizes $4$ valence electrons,leaving one lone pair on the sulphur atom.
Therefore,$H_2SO_3$ is the correct molecule.

(C) The chemical reaction is:
$S + 2H_2SO_4 \text{ (conc.)} \longrightarrow 3SO_2 + 2H_2O$
Here,$X = SO_2$ (gas) and $Y = H_2O$ (liquid).
Both $SO_2$ and $H_2O$ are triatomic molecules.
In $SO_2$,the central atom is sulphur $(S)$. Sulphur has $6$ valence electrons. It forms two double bonds with two oxygen atoms,using $4$ electrons. Thus,it has $6 - 4 = 2$ electrons remaining,which form $1$ lone pair.
In $H_2O$,the central atom is oxygen $(O)$. Oxygen has $6$ valence electrons. It forms two single bonds with two hydrogen atoms,using $2$ electrons. Thus,it has $6 - 2 = 4$ electrons remaining,which form $2$ lone pairs.
Therefore,the number of lone pairs on the central atoms of $X$ and $Y$ are $1$ and $2$ respectively.

(B) Molecular bromine $(Br_2)$ reacts with $3$ moles of molecular fluorine $(F_2)$ to produce $2$ moles of $BrF_3$.
$BrF_3$ is an interhalogen compound.
The chemical reaction is:
$Br_{2(g)} + 3F_{2(g)} \longrightarrow 2BrF_{3(g)}$
In $BrF_3$,the central bromine atom has $7$ valence electrons. It forms $3$ covalent bonds with $3$ fluorine atoms,using $3$ electrons.
Remaining electrons = $7 - 3 = 4$ electrons,which form $2$ lone pairs.
Thus,the total number of lone pairs at the central bromine atom is $2$.

(B) In $SF_6$,the sulphur atom is $sp^3d^2$ hybridized,resulting in an octahedral geometry.
Six fluorine atoms surround the central sulphur atom,creating a highly crowded environment.
This steric hindrance prevents the attack of any nucleophile or electrophile on the sulphur atom,making $SF_6$ kinetically inert.

(B) The reaction of fluorine with dilute $NaOH$ is given by:
$2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$
Thus,the gaseous product $A$ is oxygen difluoride $(OF_2)$.
In $OF_2$,the central oxygen atom is $sp^3$ hybridized with two bond pairs and two lone pairs.
Due to the high electronegativity of fluorine atoms,the lone pair-lone pair repulsion is significant,which compresses the bond angle.
The bond angle in $OF_2$ is $103^{\circ}$.

(A) To determine the number of lone pairs on the central $Xe$ atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - M)$,where $V$ is the number of valence electrons of the central atom ($8$ for $Xe$) and $M$ is the number of monovalent atoms bonded to it (Oxygen is divalent,so it does not contribute to $M$).
$(1) XeO_3$: $V=8$,$M=0$. $\text{Lone pairs} = \frac{1}{2} (8 - 0) = 4$ electrons,which is $1$ lone pair.
$(2) XeF_6$: $V=8$,$M=6$. $\text{Lone pairs} = \frac{1}{2} (8 - 6) = 2$ electrons,which is $1$ lone pair.
$(3) XeF_2$: $V=8$,$M=2$. $\text{Lone pairs} = \frac{1}{2} (8 - 2) = 6$ electrons,which is $3$ lone pairs.
$(4) XeF_4$: $V=8$,$M=4$. $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons,which is $2$ lone pairs.
$(5) XeOF_4$: $V=8$,$M=4$ (Oxygen is divalent). $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons,which is $1$ lone pair.
Comparing the results,$XeO_3$,$XeF_6$,and $XeOF_4$ all have $1$ lone pair. Among the given options,the pair $(XeO_3, XeF_6)$ matches.

(A) In $CH_3OH$ (methanol),the oxygen atom is $sp^3$ hybridized with two lone pairs. Due to the repulsion between lone pairs,the bond angle is compressed to less than the tetrahedral angle of $109^{\circ} 28'$.
In $CH_3OCH_3$ (dimethyl ether),the oxygen atom is also $sp^3$ hybridized,but the presence of two bulky methyl groups causes steric repulsion between them,which increases the bond angle to greater than $109^{\circ} 28'$.
Therefore,$X < 109^{\circ} 28'$ and $Y > 109^{\circ} 28'$.

(B) To determine the presence of lone pairs on the central atom,we calculate the number of lone pairs using the formula: $\text{Lone pair} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (monovalent atoms attached).
$i$. $SnCl_2$ ($Sn$ has $2$ lone pairs),$NH_3$ ($N$ has $1$ lone pair),$SF_4$ ($S$ has $1$ lone pair).
$ii$. $HgCl_2$ ($Hg$ has $0$ lone pairs),$SO_3$ ($S$ has $0$ lone pairs),$SF_6$ ($S$ has $0$ lone pairs).
$iii$. $BeCl_2$ ($Be$ has $0$ lone pairs),$BF_3$ ($B$ has $0$ lone pairs),$PCl_5$ ($P$ has $0$ lone pairs).
$iv$. $ClF_3$ ($Cl$ has $2$ lone pairs),$BrF_5$ ($Br$ has $1$ lone pair),$XeF_6$ ($Xe$ has $1$ lone pair).
Thus,sets $ii$ and $iii$ contain molecules where the central atom has no lone pairs.

