VSEPR Theory Questions in English

(A) According to the $VSEPR$ theory,for an $AB_{3}$ molecule to have a $T$-shaped geometry,the central atom $A$ must have $3$ bond pairs and $2$ lone pairs of electrons.
This corresponds to a steric number of $5$,which implies $sp^{3}d$ hybridization.
As shown in the structure,there are $2$ lone pairs on the central atom $A$.

(A) To determine the number of lone pairs on the central atom $(Xe)$ in each species:
$1$. $XeF_{2}$: $Xe$ has $8$ valence electrons. $2$ are used for bonding with $F$ atoms. Remaining electrons = $8-2 = 6$,which form $3$ lone pairs. Thus,$(a) - (iv)$.
$2$. $XeO_{2}F_{2}$: $Xe$ has $8$ valence electrons. $2$ are used for $F$ atoms and $4$ are used for $2$ double-bonded $O$ atoms. Remaining electrons = $8-2-4 = 2$,which form $1$ lone pair. Thus,$(b) - (ii)$.
$3$. $XeO_{3}F_{2}$: $Xe$ has $8$ valence electrons. $2$ are used for $F$ atoms and $6$ are used for $3$ double-bonded $O$ atoms. Remaining electrons = $8-2-6 = 0$. Thus,$(c) - (i)$.
$4$. $XeF_{4}$: $Xe$ has $8$ valence electrons. $4$ are used for bonding with $F$ atoms. Remaining electrons = $8-4 = 4$,which form $2$ lone pairs. Thus,$(d) - (iii)$.
Therefore,the correct matching is $a-iv, b-ii, c-i, d-iii$.

(A) To determine the shape of the given species,we use the $VSEPR$ theory:
$1$. $SO_{3}$: The central $S$ atom has $3$ bond pairs and $0$ lone pairs. The geometry is trigonal planar (non-pyramidal).
$2$. $NO_{3}^{-}$: The central $N$ atom has $3$ bond pairs and $0$ lone pairs. The geometry is trigonal planar (non-pyramidal).
$3$. $PCl_{3}$: The central $P$ atom has $3$ bond pairs and $1$ lone pair. The geometry is trigonal pyramidal.
$4$. $CO_{3}^{2-}$: The central $C$ atom has $3$ bond pairs and $0$ lone pairs. The geometry is trigonal planar (non-pyramidal).
Thus,the species $SO_{3}$,$NO_{3}^{-}$,and $CO_{3}^{2-}$ are non-pyramidal.
The total number of such species is $3$.

(A) To determine the geometry,we calculate the Steric Number $(SN)$ using the formula: $SN = \text{Number of } \sigma \text{ bonds} + \text{Number of lone pairs}$.
$1$. $PCl_{5}$: $SN = 5 + 0 = 5$. Geometry: Trigonal bipyramidal $(iv)$.
$2$. $SF_{6}$: $SN = 6 + 0 = 6$. Geometry: Octahedral $(iii)$.
$3$. $BrF_{5}$: $SN = 5 + 1 = 6$. Geometry: Square pyramidal $(i)$.
$4$. $BF_{3}$: $SN = 3 + 0 = 3$. Geometry: Trigonal planar $(ii)$.
Thus,the correct matching is $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$.

(A) The $I_{3}^{-}$ ion has a linear structure where the central iodine atom is bonded to two other iodine atoms.
The central iodine atom has $7$ valence electrons.
It forms $2$ single bonds with the other two iodine atoms,using $2$ electrons.
Including the negative charge,the total number of electrons around the central iodine atom is $7 + 1 = 8$.
After forming $2$ bonds,the remaining electrons are $8 - 2 = 6$,which form $3$ lone pairs.
Therefore,the number of lone pairs on the central $I$ atom is $3$.

(A) In gaseous triethylamine,the nitrogen atom is $sp^3$ hybridized with one lone pair. Due to the repulsion between the bulky ethyl groups,the $C-N-C$ bond angle is increased from the ideal tetrahedral angle of $109.5^\circ$ to approximately $108^\circ$ to $110^\circ$ depending on the specific experimental conditions,but $108^\circ$ is the standard accepted value in this context.

(C) The bond angle in hydrides of group $16$ elements depends on the electronegativity of the central atom.
As we move down the group,the electronegativity of the central atom decreases,which causes the bond pair electrons to move further away from the central atom.
This results in a decrease in bond pair-bond pair repulsion,leading to a smaller bond angle.
Therefore,$H_2O$ has the largest bond angle $(104.5^\circ)$ compared to $H_2S$ $(92.1^\circ)$,$H_2Se$ $(91^\circ)$,and $H_2Te$ $(90^\circ)$.

