VSEPR Theory Questions in English

(B) $1$. For $XeOF_4$: Central atom $Xe$ has $8$ valence electrons. It forms $4$ $\sigma$-bonds with $F$ atoms and $1$ $\sigma$-bond with $O$ atom (total $5$ $\sigma$-bonds). It has $1$ lone pair. Shape is square pyramidal. Total (lone pair + $\sigma$-bond) = $1 + 5 = 6$.
$2$. For $XeF_5^{\oplus}$: Central atom $Xe$ has $7$ valence electrons (after losing $1$). It forms $5$ $\sigma$-bonds with $F$ atoms. It has $1$ lone pair. Shape is square pyramidal. Total (lone pair + $\sigma$-bond) = $1 + 5 = 6$.
$3$. Since both have the same shape (square pyramidal) and the same total number of lone pairs and $\sigma$-bonds $(6)$,the correct pair is $XeOF_4$ and $XeF_5^{\oplus}$.

(A) $1$. For $SOF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ and $1$ double bond with $O$. The steric number is $5$ ($4$ sigma bonds + $1$ pi bond is not counted in $VSEPR$,but the double bond occupies one equatorial position). The geometry is trigonal bipyramidal with no lone pairs,so the shape is trigonal bipyramidal.
$2$. For $XeO_2F_2$: The central atom $Xe$ has $8$ valence electrons. It forms $2$ single bonds with $F$ and $2$ double bonds with $O$. The steric number is $4$ (bonds) + $1$ lone pair = $5$. The lone pair occupies an equatorial position,resulting in a see-saw shape.
$3$. For $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ and has $2$ lone pairs. The steric number is $3 + 2 = 5$. The lone pairs occupy equatorial positions,resulting in a $T$-shape.
Thus,the shapes are trigonal bipyramidal,see-saw,and $T$-shape.

(A) In the $XeF_4$ molecule,the central atom $Xe$ has $6$ electron pairs in its valence shell,consisting of $4$ bonding pairs and $2$ lone pairs.
According to $VSEPR$ theory,the presence of $6$ electron pairs suggests an octahedral geometry.
However,due to the presence of $2$ lone pairs,the molecule adopts a square planar geometry to minimize electron-pair repulsions.
$XeO_4$ is tetrahedral,$XeO_2F_2$ is see-saw shaped,and $XeOF_4$ is square pyramidal.

(A) Let us analyze each option:
$A$: In $NCl_3$ and $NBr_3$,the central atom is the same $(N)$. As the electronegativity of the surrounding atom decreases $(Cl > Br)$,the bond pair-bond pair repulsion decreases,leading to a smaller bond angle. Thus,$NCl_3 > NBr_3$ is correct.
$B$: In hydrides of group $16$ $(H_2O, H_2S, H_2Se, H_2Te)$,as the size of the central atom increases,the bond angle decreases. Thus,$H_2S > H_2Te$ is the correct order.
$C$: Both $N_3^-$ and $I_3^-$ are linear molecules with a bond angle of $180^{\circ}$. Thus,$N_3^- = I_3^-$.
$D$: In $NO_3^-$ ($sp^2$ hybridized,$120^{\circ}$) and $NO_2^-$ ($sp^2$ hybridized with one lone pair,$< 120^{\circ}$),the bond angle in $NO_3^-$ is greater than in $NO_2^-$. Thus,$NO_3^- > NO_2^-$.
Therefore,the correct statement is $A$.

(A) To determine the shape,we calculate the hybridization and number of lone pairs 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 $SO_3^{2-}$: $H = \frac{1}{2}(6 + 0 - 0 + 2) = 4$ ($sp^3$ hybridization). It has $3$ bond pairs and $1$ lone pair,resulting in a pyramidal shape.
$2$. For $SO_3$: $H = \frac{1}{2}(6 + 0 - 0 + 0) = 3$ ($sp^2$ hybridization). It has $3$ bond pairs and $0$ lone pairs,resulting in a trigonal planar shape.
$3$. For $CO_3^{2-}$: $H = \frac{1}{2}(4 + 0 - 0 + 2) = 3$ ($sp^2$ hybridization). It has $3$ bond pairs and $0$ lone pairs,resulting in a trigonal planar shape.
$4$. For $ClF_3$: $H = \frac{1}{2}(7 + 3 - 0 + 0) = 5$ ($sp^3d$ hybridization). It has $3$ bond pairs and $2$ lone pairs,resulting in a $T$-shaped geometry.
Therefore,the species with a pyramidal shape is $SO_3^{2-}$.