(A) According to the $VSEPR$ theory,electron pairs arrange themselves as far apart as possible to minimize repulsion.
The magnitude of different types of electronic repulsions follows the order:
$Lone pair - Lone pair > Lone pair - Bond pair > Bond pair - Bond pair$
Therefore,the correct order is $(I) > (II) > (III)$.

(C) $1$. The molecule '$X$' has $sp^3d$ hybridization and a see-saw geometry.
$2$. $sp^3d$ hybridization involves $5$ electron pairs around the central atom.
$3$. $A$ see-saw shape occurs when there are $4$ bond pairs and $1$ lone pair of electrons.
$4$. In $SF_4$,Sulfur $(S)$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair,resulting in $sp^3d$ hybridization and a see-saw shape.
$5$. $ClF_3$ has $T$-shaped geometry,$XeF_4$ has square planar geometry,and $BrF_5$ has square pyramidal geometry.
$6$. Therefore,'$X$' is $SF_4$.

(D) To determine the bond angles,we look at the hybridization and lone pairs of the central atoms:
$1$. $H_2O$: $sp^3$ hybridization,$2$ lone pairs,bond angle $\approx 104.5^\circ$.
$2$. $NH_3$: $sp^3$ hybridization,$1$ lone pair,bond angle $\approx 107^\circ$.
$3$. $O_3$: $sp^2$ hybridization,$1$ lone pair,bond angle $\approx 117^\circ$.
$4$. $SO_2$: $sp^2$ hybridization,$1$ lone pair,bond angle $\approx 119^\circ$.
Comparing these,the increasing order is $H_2O < NH_3 < O_3 < SO_2$.
Thus,the correct option is $D$.

(D) To determine the bond angles,we analyze the geometry of each molecule:
$1$. $P_4$: The phosphorus atoms are at the corners of a tetrahedron. The bond angle is $60^\circ$.
$2$. $S_6$: This molecule has a chair-like conformation with bond angles approximately $102^\circ$.
$3$. $S_8$: This molecule has a crown-shaped conformation with bond angles approximately $107^\circ$.
$4$. $O_3$: Ozone has a bent geometry with a bond angle of approximately $117^\circ$.
Comparing these values: $60^\circ (P_4) < 102^\circ (S_6) < 107^\circ (S_8) < 117^\circ (O_3)$.
In terms of designations: $B (P_4) < C (S_6) < A (S_8) < D (O_3)$.
Therefore,the correct order is $B < C < A < D$.

(D) The given reactions are part of the copper extraction process:
$1. 2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$ (where $X = Cu_2S$)
$2. 2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2$ (where $Y = SO_2$)
In the $SO_2$ molecule,the sulfur atom is $sp^2$ hybridized with one lone pair of electrons.
Due to the presence of the lone pair,the geometry is bent or angular.

(D) To determine the shape of the molecules, we use the $VSEPR$ theory:
$A. SF_4$: Hybridisation is $sp^3d$ with one lone pair, resulting in a see-saw shape $(III)$.
$B. ClF_3$: Hybridisation is $sp^3d$ with two lone pairs, resulting in a $T$-shaped geometry $(I)$.
$C. BrF_5$: Hybridisation is $sp^3d^2$ with one lone pair, resulting in a square pyramidal shape $(IV)$.
$D. XeF_4$: Hybridisation is $sp^3d^2$ with two lone pairs, resulting in a square planar shape $(II)$.
Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

(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 monovalent atoms bonded to it.
$1$. For $ClF_3$: $V = 7, N = 3$. $\text{Lone pairs} = \frac{1}{2} (7 - 3) = 2$.
$2$. For $NF_3$: $V = 5, N = 3$. $\text{Lone pairs} = \frac{1}{2} (5 - 3) = 1$.
$3$. For $SF_4$: $V = 6, N = 4$. $\text{Lone pairs} = \frac{1}{2} (6 - 4) = 1$.
$4$. For $XeF_4$: $V = 8, N = 4$. $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 2$.

Compound	Lone pairs
$ClF_3$	$2$
$NF_3$	$1$
$SF_4$	$1$
$XeF_4$	$2$

Therefore,the correct sequence is $2, 1, 1, 2$.

(B) In the planar structure of the nitric acid $(HNO_3)$ molecule,the nitrogen atom is $sp^2$ hybridized.
Based on the experimental structural data provided in the image:
The bond angle $H-O-N$ is approximately $102.2^{\circ}$.
The bond angle $O-N-O$ (between the two oxygen atoms bonded to nitrogen) is approximately $130.27^{\circ}$.
Therefore,the respective bond angles are $102^{\circ}$ and $130^{\circ}$.