(B) To determine if species are isostructural,we check their hybridization and geometry:
$1$. $ICl_4^-$ ($sp^3d^2$,square planar) and $XeF_4$ ($sp^3d^2$,square planar) are isostructural.
$2$. $ClO_3^-$ ($sp^3$,pyramidal) and $CO_3^{2-}$ ($sp^2$,trigonal planar) are not isostructural.
$3$. $IBr_2^-$ ($sp^3d$,linear) and $XeF_2$ ($sp^3d$,linear) are isostructural.
$4$. $BrO_3^-$ ($sp^3$,pyramidal) and $XeO_3$ ($sp^3$,pyramidal) are isostructural.
Thus,the pair $ClO_3^-$ and $CO_3^{2-}$ is not isostructural.

(C) To determine the maximum 'lone pair - lone pair' (lp-lp) repulsions,we analyze the number of lone pairs on the central atom for each molecule:
$1$. $IF_{5}$: Iodine has $7$ valence electrons. It forms $5$ bonds with $F$ atoms,leaving $1$ lone pair. Number of lp-lp repulsions = $0$.
$2$. $SF_{4}$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $1$ lone pair. Number of lp-lp repulsions = $0$.
$3$. $XeF_{2}$: Xenon has $8$ valence electrons. It forms $2$ bonds with $F$ atoms,leaving $3$ lone pairs. These $3$ lone pairs occupy the equatorial positions of the trigonal bipyramidal geometry,resulting in $3$ lp-lp repulsions at $90^\circ$ angles.
$4$. $ClF_{3}$: Chlorine has $7$ valence electrons. It forms $3$ bonds with $F$ atoms,leaving $2$ lone pairs. Number of lp-lp repulsions = $1$.
Comparing these,$XeF_{2}$ has the maximum number of lone pair - lone pair repulsions.

(A) To determine the number of lone pairs on the central atom:
$1$. $SF_{4}$: Sulfur $(S)$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $1$ lone pair on $S$.
$2$. $XeF_{4}$: Xenon $(Xe)$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $2$ lone pairs on $Xe$.
$3$. $CF_{4}$: Carbon $(C)$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $0$ lone pairs on $C$.
$4$. $H_{2}O$: Oxygen $(O)$ has $6$ valence electrons. It forms $2$ bonds with $H$ atoms,leaving $2$ lone pairs on $O$.
Thus,$XeF_{4}$ and $H_{2}O$ have $2$ lone pairs on their central atoms.
The total number of such species is $2$.

(C) In $PCl_5$,the phosphorus atom undergoes $sp^3d$ hybridization resulting in a trigonal bipyramidal geometry.
There are two types of $P-Cl$ bonds: three equatorial bonds and two axial bonds.
The axial bonds are longer and weaker than the equatorial bonds due to greater repulsion from the equatorial bond pairs.
Therefore,the statement that axial bonds are stronger than equatorial bonds is incorrect.

(B) To determine the shape of the molecules,we use the $VSEPR$ theory:
$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $3 + 2 = 5$ ($sp^3d$ hybridization). Due to $2$ lone pairs in equatorial positions,the shape is $T$-shaped $(A-III)$.
$2$. $BF_3$: The central atom $B$ has $3$ valence electrons. It forms $3$ bonds with $F$ atoms and has $0$ lone pairs. The steric number is $3$ ($sp^2$ hybridization). The shape is trigonal planar $(B-IV)$.
$3$. $XeF_4$: 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). Due to $2$ lone pairs in axial positions,the shape is square planar $(C-I)$.
$4$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The steric number is $4 + 1 = 5$ ($sp^3d$ hybridization). Due to $1$ lone pair in an equatorial position,the shape is see-saw $(D-II)$.
Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

(A) The central atom $S$ in $SF_{4}$ has $5$ electron pairs ($4$ bond pairs and $1$ lone pair),resulting in $sp^{3}d$ hybridization.
According to $VSEPR$ theory,the lone pair occupies the equatorial position in a trigonal bipyramidal geometry to minimize repulsion.
In the equatorial position,the lone pair experiences repulsion from two axial bond pairs at $90^{\circ}$ and two equatorial bond pairs at $120^{\circ}$.
Therefore,there are two lone pair-bond pair repulsions at $90^{\circ}$.

