VSEPR Theory Questions in English

(D) To determine if all bonds are equal,we look at the molecular geometry and hybridization:
$1$. $SiF_4$: It has $sp^3$ hybridization and a tetrahedral geometry. All four $Si-F$ bonds are equivalent.
$2$. $BF_4^-$: It has $sp^3$ hybridization and a tetrahedral geometry. All four $B-F$ bonds are equivalent.
$3$. $XeF_4$: It has $sp^3d^2$ hybridization and a square planar geometry. All four $Xe-F$ bonds are equivalent due to symmetry.
$4$. $SF_4$: It has $sp^3d$ hybridization and a see-saw geometry (derived from trigonal bipyramidal). It contains one lone pair on the sulfur atom. Due to the presence of the lone pair,the axial and equatorial $S-F$ bonds have different lengths (axial bonds are longer than equatorial bonds).
Therefore,in $SF_4$,all bonds are not equal.

(C) $1$. $BF_3$ has a trigonal planar geometry with $sp^2$ hybridization. All atoms lie in the same plane.
$2$. $ClF_3$ has a $T$-shaped geometry due to two lone pairs on the central $Cl$ atom. The fluorine atoms are in the same plane.
$3$. $CHF_3$ has a tetrahedral geometry ($sp^3$ hybridization). The fluorine atoms are not in the same plane.
$4$. Since $CHF_3$ is tetrahedral,the option $All$ is incorrect. Among the given choices,$BF_3$ and $ClF_3$ have all fluorine atoms in the same plane,but $BF_3$ is the most standard example of a planar molecule.

(B) Compounds having the same shape and the same hybridization are known as isostructural.
$XeF_2$ has $sp^3d$ hybridization and a linear shape.
$IF_2^-$ has $sp^3d$ hybridization and a linear shape.
Therefore,both are isostructural.

(A) $BF_{3}$ has a trigonal planar structure with $sp^{2}$ hybridization,where all $B-F$ bonds are equal.
$AlF_{3}$ is ionic in nature,but in its molecular form,it exhibits a symmetric structure.
$NF_{3}$ has a trigonal pyramidal geometry due to $sp^{3}$ hybridization with one lone pair,where all $N-F$ bonds are equal.
$ClF_{3}$ has a $T$-shaped geometry due to $sp^{3}d$ hybridization with two lone pairs at equatorial positions,resulting in two different bond lengths (axial and equatorial).
Therefore,all bonds in $ClF_{3}$ are not equal.
Thus,option $A$ is correct.

(B) The bond angles for $NO_2^+$,$NO_2$,and $NO_2^-$ are $180^{\circ}$,$134^{\circ}$,and $115^{\circ}$ respectively.
As we move from $NO_2^+$ to $NO_2^-$,the number of lone pairs on the central nitrogen atom increases,which causes greater repulsion and leads to a decrease in the $O-N-O$ bond angle.
$NO_2^+$ is $sp$ hybridized (linear),$NO_2$ is $sp^2$ hybridized (bent),and $NO_2^-$ is $sp^2$ hybridized (bent).
Therefore,the correct statement is the decrease of the $O-N-O$ bond angle.

(D) To determine the geometry,we calculate the number of electron pairs around the central atom using the formula: $Total \ electron \ pairs = \frac{1}{2} (V + M - C + A)$,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.
$1$. For $XeOF_2$: The central atom $Xe$ has $8$ valence electrons. $M = 2$ ($F$ atoms),$O$ is divalent (does not contribute to $M$). Total electron pairs = $\frac{1}{2}(8 + 2) = 5$. This corresponds to $sp^3d$ hybridization with $3$ bond pairs and $2$ lone pairs,resulting in a $T$-shape geometry.
$2$. For $XeOF_4$: The central atom $Xe$ has $8$ valence electrons. $M = 4$ ($F$ atoms),$O$ is divalent. Total electron pairs = $\frac{1}{2}(8 + 4) = 6$. This corresponds to $sp^3d^2$ hybridization with $5$ bond pairs and $1$ lone pair,resulting in a square pyramidal geometry.
Therefore,the correct pair is $XeOF_2$ and $XeOF_4$.