(A) $1$. In $[ONCl_2]^+$,the central atom is $N$. The hybridization of $N$ is $sp^2$ with one double bond to $O$ and two single bonds to $Cl$. The geometry is trigonal planar,and the bond angle is approximately $120^{\circ}$.
$2$. In $(OSCl_2)$,the central atom is $S$. The hybridization of $S$ is $sp^3$ with one lone pair,one double bond to $O$,and two single bonds to $Cl$. The geometry is trigonal pyramidal. Due to the presence of a lone pair on $S$,the $Cl-S-Cl$ bond angle is compressed to approximately $96^{\circ}-106^{\circ}$.
$3$. Comparing the two,$120^{\circ} > 96^{\circ}-106^{\circ}$. Therefore,the bond angle in $[ONCl_2]^+$ is greater than in $(OSCl_2)$.

(D) The structure of $O_3$ (ozone) is bent due to the presence of a lone pair on the central oxygen atom,which causes repulsion and results in a bond angle of approximately $117^{\circ}$.
The structure of $N_3^-$ (azide ion) is linear,as the central nitrogen atom is $sp$ hybridized with no lone pairs,resulting in a bond angle of $180^{\circ}$.
Therefore,$O_3$ and $N_3^-$ have bent and linear structures,respectively.

(C) $1$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The geometry is see-saw.
$2$. $CF_4$: The central atom $C$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms and has $0$ lone pairs. The geometry is tetrahedral.
$3$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The geometry is square planar.
Therefore,the shapes are different,and the number of lone pairs are $1, 0$,and $2$ respectively.

(B) $1$. In methane $(CH_4)$,the hybridization is $sp^3$ with no lone pair,resulting in a bond angle of $109.5^\circ$.
$2$. In ammonia $(NH_3)$,the hybridization is $sp^3$ with one lone pair. Due to lone pair-bond pair repulsion,the bond angle decreases to $107^\circ$.
$3$. In trimethylamine $(N(CH_3)_3)$,the hybridization is $sp^3$ with one lone pair. However,the bulky methyl groups cause steric repulsion,which increases the bond angle to approximately $108^\circ$ to $109^\circ$.
$4$. Comparing the values: $NH_3$ $(107^\circ)$ < $N(CH_3)_3$ $(108^\circ-109^\circ)$ < $CH_4$ $(109.5^\circ)$.
$5$. Therefore,the correct order of increasing bond angles is $(II) < (III) < (I)$.

(C) To determine the shape,we calculate the hybridization and number of lone pairs for each species:
$1$. $NH_4^+$: Central atom $N$ has $5$ valence electrons. $H = (5+4-1)/2 = 4$. Hybridization is $sp^3$ with $0$ lone pairs,so it is tetrahedral.
$2$. $SiCl_4$: Central atom $Si$ has $4$ valence electrons. $H = (4+4)/2 = 4$. Hybridization is $sp^3$ with $0$ lone pairs,so it is tetrahedral.
$3$. $SF_4$: Central atom $S$ has $6$ valence electrons. $H = (6+4)/2 = 5$. Hybridization is $sp^3d$ with $1$ lone pair. Due to the presence of a lone pair,its shape is see-saw,not tetrahedral.
$4$. $SO_4^{2-}$: Central atom $S$ has $6$ valence electrons. $H = (6+0+2)/2 = 4$. Hybridization is $sp^3$ with $0$ lone pairs,so it is tetrahedral.
Thus,$SF_4$ is the molecule that is not tetrahedral.

(C) To find the number of lone pairs $(LP)$ on the central atom,we use the formula: $LP = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (monovalent atoms + charge adjustment).
$1. BrF_3$: Central atom $Br$ $(V=7)$. $LP = \frac{1}{2} (7 - 3) = 2 \ LP$.
$2. BrF_4^-$: Central atom $Br$ $(V=7+1=8)$. $LP = \frac{1}{2} (8 - 4) = 2 \ LP$.
$3. XeF_5^+$: Central atom $Xe$ $(V=8-1=7)$. $LP = \frac{1}{2} (7 - 5) = 1 \ LP$.
$4. I_3^-$: Central atom $I$ $(V=7+1=8)$. $LP = \frac{1}{2} (8 - 2) = 3 \ LP$.
Comparing the values,$XeF_5^+$ has the minimum number of lone pairs $(1 \ LP)$.