(B) $1$. Calculate the total number of electrons in each species:
$CH_4$: $6 + 4(1) = 10$ electrons.
$NH_4^+$: $7 + 4(1) - 1 = 10$ electrons.
$BH_4^-$: $5 + 4(1) + 1 = 10$ electrons.
Since all species have $10$ electrons,they are isoelectronic.
$2$. Determine the geometry:
All three species have a central atom bonded to $4$ hydrogen atoms with no lone pairs on the central atom. According to $VSEPR$ theory,they all exhibit $sp^3$ hybridization and possess a tetrahedral geometry.

(C) For $BrF_{3}$ molecule:
Central atom is $Br$ (Bromine),which has $7$ valence electrons.
Number of bond pairs $= 3$ (three $Br-F$ bonds).
Number of lone pairs $= (7 - 3) / 2 = 2$.
Steric number $= 3$ (bond pairs) $+ 2$ (lone pairs) $= 5$,which corresponds to $sp^{3}d$ hybridization.
Due to the presence of $2$ lone pairs,the geometry is trigonal bipyramidal,but the shape is bent $T$-shape.

(B) To find the number of lone pairs on the central atom $(Xe)$ in each molecule:
$1$. In $XeO_3$: The $Xe$ atom has $8$ valence electrons. It forms $3$ double bonds with $3$ oxygen atoms (using $6$ electrons). Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.
$2$. In $XeOF_4$: The $Xe$ atom has $8$ valence electrons. It forms $1$ double bond with oxygen ($2$ electrons) and $4$ single bonds with fluorine atoms ($4$ electrons). Total electrons used = $6$. Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.
$3$. In $XeF_6$: The $Xe$ atom has $8$ valence electrons. It forms $6$ single bonds with fluorine atoms ($6$ electrons). Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.
Sum of lone pairs = $1 + 1 + 1 = 3$.

(C) The shapes of the given compounds are as follows:
$A$. $BrF_5$: It has $5$ bond pairs and $1$ lone pair,resulting in a square pyramidal geometry $(II)$.
$B$. $[CrF_6]^{3-}$: It has $6$ bond pairs and $0$ lone pairs,resulting in an octahedral geometry $(IV)$.
$C$. $O_3$: It has $2$ bond pairs and $1$ lone pair on the central atom,resulting in a bent shape $(I)$.
$D$. $PCl_5$: It has $5$ bond pairs and $0$ lone pairs,resulting in a trigonal bipyramidal geometry $(III)$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(C) To determine the number of non-planar species,we analyze the geometry of each:
$1$. $NO_{3}^{-}$: $sp^{2}$ hybridization,trigonal planar.
$2$. $H_{2}O_{2}$: Open book structure,non-planar.
$3$. $BF_{3}$: $sp^{2}$ hybridization,trigonal planar.
$4$. $PCl_{3}$: $sp^{3}$ hybridization,trigonal pyramidal,non-planar.
$5$. $XeF_{4}$: $sp^{3}d^{2}$ hybridization,square planar.
$6$. $SF_{4}$: $sp^{3}d$ hybridization,see-saw shape,non-planar.
$7$. $XeO_{3}$: $sp^{3}$ hybridization,trigonal pyramidal,non-planar.
$8$. $PH_{4}^{+}$: $sp^{3}$ hybridization,tetrahedral,non-planar.
$9$. $SO_{3}$: $sp^{2}$ hybridization,trigonal planar.
$10$. $[Al(OH)_{4}]^{-}$: $sp^{3}$ hybridization,tetrahedral,non-planar.
The non-planar species are: $H_{2}O_{2}, PCl_{3}, SF_{4}, XeO_{3}, PH_{4}^{+}, [Al(OH)_{4}]^{-}$.
Counting these,we get a total of $6$ non-planar species.

(B) 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).
$1$. For $SCl_2$: Sulfur $(S)$ has $6$ valence electrons. It forms $2$ bonds with $Cl$ atoms. Lone pairs = $\frac{1}{2} (6 - 2) = 2$.
$2$. For $O_3$: The central Oxygen atom has $6$ valence electrons. It forms $2$ bonds with one oxygen atom and $1$ coordinate bond with another. The central atom has $1$ lone pair.
$3$. For $ClF_3$: Chlorine $(Cl)$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (7 - 3) = 2$.
$4$. For $SF_6$: Sulfur $(S)$ has $6$ valence electrons. It forms $6$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (6 - 6) = 0$.
Thus,the number of lone pairs are $2, 1, 2$ and $0$ respectively. The correct option is $B$.