(A) $NF_3$ has a trigonal pyramidal geometry,not trigonal planar.
In $NF_3$,the $N$ atom is $sp^3$ hybridised with one lone pair and three bond pairs of electrons.
The electron pair geometry is tetrahedral,while the molecular geometry is trigonal pyramidal.
$BF_3$ is trigonal planar,$AsF_5$ is trigonal bipyramidal,and $H_2O$ is bent.
Therefore,the incorrect geometry is represented by $NF_3$.

(B) To determine the shape of a molecule,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 $CO_3^{2-}$: $H = \frac{1}{2}(4 + 0 - 0 + 2) = 3$ ($sp^2$ hybridization,triangular planar).
$2$. For $NO_3^-$: $H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$ ($sp^2$ hybridization,triangular planar).
$3$. For $SO_3$: $H = \frac{1}{2}(6 + 0 - 0 + 0) = 3$ ($sp^2$ hybridization,triangular planar).
All species in option $B$ have $sp^2$ hybridization with no lone pairs on the central atom,resulting in a triangular planar geometry.

(D) To determine the shape,we calculate the hybridization and number of lone pairs for each molecule in option $D$:
$ClF_3$: Hybridization $= 3 + \frac{1}{2}[7 - 3] = 5$ $(sp^3d)$,with $2$ lone pairs,resulting in a $T$-shaped geometry.
$XeOF_2$: Hybridization $= 3 + \frac{1}{2}[8 - 4] = 5$ $(sp^3d)$,with $2$ lone pairs,resulting in a $T$-shaped geometry.
$XeF_3^+$: Hybridization $= 3 + \frac{1}{2}[8 - 3 - 1] = 5$ $(sp^3d)$,with $2$ lone pairs,resulting in a $T$-shaped geometry.
Since all three molecules have $sp^3d$ hybridization and $2$ lone pairs,they all possess an identical $T$-shaped geometry.

(B) The bond angle depends on the hybridization of the central atom and the repulsion between lone pairs and bond pairs.
In $CH_4$,the central atom $C$ is $sp^3$ hybridized with no lone pair,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.
In $NH_3$,$H_2O$,and $PH_3$,the central atoms have lone pairs which cause bond pair-lone pair repulsion,leading to a decrease in bond angle from the ideal tetrahedral angle.
Specifically,$PH_3$ has a bond angle of approximately $93.5^{\circ}$ due to the lack of hybridization and the large size of the $P$ atom.
Therefore,$CH_4$ has the greatest bond angle among the given options.

(C) In $XeOF_4$,the central atom $Xe$ has $8$ valence electrons.
It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom,utilizing $6$ electrons for bonding.
This leaves $2$ electrons,forming $1$ lone pair.
Total electron pairs = $5$ bond pairs + $1$ lone pair = $6$ electron pairs.
With $6$ electron pairs,the hybridization is $sp^3d^2$,which corresponds to an octahedral electron geometry.
Due to the presence of $1$ lone pair,the molecular geometry is square pyramidal.

(B) In the case of $(SiH_3)_3N$,the nitrogen atom is $sp^2$ hybridized because the lone pair of electrons on the nitrogen atom is donated into the empty $d$-orbital of the silicon atom,forming a $d\pi - p\pi$ back-bonding.
This results in a trigonal planar geometry for $(SiH_3)_3N$.
In contrast,$(CH_3)_3N$,$P(CH_3)_3$,and $P(SiH_3)_3$ all have a lone pair on the central atom and exhibit a pyramidal shape due to $sp^3$ hybridization.

(B) The central atom $Xe$ in $XeO_3F_2$ has $5$ valence electrons involved in bonding ($3$ double bonds with $O$ and $2$ single bonds with $F$).
This corresponds to a steric number of $5$,which implies $sp^3d$ hybridization.
The geometry is trigonal bipyramidal,where the three oxygen atoms occupy the equatorial positions and the two fluorine atoms occupy the axial positions.