(B) In $NH_3$,the bond angle is $107^o$ due to $sp^3$ hybridization and lone pair-bond pair repulsion.
In $(CH_3)_3N$,the bond angle is approximately $108^o$ because the bulky methyl groups cause more repulsion than the smaller $H$ atoms.
In $(SiH_3)_3N$,the nitrogen atom is $sp^2$ hybridized with a bond angle of $120^o$ because the lone pair of $N$ engages in $p\pi-d\pi$ back bonding with the empty $d$ orbitals of $Si$.
Therefore,the correct order of bond angles is $(SiH_3)_3N > (CH_3)_3N > NH_3$.

(A) The given reaction is $2Se_2Cl_2 \to 3Se + SeCl_4$.
In $Se_2Cl_2$,the oxidation state of $Se$ is $+1$. In $SeCl_4$,the oxidation state of $Se$ is $+4$,and in $Se$ (elemental),it is $0$. Since the oxidation state changes,it is a redox reaction.
For $SeCl_4$,the central atom $Se$ has $6$ valence electrons. It forms $4$ bonds with $Cl$ atoms and has $1$ lone pair.
Steric number $= 4 \text{ (bond pairs)} + 1 \text{ (lone pair)} = 5$.
This corresponds to $sp^3d$ hybridization.
Due to the presence of one lone pair in the equatorial position of a trigonal bipyramidal geometry,the shape of $SeCl_4$ is see-saw.

(C) The central atom in $NH_3$ and $NF_3$ undergoes $sp^3$ hybridization,resulting in a pyramidal geometry with one lone pair and three bond pairs.
In $NH_3$,the bond pairs are closer to the nitrogen atom,leading to greater repulsion between them compared to $NF_3$,where the highly electronegative fluorine atoms pull the electron density away from the nitrogen.
Consequently,the bond angle in $NH_3$ $(107^{\circ})$ is greater than in $NF_3$ $(102^{\circ})$.
In $PH_3$,the central atom $P$ is larger and less electronegative,and the molecule involves almost pure $p$-orbitals for bonding,resulting in a bond angle close to $92^{\circ}$.
Thus,the correct order of bond angle is $NH_3 > NF_3 > PH_3$.

(A) To determine the electron geometry,we calculate the steric number $(SN)$ using the formula: $SN = \text{Number of bonding pairs} + \text{Number of lone pairs}$.
$A$. For $ICl_2^-$,the central atom $I$ has $7$ valence electrons. It forms $2$ bonds with $Cl$ and has $3$ lone pairs. $SN = 2 + 3 = 5$. The electron geometry is Trigonal bipyramidal. This statement is correct.
$B$. For $NH_3$,the central atom $N$ has $5$ valence electrons. It forms $3$ bonds with $H$ and has $1$ lone pair. $SN = 3 + 1 = 4$. The electron geometry is Tetrahedral. This statement is correct.
$C$. For $SnF_4$,the central atom $Sn$ has $4$ valence electrons. It forms $4$ bonds with $F$ and has $0$ lone pairs. $SN = 4 + 0 = 4$. The electron geometry is Tetrahedral. This statement is correct.
$D$. For $PBr_{5(g)}$,the central atom $P$ has $5$ valence electrons. It forms $5$ bonds with $Br$ and has $0$ lone pairs. $SN = 5 + 0 = 5$. The electron geometry is Trigonal bipyramidal. This statement is correct.
Wait,re-evaluating the options: All provided options describe the correct electron geometries. However,if the question implies a specific error in geometry naming or structure,$ICl_2^-$ is indeed trigonal bipyramidal (linear molecular geometry). Given the standard options,all are technically correct. If forced to choose an error,there might be a typo in the question source. Assuming the question is valid,all statements are correct.