(C) $SnH_4$ (stannane) is a molecular hydride formed by the reaction of $Sn$ with hydrogen,which is true.
$SnH_4$ has a tetrahedral geometry with $sp^3$ hybridization,which means it is a non-planar molecule.
Therefore,Statement $I$ is true and Statement $II$ is false.

(D) The bond angles for the given species are as follows:
$1$. $PH_3$: The bond angle is approximately $93.5^\circ$ due to the absence of hybridization and the presence of a lone pair.
$2$. $NH_3$: The bond angle is $107^\circ$ due to $sp^3$ hybridization with one lone pair.
$3$. $NH_4^+$: The bond angle is $109.5^\circ$ due to $sp^3$ hybridization with no lone pair, resulting in a perfect tetrahedral geometry.
$4$. $BF_3$: This molecule does not contain $H - X - H$ bonds, so it cannot be compared in this series.
Comparing the bond angles of the hydrides: $PH_3 (93.5^\circ) < NH_3 (107^\circ) < NH_4^+ (109.5^\circ)$.
Therefore, the correct order is $PH_3 < NH_3 < NH_4^+$.

(D) $S$ has $6$ valence electrons.
It forms $4$ bond pairs with $Cl$ atoms and has $1$ lone pair remaining.
According to $VSEPR$ theory,a molecule with $4$ bond pairs and $1$ lone pair ($AX_4E$ type) adopts a see-saw geometry.

(B) The bond angle in a molecule is affected by the number of lone pairs and the electronegativity of the central atom.
In $CH_4$ (methane),the central carbon atom has $4$ bonding pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.
In $NH_3$ (ammonia),the central nitrogen atom has $3$ bonding pairs and $1$ lone pair. The lone pair-bond pair repulsion reduces the bond angle to $107.1^{\circ}$.
In $H_2O$ (water),the central oxygen atom has $2$ bonding pairs and $2$ lone pairs. The greater lone pair-lone pair repulsion further reduces the bond angle to $104.5^{\circ}$.
Therefore,the bond angles are $109.5^{\circ}, 107.1^{\circ}, 104.5^{\circ}$ respectively.

(C) The central atom $Xe$ has $8$ valence electrons.
In $XeF_2$,$Xe$ forms $2$ covalent bonds with fluorine atoms,leaving $8 - 2 = 6$ electrons,which corresponds to $3$ lone pairs.
In $XeF_4$,$Xe$ forms $4$ covalent bonds with fluorine atoms,leaving $8 - 4 = 4$ electrons,which corresponds to $2$ lone pairs.
Therefore,the number of lone pairs on $Xe$ in $XeF_2$ and $XeF_4$ are $3$ and $2$,respectively.

(C)
Central atom $Cl$ has $7$ valence electrons,out of which it forms $3$ bond pairs with fluorine and $2$ pairs of electrons remain non-bonded (lone pairs).
According to $VSEPR$ theory,the molecule has $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
To minimize repulsion,the two lone pairs occupy the equatorial positions,resulting in a $T$-shaped molecular geometry.

(A) The molecule $PCl_3F_2$ has a trigonal bipyramidal geometry.
According to Bent's rule,more electronegative atoms prefer to occupy the axial positions,while larger and less electronegative atoms prefer the equatorial positions to minimize repulsion.
In $PCl_3F_2$,the $F$ atoms are more electronegative than $Cl$ atoms.
Therefore,the $F$ atoms occupy the axial positions,and the $Cl$ atoms occupy the equatorial positions.
This corresponds to the structure shown in option $A$.

(C) The central chlorine atom $(Cl)$ has $7$ valence electrons.
It forms $3$ bond pairs with fluorine atoms and retains $2$ lone pairs of electrons.
According to $VSEPR$ theory,the molecule adopts a trigonal bipyramidal electron geometry with the two lone pairs occupying equatorial positions to minimize repulsion.
Consequently,the molecular geometry of $ClF_3$ is $T$-shaped.

(B) The bond angle depends on the hybridization and the number of lone pairs of electrons on the central atom.
$1$. $CH_4$: $sp^3$ hybridization,tetrahedral geometry,bond angle = $109.5^{\circ}$.
$2$. $CO_2$: $sp$ hybridization,linear geometry,bond angle = $180^{\circ}$.
$3$. $H_2O$: $sp^3$ hybridization with $2$ lone pairs,bent geometry,bond angle = $104.5^{\circ}$.
$4$. $SO_2$: $sp^2$ hybridization with $1$ lone pair,bent geometry,bond angle $\approx 119^{\circ}$.
Comparing these,$CO_2$ has the highest bond angle of $180^{\circ}$.