(C) To determine the shape,we calculate the steric number of the central atom $Xe$ in $XeO_3F_2$.
$Xe$ has $8$ valence electrons.
It forms $3$ double bonds with $O$ atoms and $2$ single bonds with $F$ atoms.
Total electron pairs around $Xe = 3 + 2 = 5$.
$A$ steric number of $5$ corresponds to $sp^3d$ hybridization,which results in a trigonal bipyramidal geometry.
Thus,$XeO_3F_2$ has a trigonal bipyramidal shape.

(A) The central iodine atom in $IF_6^-$ has $7$ valence electrons.
It forms $6$ bonds with fluorine atoms and has $1$ lone pair of electrons,resulting in a total of $7$ electron pairs ($sp^3d^3$ hybridization).
According to $VSEPR$ theory,the presence of this lone pair causes a distortion in the octahedral geometry,leading to a distorted octahedral shape.

(D) Species are isostructural if they have the same hybridization and geometry.
In the set $BF_4^-, CCl_4, NH_4^+, PCl_4^+$,all central atoms $(B, C, N, P)$ are $sp^3$ hybridized and have a tetrahedral geometry.
Thus,all these species are isostructural.

(A) trigonal pyramidal molecular shape occurs when the central atom has three bond pairs and one lone pair of electrons.
$1$. $NI_3$: Nitrogen has $5$ valence electrons. It forms $3$ bonds with $I$ atoms and has $1$ lone pair. Shape: Trigonal pyramidal.
$2$. $I_3^-$: The central $I$ atom has $3$ lone pairs and $2$ bond pairs. Shape: Linear.
$3$. $SO_3^{2-}$: Sulfur has $6$ valence electrons. It forms $3$ bonds with $O$ atoms (one double,two single) and has $1$ lone pair. Shape: Trigonal pyramidal.
$4$. $NO_3^-$: Nitrogen has $5$ valence electrons. It forms $3$ bonds with $O$ atoms and has no lone pairs. Shape: Trigonal planar.
Thus,$NI_3$ $(I)$ and $SO_3^{2-}$ $(III)$ have a trigonal pyramidal shape.

(B) For $ICl_5$: The central iodine atom has $7$ valence electrons. It forms $5$ bonds with $Cl$ atoms and has $1$ lone pair. The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization. Due to $5$ bond pairs and $1$ lone pair,the geometry is octahedral,but the shape is square pyramidal.
For $ICl_4^-$: The central iodine atom has $7$ valence electrons,plus $1$ from the negative charge,totaling $8$ electrons. It forms $4$ bonds with $Cl$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridization. Due to $4$ bond pairs and $2$ lone pairs,the geometry is octahedral,but the shape is square planar.

(D) For $CH_4$ (tetrahedral),there are $6$ bond angles of $109^\circ 28'$.
For $SF_6$ (octahedral),there are $12$ bond angles of $90^\circ$.
For $IF_7$ (pentagonal bipyramidal),there are $10$ bond angles of $90^\circ$ (between axial and equatorial bonds).
For $IF_5$ (square pyramidal),the structure has $1$ lone pair. The bond angles between the equatorial $F$ atoms and the axial $F$ atom are $90^\circ$. There are $4$ such angles. The angles between equatorial $F$ atoms are also $90^\circ$. There are $4$ such angles. Total $90^\circ$ angles = $8$. However,due to the lone pair,the bond angles are distorted from $90^\circ$. Thus,$IF_5$ is not correctly matched.

(A) To determine the bond angle,we analyze the hybridization and geometry of each molecule:
$1$. $BeCl_2$: It has $sp$ hybridization and a linear geometry with a bond angle of $180^{\circ}$.
$2$. $NO_2$: It has $sp^2$ hybridization with one unpaired electron on the central nitrogen atom. The bond angle is approximately $134^{\circ}$.
$3$. $SO_2$: It has $sp^2$ hybridization with one lone pair on the central sulfur atom. The lone pair-bond pair repulsion is greater than the single electron-bond pair repulsion,leading to a smaller bond angle of approximately $119^{\circ}$.
Therefore,the decreasing order of bond angle is $BeCl_2 (180^{\circ}) > NO_2 (134^{\circ}) > SO_2 (119^{\circ})$.