(A) The structure of $I_2Cl_6$ is a dimer of $ICl_3$,which exists as a planar molecule.
Each $I$ atom is $sp^3d$ hybridized with two lone pairs,making the geometry around each $I$ atom square planar (if we consider the bridging atoms).
There are $2$ bridging $Cl$ atoms and $4$ terminal $Cl$ atoms.
Regarding the angles: $\theta_1$ is the $Cl-I-Cl$ angle involving bridging atoms,$\theta_2$ is the $Cl-I-Cl$ angle for terminal atoms,and $\theta_3$ is the $Cl-I-Cl$ angle between a terminal and a bridging atom.
The correct order of angles is $\theta_3 > \theta_2 > \theta_1$.
There are $20$ lone pairs in total ($3$ on each terminal $Cl$,$2$ on each bridging $Cl$,and $2$ on each $I$ atom: $4 \times 3 + 2 \times 2 + 2 \times 2 = 12 + 4 + 4 = 20$).
Thus,statement $(A)$ is the only correct statement among the options provided.

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

(D) The central atom $Xe$ has $8$ valence electrons.
For $XeF_5^-$,the number of electron pairs is calculated as: $\frac{1}{2} (8 + 5 + 1) = 7$.
Out of these $7$ electron pairs,$5$ are bond pairs and $2$ are lone pairs.
According to $VSEPR$ theory,the steric number $7$ corresponds to pentagonal bipyramidal geometry.
With $5$ bond pairs and $2$ lone pairs,the lone pairs occupy the axial positions to minimize repulsion,resulting in a pentagonal planar molecular geometry.

(B) To determine the structure of $IF_6^-$,we calculate the number of electron pairs around the central iodine atom $(I)$.
The central atom $I$ has $7$ valence electrons.
Adding $1$ electron from the negative charge gives $8$ electrons.
It forms $6$ bonds with fluorine atoms,using $6$ electrons.
This leaves $2$ electrons,which form $1$ lone pair.
Total electron pairs = $6$ (bond pairs) + $1$ (lone pair) = $7$ electron pairs.
According to $VSEPR$ theory,$7$ electron pairs correspond to a pentagonal bipyramidal electron geometry.
With $6$ bond pairs and $1$ lone pair,the molecular geometry is distorted octahedral.

(D) $SF_4$ has a see-saw shape (distorted tetrahedral).
$SiF_4$ has a tetrahedral shape.
$XeF_4$ has a square planar shape.
Therefore,the set in option $D$ is incorrect because not all of them are tetrahedral.

(D) In a pentagonal bipyramidal geometry (e.g.,$IF_7$),there are two sets of positions: equatorial and axial.
$1$. The five equatorial bonds are arranged in a pentagon,resulting in bond angles of $360^o / 5 = 72^o$ between adjacent equatorial bonds.
$2$. The two axial bonds are perpendicular to the equatorial plane,resulting in bond angles of $90^o$ between axial and equatorial bonds.
$3$. The angle between the two axial bonds is $180^o$.
Therefore,the bond angles present are $72^o, 90^o,$ and $180^o$.

(A) The $PF_3Cl_2$ molecule has a trigonal bipyramidal geometry.
According to Bent's rule,more electronegative atoms prefer to occupy the axial positions where the bond angle is $90^{\circ}$,while less electronegative atoms prefer the equatorial positions where the bond angle is $120^{\circ}$.
Here,the electronegativity of $F$ is greater than that of $Cl$ $(F > Cl)$.
Therefore,the three $F$ atoms should occupy the equatorial positions,and the two $Cl$ atoms should occupy the axial positions.
This corresponds to the structure shown in option $A$.

(C) $1$. $NO_2^+$ is linear ($sp$ hybridization),while $NO_2^-$ is bent ($sp^2$ hybridization).
$2$. $PCl_5$ is trigonal bipyramidal ($sp^3d$ hybridization),while $BrF_5$ is square pyramidal ($sp^3d^2$ hybridization).
$3$. $XeF_4$ has $sp^3d^2$ hybridization with two lone pairs,resulting in a square planar shape. $ICl_4^-$ also has $sp^3d^2$ hybridization with two lone pairs,resulting in a square planar shape.
Therefore,$XeF_4$ and $ICl_4^-$ have an identical shape.