(D) $BrF_5$ has a square pyramidal geometry with one lone pair on the $Br$ atom. Due to the lone pair-bond pair repulsion,the equatorial $F$ atoms are pushed slightly away from the lone pair,making the $F_{axial}-Br-F_{equatorial}$ bond angles less than $90^{\circ}$ (approximately $84.8^{\circ}$),while the $F_{equatorial}-Br-F_{equatorial}$ angles in the square plane are $90^{\circ}$. Thus,the $F-Br-F$ bond angles are non-identical.
$PCl_5$ has a trigonal bipyramidal geometry. It has two types of $Cl-P-Cl$ bond angles: equatorial-equatorial $(120^{\circ})$ and axial-equatorial $(90^{\circ})$. Thus,the $Cl-P-Cl$ bond angles are non-identical.

(C) The bond angle in a molecule is influenced by the hybridization of the central atom and the presence of lone pairs.
$BCl_3$ has $sp^2$ hybridization with a trigonal planar geometry and a bond angle of $120^{\circ}$.
$SO_3$ also has $sp^2$ hybridization with a trigonal planar geometry and a bond angle of $120^{\circ}$.
$POCl_3$ has $sp^3$ hybridization with a tetrahedral geometry,where the bond angles are approximately $109.5^{\circ}$.
$ClF_3$ has $sp^3d$ hybridization with a $T$-shaped geometry due to two lone pairs on the central $Cl$ atom,resulting in bond angles slightly less than $90^{\circ}$ and $180^{\circ}$.
Comparing these,$BCl_3$ and $SO_3$ have the largest bond angles of $120^{\circ}$. However,in the context of standard chemistry problems of this type,$BCl_3$ is the classic example of a $120^{\circ}$ bond angle.

(A) The central atom in $SOF_4$ is sulphur $(S)$. The valence shell configuration of $S$ is $3s^2 3p^4$. It forms four $S-F$ bonds and one $S=O$ double bond. The total number of electron pairs around the central atom is $5$ (four single bonds and one double bond),which corresponds to $sp^3d$ hybridization. According to the $VSEPR$ theory,the geometry is trigonal bipyramidal. To minimize repulsion,the double-bonded oxygen atom occupies the equatorial position in the trigonal plane,while the fluorine atoms occupy the two axial and two equatorial positions.

(A) $IF_7$: Central atom $I$ has $7$ valence electrons,all used in bonding. Lone pairs = $0$ $(IV)$.
$ICl_4^-$: Central atom $I$ has $7$ valence electrons + $1$ (negative charge) = $8$. $4$ electrons used in bonding,leaving $4$ electrons ($2$ lone pairs) $(III)$.
$XeF_6$: Central atom $Xe$ has $8$ valence electrons. $6$ used in bonding,leaving $2$ electrons ($1$ lone pair) $(II)$.
$XeF_2$: Central atom $Xe$ has $8$ valence electrons. $2$ used in bonding,leaving $6$ electrons ($3$ lone pairs) $(I)$.
Correct matching: $A-IV, B-III, C-II, D-I$.

(D) $OF_2$ has $sp^3$ hybridization with two lone pairs on the oxygen atom.
Due to the presence of two lone pairs,the molecule adopts a bent $V$-shaped geometry.
The $FOF$ bond angle is $103^{\circ}$,which is less than $104.5^{\circ}$ (the bond angle of $H_2O$).
Since fluorine is more electronegative than oxygen,the oxidation state of $O$ in $OF_2$ is $+2$.
Therefore,statements $A$,$B$,and $D$ are correct.

(D) To determine the geometry of the molecules,we use the $VSEPR$ theory:
$1$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $6$ ($sp^3d^2$ hybridization),resulting in a square planar geometry $(A-II)$.
$2$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The steric number is $5$ ($sp^3d$ hybridization),resulting in a see-saw geometry $(B-I)$.
$3$. $NH_4^{+}$: The central atom $N$ has $5$ valence electrons. It forms $4$ bonds with $H$ atoms and has no lone pairs. The steric number is $4$ ($sp^3$ hybridization),resulting in a tetrahedral geometry $(C-IV)$.
$4$. $BrF_3$: The central atom $Br$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $5$ ($sp^3d$ hybridization),resulting in a bent $T$-shaped geometry $(D-III)$.
Thus,the correct matching is $A-II, B-I, C-IV, D-III$.