(B) In $PF_5$,the molecule undergoes rapid Berry pseudorotation at room temperature,which makes all five $P-F$ bonds appear equivalent on the $NMR$ timescale.
In $PCl_5$,the larger size of the $Cl$ atoms and the steric hindrance prevent this rapid pseudorotation,resulting in distinct axial and equatorial bond lengths ($3$ equatorial bonds are shorter than $2$ axial bonds).

(A) $1$. $XeF_4$: It has $4$ bond pairs and $2$ lone pairs,resulting in square planar geometry with bond angles of $90^o$. Thus,$I-C$.
$2$. $I_3^-$: It has $2$ bond pairs and $3$ lone pairs on the central atom,resulting in a linear geometry. Thus,$II-D$.
$3$. $XeO_2F_2$: It has $4$ bond pairs and $1$ lone pair,resulting in a see-saw shape. Thus,$III-A$.
$4$. $SO_4^{2-}$: It has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry. Thus,$IV-B$.
Therefore,the correct matching is $I-C, II-D, III-A, IV-B$.

(D) In $POCl_3$,the central $P$ atom is bonded to one $O$ atom via a double bond and three $Cl$ atoms via single bonds.
$1$. The Lewis structure of $POCl_3$ follows the octet rule for all atoms (including $P$ which expands its octet to $10$ electrons),so it does not violate the octet rule in the context of hypervalency.
$2$. The geometry is irregular tetrahedral due to the presence of different types of bonds ($P=O$ and $P-Cl$).
$3$. According to $VSEPR$ theory,the repulsion between a double bond $(P=O)$ and a single bond $(P-Cl)$ is greater than the repulsion between two single bonds ($P-Cl$ and $P-Cl$).
$4$. Due to the greater repulsion from the $P=O$ bond,the $Cl-P=O$ bond angle ($109.5^{\circ}$ approximately) is actually greater than the $Cl-P-Cl$ bond angle $(103.5^{\circ})$.
Therefore,the statement in option $D$ is incorrect.

(D) 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,resulting in a linear geometry with a bond angle of $180^{\circ}$.
$2$. $NO_3^-$: The nitrogen atom is $sp^2$ hybridized with $0$ lone pairs,resulting in a trigonal planar geometry with a bond angle of $120^{\circ}$.
$3$. $NO_2^-$: The nitrogen atom is $sp^2$ hybridized with $1$ lone pair,resulting in a bent geometry. Due to the lone pair-bond pair repulsion,the bond angle is approximately $115^{\circ}$.
Comparing these,the order of increasing bond angle is $NO_2^- (115^{\circ}) < NO_3^- (120^{\circ}) < NO_2^+ (180^{\circ})$.
Thus,the correct sequence is $NO_2^-, NO_3^-, NO_2^+$.

(A) In $CF_4$,$CCl_4$,and $CBr_4$,all are tetrahedral with bond angles of $109.5^\circ$. Thus,the order $\alpha > \beta > \gamma$ is incorrect as they are equal.
In $NCl_3$,$PCl_3$,and $AsCl_3$,the bond angle decreases as the electronegativity of the central atom decreases $(N > P > As)$,so $NCl_3 > PCl_3 > AsCl_3$ is correct.
In $SOF_2$,the $S=O$ double bond exerts more repulsion than the $S-F$ single bonds,making $\angle FSO > \angle FSF$ correct.
In $NO_2^+$,$NO_2$,and $NO_2^-$,the bond angles are $180^\circ$,$134^\circ$,and $115^\circ$ respectively,so $NO_2^+ > NO_2 > NO_2^-$ is correct.
Therefore,option $A$ is the incorrect order.

(A) To determine the bond angle,we consider the steric repulsion between the groups attached to the oxygen atom.
$1$. In $CH_3-O-CH_3$ (dimethyl ether),there are two bulky methyl groups attached to the oxygen. The steric repulsion between these two methyl groups increases the bond angle to approximately $111.7^\circ$.
$2$. In $Ph-OH$ (phenol),the oxygen is attached to a phenyl group and a hydrogen atom. The resonance effect of the phenyl ring and the steric bulk of the phenyl group result in a bond angle of approximately $109^\circ$.
$3$. In $CH_3-OH$ (methanol),the oxygen is attached to a methyl group and a hydrogen atom. The steric repulsion is less than in the other two cases,resulting in a bond angle of approximately $108.9^\circ$.
Therefore,the correct order of bond angle is $CH_3-O-CH_3 > Ph-OH > CH_3-OH$.