(B) The molecular shapes are determined by $VSEPR$ theory:
$(a) \ XeO_6^{4-}$ has $6$ bonding pairs and $0$ lone pairs,resulting in an Octahedral shape $(S)$.
$(b) \ ClO_2$ has $1$ lone pair and $2$ bond pairs on the central $Cl$ atom,resulting in a $V$-shape $(Q)$.
$(c) \ NH_4^+$ has $4$ bonding pairs and $0$ lone pairs,resulting in a Tetrahedral shape $(P)$.
$(d) \ XeO_3F_2$ has $5$ bonding pairs and $0$ lone pairs,resulting in a Trigonal Bipyramidal shape $(R)$.
Therefore,the correct matching is $a-S, b-Q, c-P, d-R$.

(D) To be isostructural,planar,and non-polar,the species must have the same hybridization and geometry.
$1$. $I_3^-$: Central atom $I$ has $7$ valence electrons + $1$ (negative charge) + $2$ (from $2$ $I$ atoms) = $10$ electrons,divided by $2$ gives $5$ electron pairs ($sp^3d$ hybridization). It has $3$ lone pairs in the equatorial position,resulting in a linear geometry $(180^{\circ})$,which is planar and non-polar.
$2$. $HgCl_2$: $Hg$ has $2$ valence electrons,forming $2$ bonds with $Cl$. It has $sp$ hybridization,resulting in a linear geometry $(180^{\circ})$,which is planar and non-polar.
$3$. $XeF_2$: Central atom $Xe$ has $8$ valence electrons + $2$ (from $2$ $F$ atoms) = $10$ electrons,divided by $2$ gives $5$ electron pairs ($sp^3d$ hybridization). It has $3$ lone pairs in the equatorial position,resulting in a linear geometry $(180^{\circ})$,which is planar and non-polar.
All three species are linear,planar,and non-polar.

(C) To determine the planar structure,we analyze the hybridization and geometry of each molecule:
$1$. $O_2SF_2$: Sulfur is $sp^3$ hybridized with a tetrahedral geometry,which is non-planar.
$2$. $OSF_2$: Sulfur is $sp^3$ hybridized with a trigonal pyramidal geometry (due to one lone pair),which is non-planar.
$3$. $XeF_4$: Xenon 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). According to $VSEPR$ theory,the $2$ lone pairs occupy axial positions to minimize repulsion,resulting in a square planar geometry.
$4$. $ClO_4^-$: Chlorine is $sp^3$ hybridized with a tetrahedral geometry,which is non-planar.
Therefore,$XeF_4$ is the molecule with a square planar structure.

(A) For $XeF_3^+$: The central atom $Xe$ has $8$ valence electrons. $XeF_3^+$ has $8 - 3 + 1 = 6$ electrons,which corresponds to $3$ bond pairs and $2$ lone pairs. The steric number is $5$,indicating $sp^3d$ hybridization. With $2$ lone pairs in the equatorial positions,the shape is $T$-shaped.
For $SNF_3$: The central atom $S$ is bonded to $N$ (triple bond) and $3$ $F$ atoms (single bonds). The total number of electron pairs around $S$ is $4$ ($1$ triple bond + $3$ single bonds). The steric number is $4$,indicating $sp^3$ hybridization. The geometry is tetrahedral.

(A) The shapes of the given species are determined using $VSEPR$ theory:
$A$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The geometry is trigonal bipyramidal with one equatorial position occupied by a lone pair,resulting in a See-Saw shape $(A-3)$.
$B$. $BrF_3$: Bromine has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The geometry is trigonal bipyramidal with two equatorial positions occupied by lone pairs,resulting in a Bent $T$-shaped structure $(B-4)$.
$C$. $BrO_3^-$: Bromine has $7$ valence electrons. It forms $3$ double bonds with $O$ atoms and has $1$ lone pair. The geometry is tetrahedral with one position occupied by a lone pair,resulting in a Pyramidal shape $(C-2)$.
$D$. $NH_4^+$: Nitrogen has $5$ valence electrons. It forms $4$ bonds with $H$ atoms and has no lone pairs. The geometry is Tetrahedral $(D-1)$.
Thus,the correct matching is $A-3, B-4, C-2, D-1$.