(C) To determine the shape of the given species,we use the $VSEPR$ theory:
$1$. $XeF_2$: $sp^3d$ hybridization,$3$ lone pairs on $Xe$,shape is linear.
$2$. $I_3^{+}$: $sp^3$ hybridization,$2$ lone pairs on central $I$,shape is bent ($V$-shape).
$3$. $C_3O_2$: $O=C=C=C=O$,central carbon is $sp$ hybridized,shape is linear.
$4$. $I_3^{-}$: $sp^3d$ hybridization,$3$ lone pairs on central $I$,shape is linear.
$5$. $CO_2$: $O=C=O$,$sp$ hybridization,shape is linear.
$6$. $SO_2$: $sp^2$ hybridization,$1$ lone pair on $S$,shape is bent ($V$-shape).
$7$. $BeCl_2$: $sp$ hybridization,no lone pairs on $Be$,shape is linear.
$8$. $BCl_2^{\Theta}$: $sp^2$ hybridization,$1$ lone pair on $B$,shape is bent ($V$-shape).
The linear species are: $XeF_2, C_3O_2, I_3^{-}, CO_2, BeCl_2$.
Total number of linear species = $5$.

(A) $1$. Determine the hybridization and geometry of each species:
- $CO_2$: $sp$ hybridization,linear,bond angle = $180^{\circ}$.
- $NH_4^{+}$: $sp^3$ hybridization,tetrahedral,bond angle = $109.5^{\circ}$.
- $NH_3$: $sp^3$ hybridization,trigonal pyramidal,one lone pair,bond angle = $107^{\circ}$.
- $H_2O$: $sp^3$ hybridization,bent,two lone pairs,bond angle = $104.5^{\circ}$.
$2$. Comparing the bond angles: $104.5^{\circ} (H_2O) < 107^{\circ} (NH_3) < 109.5^{\circ} (NH_4^{+}) < 180^{\circ} (CO_2)$.
$3$. The correct order is $H_2O < NH_3 < NH_4^{+} < CO_2$.

(B) $NH_3$: $sp^3$ hybridization with $1$ lone pair $(LP)$ results in a Trigonal pyramidal geometry $(a-iii)$.
$ClF_3$: $sp^3d$ hybridization with $2$ lone pairs $(LP)$ results in a $T$-shape geometry $(b-iv)$.
$PCl_5$: $sp^3d$ hybridization with $0$ lone pairs $(LP)$ results in a Trigonal bipyramidal geometry $(c-ii)$.
$BrF_5$: $sp^3d^2$ hybridization with $1$ lone pair $(LP)$ results in a Square pyramidal geometry $(d-i)$.
Therefore,the correct matching is $a-iii, b-iv, c-ii, d-i$.

(D) To determine the number of species with a square pyramidal structure,we analyze the hybridization and geometry of each:
$1$. $PF_{5}$: $sp^{3}d$ hybridization,trigonal bipyramidal geometry.
$2$. $BrF_{4}^{-}$: $sp^{3}d^{2}$ hybridization with $2$ lone pairs,square planar geometry.
$3$. $IF_{5}$: $sp^{3}d^{2}$ hybridization with $1$ lone pair,square pyramidal geometry.
$4$. $BrF_{5}$: $sp^{3}d^{2}$ hybridization with $1$ lone pair,square pyramidal geometry.
$5$. $XeOF_{4}$: $sp^{3}d^{2}$ hybridization with $1$ lone pair,square pyramidal geometry.
$6$. $ICl_{4}^{-}$: $sp^{3}d^{2}$ hybridization with $2$ lone pairs,square planar geometry.
The species with a square pyramidal structure are $IF_{5}$,$BrF_{5}$,and $XeOF_{4}$.
Therefore,the total number of such species is $3$.