(C) The $VSEPR$ theory states that the electron pairs (both bonding and lone pairs) surrounding the central atom arrange themselves in space to minimize the electrostatic repulsions between them.
This arrangement determines the geometry and shape of the molecule.
For example,in $PF_5$,the five electron pairs arrange themselves in a trigonal bipyramidal geometry to minimize repulsion,with bond angles of $120^{\circ}$ and $90^{\circ}$.

(A) 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.

Species	Hybridization	Geometry
$NO_3^{-}$	$sp^2$	Trigonal planar (Planar)
$AsO_3^{3-}$	$sp^3$	Pyramidal (Non-planar)
$CO_3^{2-}$	$sp^2$	Trigonal planar (Planar)
$ClO_3^{-}$	$sp^3$	Pyramidal (Non-planar)
$SO_3^{2-}$	$sp^3$	Pyramidal (Non-planar)

Thus,the non-planar species are $AsO_3^{3-} (II)$,$ClO_3^{-} (IV)$,and $SO_3^{2-} (V)$.

(B) In $PX_3$ molecules,the central atom $P$ is $sp^3$ hybridized with one lone pair.
As the electronegativity of the surrounding halogen atom decreases $(F > Cl > Br > I)$,the bond pair electrons move further away from the central atom.
According to the $VSEPR$ theory,as the size of the surrounding atom increases,the bond pair-bond pair repulsion decreases,but the bond angle increases due to the decrease in electronegativity of the surrounding atoms.
Therefore,the correct order of bond angle is $PI_3 > PBr_3 > PCl_3 > PF_3$.

(C) To determine if a molecule is planar and non-polar,we analyze its geometry and dipole moment:
$1$. $XeO_4$: It has a tetrahedral geometry ($sp^3$ hybridization). It is non-polar but non-planar.
$2$. $SF_4$: It has a see-saw geometry ($sp^3d$ hybridization) due to one lone pair. It is polar and non-planar.
$3$. $XeF_4$: It has a square planar geometry ($sp^3d^2$ hybridization) with two lone pairs occupying axial positions. The dipole moments cancel out,making it non-polar and planar.
$4$. $CF_4$: It has a tetrahedral geometry ($sp^3$ hybridization). It is non-polar but non-planar.
Therefore,$XeF_4$ is both planar and non-polar.

(B) Isostructural species are those that have the same hybridization and the same geometry.
$1$. $[NF_3]$ is pyramidal ($sp^3$ hybridization with one lone pair) and $[BF_3]$ is trigonal planar ($sp^2$ hybridization).
$2$. $[BF_4^-]$ has $sp^3$ hybridization and tetrahedral geometry. $[NH_4^+]$ also has $sp^3$ hybridization and tetrahedral geometry. Thus,they are isostructural.
$3$. $[BCl_3]$ is trigonal planar $(sp^2)$,while $[BrCl_3]$ is $T$-shaped ($sp^3d$ with two lone pairs).
$4$. $[NH_3]$ is pyramidal $(sp^3)$,while $[NO_3^-]$ is trigonal planar $(sp^2)$.
Therefore,the correct pair is $[BF_4^- \text{ and } NH_4^+]$.

(A) To determine the shape of the species,we use the $VSEPR$ theory based on the number of bonding pairs and lone pairs of electrons.
$1$. $PCl_3$ has $3$ bond pairs and $1$ lone pair on $P$,resulting in a trigonal pyramidal shape. $NH_3$ also has $3$ bond pairs and $1$ lone pair on $N$,resulting in a trigonal pyramidal shape. Thus,they have similar shapes.
$2$. $CF_4$ is tetrahedral ($4$ bond pairs,$0$ lone pairs),while $SF_4$ is see-saw shaped ($4$ bond pairs,$1$ lone pair).
$3$. $PbCl_2$ is bent (angular) due to the presence of a lone pair on $Pb$,while $CO_2$ is linear.
$4$. $PF_5$ is trigonal bipyramidal,while $IF_5$ is square pyramidal.
Therefore,the correct pair is $PCl_3$ and $NH_3$.