(B) The number of orbitals involved in hybridization $(H)$ is calculated as $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the charge of the cation,and $A$ is the charge of the anion.
For $[XeF_3]^+$,$H = \frac{1}{2} [8 + 3 - 1] = 5$,which corresponds to $sp^3d$ hybridization with $2$ lone pairs. This results in a $T$-shaped geometry.
For $[XeF_5]^+$,$H = \frac{1}{2} [8 + 5 - 1] = 6$,which corresponds to $sp^3d^2$ hybridization with $1$ lone pair. This results in a square pyramidal geometry.
Therefore,the correct option is $B$.

(D) All the given molecules ($BF_3$,$BCl_3$,and $BBr_3$) have the same central atom $(B)$ and the same hybridization $(sp^2)$.
According to $VSEPR$ theory,all these molecules have a trigonal planar geometry with an ideal bond angle of $120^{\circ}$.
Since the central atom is the same and the geometry is identical,the bond angle remains the same in all these molecules.

(A) $SeF_4$ has a see-saw shape due to $sp^3d$ hybridization with one lone pair,whereas $CH_4$ has a tetrahedral shape due to $sp^3$ hybridization with no lone pairs. Therefore,they do not have the same shape. The statement in option $A$ is incorrect.

(B) To determine the shape,we calculate the hybridization of the central atom:
$1$. $N_3^-$: The central $N$ atom is $sp$ hybridized,resulting in a linear shape.
$2$. $NO_3^-$: The central $N$ atom has $3$ bonding pairs and $0$ lone pairs,making it $sp^2$ hybridized with a triangular planar shape.
$3$. $NO_2^-$: The central $N$ atom has $2$ bonding pairs and $1$ lone pair,making it $sp^2$ hybridized with a bent (angular) shape.
$4$. $CO_2$: The central $C$ atom is $sp$ hybridized,resulting in a linear shape.
Therefore,$NO_3^-$ is the species with a triangular planar shape.

(A) The bond angles for the given molecules are as follows:
$OF_2$: $103.2^{\circ}$
$H_2O$: $104.5^{\circ}$
$NH_3$: $107^{\circ}$
$Cl_2O$: $111^{\circ}$
In $OF_2$,the high electronegativity of $F$ pulls the bonding electron pairs away from the central $O$ atom,reducing the bond angle due to lone pair-lone pair repulsion.
In $Cl_2O$,the large size of $Cl$ atoms causes significant steric repulsion,which increases the bond angle to $111^{\circ}$,making it the largest among these.
Thus,the correct order is $OF_2 < H_2O < NH_3 < Cl_2O$.

(B) According to the $VSEPR$ theory,the molecular geometry depends on the number of bonding pairs and lone pairs around the central atom $A$.
For a molecule with no lone pairs on the central atom:
$1$. $AX_3$ has $3$ bonding pairs,resulting in a trigonal planar geometry.
$2$. $AX_4$ has $4$ bonding pairs,resulting in a tetrahedral geometry.
$3$. $AX_5$ has $5$ bonding pairs,resulting in a trigonal bipyramidal geometry.
$4$. $AX_6$ has $6$ bonding pairs,resulting in an octahedral geometry.
Therefore,the molecule $AX_5$ with no lone pair on $A$ has a trigonal bipyramidal structure.

(A) The bond angle in hydrides of group $15$ elements depends on the electronegativity of the central atom.
As the electronegativity of the central atom decreases down the group $(N > P > As > Sb)$,the bond pair electrons move further away from the central atom.
This results in a decrease in the bond pair-bond pair repulsion,leading to a decrease in the bond angle.
Therefore,the correct order of bond angles is $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ})$.

(B) In a trigonal bipyramidal geometry ($sp^3d$ hybridization),the central atom has $5$ electron pairs.
When some of these positions are occupied by lone pairs,the molecular geometry changes:
$1$ lone pair results in a See-Saw shape.
$2$ lone pairs result in a $T$-shape.
$3$ lone pairs result in a Linear shape.
Trigonal planar geometry is associated with $sp^2$ hybridization ($3$ electron pairs) and cannot be derived from a trigonal bipyramidal arrangement.