(A) To determine the number of lone pairs on the central atom $Xe$,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - M - C + A)$,where $V$ is the number of valence electrons of $Xe$ $(8)$,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1$. $XeF_5^{+}$: $\frac{1}{2} (8 - 5 - 1 + 0) = 1$ lone pair.
$2$. $XeO_3$: $\frac{1}{2} (8 - 0 - 0 + 0) = 4$ electron pairs,$3$ are bonding,so $1$ lone pair.
$3$. $XeO_2F_2$: $\frac{1}{2} (8 - 2 - 0 + 0) = 3$ bonding pairs,$1$ lone pair.
$4$. $XeF_5^{-}$: $\frac{1}{2} (8 - 5 - 0 + 1) = 2$ lone pairs.
$5$. $XeO_3F_2$: $\frac{1}{2} (8 - 2 - 0 + 0) = 5$ bonding pairs,$0$ lone pairs.
$6$. $XeOF_4$: $\frac{1}{2} (8 - 4 - 0 + 0) = 5$ bonding pairs,$1$ lone pair.
$7$. $XeF_4$: $\frac{1}{2} (8 - 4 - 0 + 0) = 4$ bonding pairs,$2$ lone pairs.
The species with a single lone pair are $XeF_5^{+}$,$XeO_3$,$XeO_2F_2$,and $XeOF_4$. Thus,the total count is $4$.

(B) To determine the shape,we look at the hybridization and lone pairs on the central atom:
$N_3^{-}$: $sp$ hybridized,linear geometry.
$NO_2^{-}$: $sp^2$ hybridized with one lone pair,bent shape.
$I_3^{-}$: $sp^3d$ hybridized with three lone pairs on the central $I$ atom,linear geometry.
$O_3$: $sp^2$ hybridized with one lone pair,bent shape.
$SO_2$: $sp^2$ hybridized with one lone pair,bent shape.
Thus,the bent-shaped molecules are $NO_2^{-}$,$O_3$,and $SO_2$.
The total number of bent-shaped molecules is $3$.

(B) In $IF_5$,the central iodine atom has $7$ valence electrons. It forms $5$ single bonds with fluorine atoms,leaving $2$ electrons,which form $1$ lone pair.
In $IF_7$,the central iodine atom has $7$ valence electrons. It forms $7$ single bonds with fluorine atoms,leaving $0$ electrons,which means $0$ lone pairs.
The sum of lone pairs is $1 + 0 = 1$.

(A) To determine the shape of the given species,we use the $VSEPR$ theory:
$1$. $H_3O^+$: Oxygen has $6$ valence electrons. It forms $3$ bonds with $H$ and has $1$ lone pair. Total electron pairs = $4$ ($sp^3$ hybridization). Due to $1$ lone pair,the shape is Pyramidal $(III)$.
$2$. Acetylide anion $(C_2^{2-})$: The structure is $[C \equiv C]^{2-}$. The carbon atoms are $sp$ hybridized,resulting in a Linear $(II)$ shape.
$3$. $NH_4^+$: Nitrogen has $5$ valence electrons. It forms $4$ bonds with $H$ and has $0$ lone pairs. Total electron pairs = $4$ ($sp^3$ hybridization). The shape is Tetrahedral $(I)$.
$4$. $ClO_2^-$: Chlorine has $7$ valence electrons. It forms $2$ bonds with $O$ and has $2$ lone pairs. Total electron pairs = $4$ ($sp^3$ hybridization). Due to $2$ lone pairs,the shape is Bent $(IV)$.
Matching the results: $A-III, B-II, C-I, D-IV$.

(A) To find the number of lone pairs on the central atom,we determine the hybridization and geometry for each species:
$1$. For $ClO_3^-$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ double bonds with $O$ atoms and has $1$ negative charge. The number of lone pairs = $(7 + 1 - 6)/2 = 1$ lone pair.
$2$. For $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms. The number of lone pairs = $(8 - 4)/2 = 2$ lone pairs.
$3$. For $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ single bonds with $F$ atoms. The number of lone pairs = $(6 - 4)/2 = 1$ lone pair.
$4$. For $I_3^-$: The central atom $I$ has $7$ valence electrons. It forms $2$ single bonds with $I$ atoms and has $1$ negative charge. The number of lone pairs = $(7 + 1 - 2)/2 = 3$ lone pairs.
Comparing the number of lone pairs: $ClO_3^-$ $(1)$,$XeF_4$ $(2)$,$SF_4$ $(1)$,$I_3^-$ $(3)$.
The maximum number of lone pairs is $3$.

(B) Statement $I$ is correct: Both $SO_2$ and $H_2O$ have a bent or $V$-shaped molecular geometry.
Statement $II$ is incorrect: The bond angle in $SO_2$ is approximately $119.5^{\circ}$ ($sp^2$ hybridization),whereas the bond angle in $H_2O$ is approximately $104.5^{\circ}$ ($sp^3$ hybridization).
Therefore,the bond angle of $SO_2$ is greater than that of $H_2O$ due to the presence of double bonds and different hybridization states.
Thus,Statement $I$ is correct but Statement $II$ is incorrect.