(C) The central atom $Xe$ has $8$ valence electrons.
For $XeF^{+}_3$,the number of valence electrons is $8 + 3(7) - 1 = 28$.
The number of electron pairs around $Xe$ is calculated as:
$Steric \ number = \frac{1}{2} (V + M - C + A) = \frac{1}{2} (8 + 3 - 1) = 5$.
With $5$ electron pairs,the hybridization is $sp^3d$.
There are $3$ bond pairs and $2$ lone pairs on the $Xe$ atom.
According to $VSEPR$ theory,the $2$ lone pairs occupy the equatorial positions to minimize repulsion.
This results in a bent $T$-shape for the $XeF^{+}_3$ ion.

(C) In the $XeF_n$ molecule,the possible values for $n$ are $2, 4, 6, 8$.
$XeF_2$ has a linear shape.
$XeF_4$ has a square planar shape.
$XeF_6$ has a distorted octahedral (capped octahedral) shape.
Therefore,a trigonal planar shape is not possible for any $XeF_n$ molecule.

(D) $BeCl_2$ has a linear structure due to $sp$ hybridization.
$ICl_2^-$ has $sp^3d$ hybridization with $3$ lone pairs on the central atom,resulting in a linear geometry.
$C_2H_2$ $(HC \equiv CH)$ has $sp$ hybridization and is linear.
$XeF_2$ has $sp^3d$ hybridization with $3$ lone pairs on the central atom,resulting in a linear geometry.
$GeCl_2$ has $sp^2$ hybridization with $1$ lone pair on the central atom,resulting in a bent (angular) geometry.
Therefore,$BeCl_2$ is not isostructural with $GeCl_2$.

(B) For $[XeF_3^+]$: The central atom $Xe$ has $8$ valence electrons. It forms $3$ bonds with $F$ atoms and has $1$ positive charge,leaving $8 - 3 - 1 = 4$ electrons,which form $2$ lone pairs. The total number of electron pairs is $3 + 2 = 5$,corresponding to $sp^3d$ hybridization. With $3$ bond pairs and $2$ lone pairs,the geometry is trigonal bipyramidal,and the shape is bent $T$-shape.
For $[XeF_5^+]$: The central atom $Xe$ has $8$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ positive charge,leaving $8 - 5 - 1 = 2$ electrons,which form $1$ lone pair. The total number of electron pairs is $5 + 1 = 6$,corresponding to $sp^3d^2$ hybridization. With $5$ bond pairs and $1$ lone pair,the geometry is octahedral,and the shape is square pyramidal.

(D) To determine the number of atoms in the same plane,we analyze the geometry of each molecule:
$1$. $XeF_2O_2$: It has a trigonal bipyramidal geometry with two lone pairs (or distorted structure). The maximum number of atoms in a single plane is $3$.
$2$. $PCl_5$: It has a trigonal bipyramidal geometry. The equatorial plane contains the central $P$ atom and $3$ $Cl$ atoms,totaling $4$ atoms in the same plane.
$3$. $AsH_4^+$: It has a tetrahedral geometry. Any $3$ atoms define a plane,but only $3$ atoms can lie in the same plane simultaneously.
$4$. $XeF_4$: It has a square planar geometry. The central $Xe$ atom and the $4$ $F$ atoms all lie in the same plane,totaling $5$ atoms.
Therefore,$XeF_4$ has the maximum number of atoms $(5)$ in the same plane.

(D) 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 (assuming single bonds).
$1$. For $SF_4$: Sulfur $(S)$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{1}{2} (6 - 4) = 1$.
$2$. For $CF_4$: Carbon $(C)$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{1}{2} (4 - 4) = 0$.
$3$. For $XeF_4$: Xenon $(Xe)$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms. $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 2$.
Thus,the number of lone pairs are $1$,$0$,and $2$ respectively. The correct option is $D$.