(A) 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 and $N$ is the number of bonding electrons.
$1$. For $XeF_2$: $Xe$ has $8$ valence electrons. It forms $2$ bonds. Lone pairs = $\frac{1}{2} \times (8 - 2) = 3$.
$2$. For $SF_4$: $S$ has $6$ valence electrons. It forms $4$ bonds. Lone pairs = $\frac{1}{2} \times (6 - 4) = 1$.
$3$. For $ClF_3$: $Cl$ has $7$ valence electrons. It forms $3$ bonds. Lone pairs = $\frac{1}{2} \times (7 - 3) = 2$.
$4$. For $H_2O$: $O$ has $6$ valence electrons. It forms $2$ bonds. Lone pairs = $\frac{1}{2} \times (6 - 2) = 2$.
Comparing these,$XeF_2$ has the maximum number of lone pairs $(3)$.

(C) According to the $VSEPR$ theory,the trigonal bipyramidal electronic geometry (where the steric number is $5$) can result in various molecular shapes depending on the number of lone pairs present on the central atom:
$1$. $0$ lone pairs: Trigonal bipyramidal
$2$. $1$ lone pair: See-saw
$3$. $2$ lone pairs: $T$-shaped
$4$. $3$ lone pairs: Linear
$A$ tetrahedral shape arises from a tetrahedral electronic geometry (steric number $4$),not from a trigonal bipyramidal geometry.

(D) In $PF_3$ $(sp^3)$,$BF_3$ $(sp^2)$,and $CCl_4$ $(sp^3)$,all bond lengths are equivalent due to the high symmetry of the molecules.
In $SF_4$,the sulfur atom undergoes $sp^3d$ hybridization with one lone pair.
Due to the presence of the lone pair and the trigonal bipyramidal geometry,the axial and equatorial $S-F$ bonds have different lengths,making them unequal.

(D) The $ClF_3$ molecule has $sp^3d$ hybridization with $2$ lone pairs and $3$ bond pairs. According to $VSEPR$ theory,lone pairs occupy equatorial positions to minimize repulsion. If the lone pairs were to occupy axial positions,the molecule would adopt a $T$-shaped geometry with a $120^o$ bond angle between the equatorial fluorine atoms. In this hypothetical configuration,the molecule would be non-polar due to the symmetric arrangement of the polar $Cl-F$ bonds,and the shape would be trigonal planar. Since all these statements describe the consequences of the lone pairs occupying axial positions,the correct option is $D$.

(D) The molecule $XeF_4$ has a square planar geometry.
In this structure,the four $Xe-F$ bonds are arranged symmetrically,and the two lone pairs on the Xenon atom are positioned above and below the plane.
Due to this highly symmetrical arrangement,the dipole moments of the bonds cancel each other out,resulting in a net dipole moment of $\mu = 0$.

(D) The order $PH_4^+ > NH_3$ regarding bond angle cannot be explained by back bonding.
In $PH_4^+$,the central atom $P$ is $sp^3$ hybridized with a bond angle of $109.5^o$.
In $NH_3$,the central atom $N$ is $sp^3$ hybridized with one lone pair,resulting in a bond angle of $107^o$ due to $lp-bp$ repulsion.
This difference is due to hybridization and $VSEPR$ theory,not back bonding.

(D) $XeF_2$ has a linear geometry with a bond angle of $180^{\circ}$.
$BeF_3^-$,$BF_3$,and $CH_3^+$ are all $sp^2$ hybridized with $0$ lone pairs,resulting in a bond angle of $120^{\circ}$.
$CH_4$,$SiCl_4$,and $BF_4^-$ are all $sp^3$ hybridized with $0$ lone pairs,resulting in a bond angle of $109.5^{\circ}$.
$CCl_4$ has a bond angle of $109.5^{\circ}$ and $H_2O$ has a bond angle of $104.5^{\circ}$.
In $PH_3$ and $PCl_3$,the bond angle depends on the electronegativity of the central atom and the surrounding atoms. $PH_3$ has a bond angle of $\approx 93^{\circ}$ and $PCl_3$ has a bond angle of $\approx 100^{\circ}$.
Therefore,the order $PH_3 > PCl_3$ is incorrect.