(A) To determine the number of lone pairs on the central atom,we calculate the number of lone pairs using the formula: $\text{Lone pair} = \frac{V - B}{2}$,where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons (shared with surrounding atoms).
$1$. $O_3$: Central $O$ has $1$ lone pair.
$2$. $H_2O$: Central $O$ has $2$ lone pairs.
$3$. $SF_4$: Central $S$ has $1$ lone pair.
$4$. $ClF_3$: Central $Cl$ has $2$ lone pairs.
$5$. $NH_3$: Central $N$ has $1$ lone pair.
$6$. $BrF_5$: Central $Br$ has $1$ lone pair.
$7$. $XeF_4$: Central $Xe$ has $2$ lone pairs.
The compounds with exactly one lone pair on the central atom are $O_3$,$SF_4$,$NH_3$,and $BrF_5$.
Thus,the total count is $4$.

(D) The molecular shapes are determined by $VSEPR$ theory:
$1$. $BrF_5$: Central atom $Br$ has $5$ bond pairs and $1$ lone pair,resulting in a square pyramidal shape $(A-IV)$.
$2$. $H_2O$: Central atom $O$ has $2$ bond pairs and $2$ lone pairs,resulting in a bent shape $(B-III)$.
$3$. $ClF_3$: Central atom $Cl$ has $3$ bond pairs and $2$ lone pairs,resulting in a $T$-shape $(C-I)$.
$4$. $SF_4$: Central atom $S$ has $4$ bond pairs and $1$ lone pair,resulting in a see-saw shape $(D-II)$.
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(C) To determine the shape,we calculate the steric number using the formula: $Steric \ Number = \frac{1}{2} \times (V + M - C + A)$,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.
For $BrF_5$: $V = 7$ (for $Br$),$M = 5$ (for $F$). $Steric \ Number = \frac{1}{2} \times (7 + 5) = 6$.
$A$ steric number of $6$ corresponds to $sp^3d^2$ hybridization with an octahedral electron geometry.
Since there are $5$ bond pairs and $1$ lone pair,the molecular geometry is square pyramidal.

(A) The geometries of the given species are as follows:
$PF_5$: Trigonal bipyramidal ($sp^3d$ hybridization).
$BrF_5$: Square pyramidal ($sp^3d^2$ hybridization).
$PCl_5$: Trigonal bipyramidal ($sp^3d$ hybridization).
$[PtCl_4]^{2-}$: Square planar ($dsp^2$ hybridization).
$BF_3$: Trigonal planar ($sp^2$ hybridization).
$Fe(CO)_5$: Trigonal bipyramidal ($dsp^3$ hybridization).
Thus,the species with trigonal bipyramidal shape are $PF_5, PCl_5$,and $Fe(CO)_5$.
The total count is $3$.

(C) To determine the geometry,we analyze the structure of each species:
$1$. $SO_3^{2-}$ (Sulfite ion): The central $S$ atom has $3$ bond pairs and $1$ lone pair. According to $VSEPR$ theory,this results in a trigonal pyramidal geometry.
$2$. $SO_4^{2-}$ (Sulfate ion): The central $S$ atom has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry.
$3$. $S_2O_3^{2-}$ (Thiosulfate ion): The central $S$ atom is bonded to $3$ oxygen atoms and $1$ sulfur atom,with $0$ lone pairs,resulting in a tetrahedral geometry.
$4$. $S_2O_7^{2-}$ (Pyrosulfate ion): Each central $S$ atom is bonded to $4$ oxygen atoms with $0$ lone pairs,resulting in a tetrahedral geometry around each sulfur atom.
Thus,only $SO_3^{2-}$ has a pyramidal geometry. The total number of such species is $1$.

$(A)$ The molecular geometries of the given interhalogen compounds are determined by the $VSEPR$ theory:
$1. ICl$: It has $2$ bond pairs and $3$ lone pairs on the central atom $I$, resulting in a linear geometry $(i)$.
$2. ICl_3$: It has $3$ bond pairs and $2$ lone pairs on the central atom $I$, resulting in a $T$-shape geometry $(ii)$.
$3. ClF_5$: It has $5$ bond pairs and $1$ lone pair on the central atom $Cl$, resulting in a square pyramidal geometry $(iii)$.
$4. IF_7$: It has $7$ bond pairs and $0$ lone pairs on the central atom $I$, resulting in a pentagonal bipyramidal geometry $(iv)$.
Therefore, the correct matching is $A-i, B-ii, C-iii, D-iv$.