(A) The central iodine atom in $I_3^-$ undergoes $sp^3d$ hybridization.
It has $3$ lone pairs of electrons and $2$ bond pairs of electrons around the central iodine atom.
According to $VSEPR$ theory,the presence of $3$ lone pairs in the equatorial positions results in a trigonal bipyramidal electron geometry and a linear molecular shape.

(C) In $PCl_5$,the phosphorus atom undergoes $sp^3d$ hybridization,resulting in a trigonal bipyramidal shape.
Due to the presence of three equatorial and two axial bonds,the bond lengths are not all equal.
The three equatorial $P-Cl$ bonds are equal to each other,and the two axial $P-Cl$ bonds are equal to each other,but the axial bonds are longer than the equatorial bonds due to greater repulsion.
Therefore,the statement that it has a regular geometry is incorrect,as a regular geometry implies all bond lengths and angles are identical.

(D) The molecular shape of $SF_4$ is see-saw with $1$ lone pair on $S$ atom.
$CF_4$ has a tetrahedral geometry with $0$ lone pairs on $C$ atom.
$XeF_4$ has a square planar structure with $2$ lone pairs on $Xe$ atom.
Since the shapes and the number of lone pairs are different for all three,the correct option is $D$.

(C) To determine the number of lone pairs $(l.p.)$ on the central atom $Xe$,we use the formula: $l.p. = \frac{1}{2} (V - M - C + A)$,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 $XeOF_4$: $Xe$ has $8$ valence electrons. $O$ is divalent (contributes $0$ to $M$),$F$ is monovalent ($4$ atoms). $l.p. = \frac{1}{2} (8 - 4) = 2$ electrons,which is $1$ lone pair.
$2$. For $XeO_2F_2$: $Xe$ has $8$ valence electrons. $O$ is divalent ($0$ to $M$),$F$ is monovalent ($2$ atoms). $l.p. = \frac{1}{2} (8 - 2) = 3$ lone pairs (Note: In the provided structure,it shows $1$ lone pair due to double bonds with $O$; calculating based on $VSEPR$ theory for $XeO_2F_2$,$Xe$ has $8$ valence $e^-$,$2$ $O$ atoms use $4$ $e^-$,$2$ $F$ atoms use $2$ $e^-$,remaining $2$ $e^-$ form $1$ lone pair).
$3$. For $XeF_3^-$: $Xe$ has $8$ valence electrons,$3$ $F$ atoms,and $1$ negative charge. $l.p. = \frac{1}{2} (8 - 3 + 1) = 3$ lone pairs.
$4$. For $XeO_3$: $Xe$ has $8$ valence electrons,$3$ $O$ atoms (divalent). $l.p. = \frac{1}{2} (8 - 0) = 4$ electrons,which is $1$ lone pair.
Comparing the lone pairs: $XeF_3^-$ has $3$ lone pairs,which is the maximum.

(C) To determine the shapes,we calculate the hybridization and number of lone pairs for each species using the $VSEPR$ theory:
$1$. $NO^{+}_2$: $N$ is central,$sp$ hybridized,linear shape. $NO^{-}_2$: $N$ is central,$sp^2$ hybridized,bent shape.
$2$. $PCl_5$: $sp^3d$ hybridized,trigonal bipyramidal. $BrF_5$: $sp^3d^2$ hybridized,square pyramidal.
$3$. $XeF_4$: $Xe$ has $8$ valence electrons,$4$ bond pairs,$2$ lone pairs,$sp^3d^2$ hybridization,square planar. $ICl^{-}_4$: $I$ has $7+1=8$ valence electrons,$4$ bond pairs,$2$ lone pairs,$sp^3d^2$ hybridization,square planar.
$4$. $TeCl_4$: $sp^3d$ hybridized,see-saw shape. $XeO_4$: $sp^3$ hybridized,tetrahedral shape.
Thus,$XeF_4$ and $ICl^{-}_4$ both have a square planar geometry.