(C) To determine the number of lone pairs on the central atom:
$1$. $C_2H_2$ (Acetylene): $H-C \equiv C-H$. The central carbon atoms have $0$ lone pairs.
$2$. $HCN$: $H-C \equiv N$. The central carbon atom has $0$ lone pairs.
$3$. $ICl_2^-$: The central iodine atom has $7$ valence electrons,plus $1$ from the negative charge,minus $2$ used for bonding with $Cl$ atoms,leaving $6$ electrons,which equals $3$ lone pairs.
$4$. $CS_2$: $S=C=S$. The central carbon atom has $0$ lone pairs.
Thus,$ICl_2^-$ is linear and has the maximum number of lone pairs $(3)$.

(B) The bond angle depends on the hybridization and the presence of lone pairs on the central atom.
$NH_4^+$ has $sp^3$ hybridization with $0$ lone pairs,resulting in a perfect tetrahedral geometry with a bond angle of $109.5^\circ$.
$NH_3$ has $sp^3$ hybridization with $1$ lone pair,resulting in a bond angle of $107^\circ$.
$PH_3$ has a bond angle of approximately $93.5^\circ$ due to the lack of hybridization (Drago's rule).
$SCl_2$ has $sp^3$ hybridization with $2$ lone pairs,resulting in a bond angle of approximately $103^\circ$.
Therefore,the bond angle is maximum in $NH_4^+$.

(B) To determine the $O-N-O$ bond angle,we look at the hybridization and geometry of the molecules:
$1$. $NO_2^+$: The nitrogen atom is $sp$-hybridized with a linear geometry. The bond angle is $180^{\circ}$.
$2$. $NO_2$: The nitrogen atom is $sp^2$-hybridized with one unpaired electron. The bond angle is approximately $134^{\circ}$.
$3$. $NO_2^-$: The nitrogen atom is $sp^2$-hybridized with one lone pair. The bond angle is approximately $115^{\circ}$.
$4$. $NO_3^-$: The nitrogen atom is $sp^2$-hybridized with no lone pairs,resulting in a trigonal planar geometry. The bond angle is $120^{\circ}$.
Comparing these values,the $O-N-O$ bond angle is highest in $NO_2^+$.

(A) In $SO_2Cl_2$,the central sulfur atom is $sp^3$ hybridized.
According to $VSEPR$ theory,the repulsion between bond pairs follows the order: $double-double > double-single > single-single$.
Here,$x$ is the angle between two $S=O$ double bonds,and $y$ is the angle between two $S-Cl$ single bonds.
Since the repulsion between the two $S=O$ double bonds is greater than the repulsion between the two $S-Cl$ single bonds,the bond angle $x$ will be greater than the bond angle $y$ $(x > y)$.

(B) The $ICl_{4}^{-}$ ion has a square planar geometry with $sp^{3}d^{2}$ hybridization.
There are $4$ bond pairs ($I-Cl$ bonds) and $2$ lone pairs on the central iodine atom.
The two lone pairs occupy the axial positions (above and below the plane).
Each lone pair is at a $90^{o}$ angle to all $4$ bond pairs in the square plane.
Therefore,the total number of bond pair-lone pair repulsions at $90^{o}$ is $2 \times 4 = 8$.

(B) In the given molecules,$x$ is the $O=S=O$ bond angle in $SO_2F_4$ (or similar structure) and $y$ is the $O=I-F$ bond angle in $IOF_4^-$.
According to $VSEPR$ theory,multiple bonds exert greater repulsion than single bonds.
However,the bond angle is also influenced by the electronegativity of the central atom and the surrounding atoms.
In $SO_2F_4$,the $S=O$ double bonds exert more repulsion than the $S-F$ single bonds,pushing the $F-S-F$ or $O-S-O$ angles.
Comparing the structures,the repulsion from the double bond in the sulfur compound is significant,but due to the larger size of the Iodine atom and the specific geometry of the $IOF_4^-$ ion,the bond angle $x$ is found to be greater than $y$ $(x > y)$.

(B) The molecule has a total of $5$ bond pairs and $2$ lone pairs,resulting in a steric number of $7$.
This corresponds to $sp^{3}d^{3}$ hybridization.
According to $VSEPR$ theory,for $AB_5L_2$ type molecules,the $5$ bond pairs occupy the equatorial positions of a pentagonal bipyramid,while the $2$ lone pairs occupy the axial positions to minimize repulsion.
Therefore,the shape of the molecule is pentagonal planar.