(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 $6$ ($sp^3d^2$ hybridization),resulting in a square planar geometry.
For $XeF_5^-$: The central atom $Xe$ has $8$ valence electrons,plus $1$ from the negative charge,forming $5$ bonds with $F$ atoms and having $2$ lone pairs. The steric number is $7$ ($sp^3d^3$ hybridization),resulting in a pentagonal planar geometry.
For $SnCl_2$: The central atom $Sn$ has $4$ valence electrons. It forms $2$ bonds with $Cl$ atoms and has $1$ lone pair. The steric number is $3$ ($sp^2$ hybridization),resulting in an angular or bent shape.

(A) The shape of $XeO_3$ is trigonal pyramidal due to the presence of one lone pair on the $Xe$ atom.
Therefore,the description 'Trigonal bipyramidal' for $XeO_3$ is incorrect.
Thus,option $A$ is not correctly matched.

(B) To determine the non-planar species,we analyze the hybridization and geometry of each ion based on $VSEPR$ theory:

Species	Hybridization and Geometry
$NO_3^-$	$sp^2$,Trigonal planar
$AsO_3^{3-}$	$sp^3$,Pyramidal (Non-planar)
$CO_3^{2-}$	$sp^2$,Trigonal planar
$ClO_3^-$	$sp^3$,Pyramidal (Non-planar)
$SO_3^{2-}$	$sp^3$,Pyramidal (Non-planar)
$BO_3^{3-}$	$sp^2$,Trigonal planar

Species with $sp^3$ hybridization and one lone pair on the central atom,such as $AsO_3^{3-}$,$ClO_3^-$,and $SO_3^{2-}$,exhibit a pyramidal geometry,which is non-planar. Therefore,the correct option is $B$.

(B) The central atom in $\ddot{N}H_3$ is nitrogen,which is $sp^3$ hybridized.
According to $VSEPR$ theory,the $sp^3$ hybridization leads to a tetrahedral electron geometry.
In the ammonia molecule,the nitrogen atom is located at the centre of the tetrahedron,while the three hydrogen atoms occupy three of the four vertices,with the fourth vertex occupied by a lone pair of electrons.

(A) In a trigonal bipyramidal electron geometry,the central atom has $5$ electron pairs.
According to $VSEPR$ theory,lone pairs occupy equatorial positions to minimize repulsion.
Possible molecular geometries derived from this include:
$1$. $4$ bonding pairs + $1$ lone pair = See-saw
$2$. $3$ bonding pairs + $2$ lone pairs = $T$-shaped
$3$. $2$ bonding pairs + $3$ lone pairs = Linear
Trigonal planar geometry arises from a trigonal planar electron geometry ($3$ electron pairs).
Therefore,trigonal planar is the least likely geometry to result from a trigonal bipyramidal electron geometry.

(B) $(I)$ $T$: According to $VSEPR$ theory,the repulsion order is $lp-lp > lp-bp > bp-bp$.
$(II)$ $F$: As the number of lone pairs increases,the bond angle decreases due to increased repulsion.
$(III)$ $T$: $H_2O$ has $2$ lone pairs on $O$,and $NH_3$ has $1$ lone pair on $N$.
$(IV)$ $F$: The structures of xenon fluorides and oxyfluorides are well-explained by $VSEPR$ theory.
Therefore,the correct order is $T, F, T, F$.

(A) To determine the geometry of the given species,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.
$A) \ CO_3^{2-}$: $H = \frac{1}{2} (4 + 0 - 0 + 2) = 3$. The hybridization is $sp^2$,which corresponds to a trigonal planar geometry.
$B) \ SO_3^{2-}$: $H = \frac{1}{2} (6 + 0 - 0 + 2) = 4$. The hybridization is $sp^3$ with one lone pair,resulting in a trigonal pyramidal geometry.
$C) \ ClO_3^{-}$: $H = \frac{1}{2} (7 + 0 - 0 + 1) = 4$. The hybridization is $sp^3$ with one lone pair,resulting in a trigonal pyramidal geometry.
$D) \ BF_4^{-}$: $H = \frac{1}{2} (3 + 4 - 0 + 1) = 4$. The hybridization is $sp^3$ with no lone pairs,resulting in a tetrahedral geometry.
Thus,only $CO_3^{2-}$ is planar.