VSEPR Theory Questions in English

(A) To determine the geometry of the $IBr^{-}_{2}$ ion,we calculate the steric number $(X)$:
$X = \frac{1}{2} [V + M - C + A]$
Where $V$ is the number of valence electrons of the central atom $(I = 7)$,
$M$ is the number of monovalent atoms attached $(Br = 2)$,
$C$ is the cationic charge $(0)$,
$A$ is the anionic charge $(1)$.
$X = \frac{1}{2} [7 + 2 + 1] = \frac{10}{2} = 5$.
$A$ steric number of $5$ corresponds to $sp^{3}d$ hybridization,which has a trigonal bipyramidal electron geometry.
For $IBr^{-}_{2}$,there are $2$ bond pairs and $3$ lone pairs.
According to $VSEPR$ theory,the $3$ lone pairs occupy the equatorial positions to minimize repulsion,resulting in a linear molecular geometry.

(C) To determine the shape of the $ClF_3$ molecule,we calculate the number of electron pairs around the central atom $Cl$ using the formula: $X = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom $(Cl = 7)$,$M$ is the number of monovalent atoms $(F = 3)$,$C$ is the cationic charge $(0)$,and $A$ is the anionic charge $(0)$.
$X = \frac{1}{2} [7 + 3] = 5$.
This indicates $sp^3d$ hybridization.
Out of $5$ electron pairs,$3$ are bond pairs (forming $3$ $\sigma$-bonds with $F$ atoms) and $2$ are lone pairs.
According to $VSEPR$ theory,the $2$ lone pairs occupy the equatorial positions to minimize repulsion,resulting in a $T$-shaped molecular geometry.

(A) $NH_3$ has a trigonal pyramidal shape due to the presence of one lone pair on the central nitrogen atom.
$SO_3^{2-}$ has a central sulfur atom with one lone pair and three bond pairs,resulting in a trigonal pyramidal shape.
$CO_3^{2-}$ has a trigonal planar shape.
$NO_3^{-}$ has a trigonal planar shape.
$SO_3$ has a trigonal planar shape.
Therefore,$SO_3^{2-}$ has the same shape as $NH_3$.

(A) For $CH^{+}_3$ (methyl carbocation):
The carbon atom is $sp^2$ hybridized with three bond pairs and no lone pair.
Its geometry is trigonal planar,meaning all atoms lie in the same plane.
For $CH^{-}_3$ (methyl carbanion):
The carbon atom is $sp^3$ hybridized with three bond pairs and one lone pair.
Its geometry is trigonal pyramidal,meaning the atoms do not lie in the same plane.
Therefore,only $CH^{+}_3$ has all atoms in the same plane.

(D) To determine the number of bond pairs and lone pairs for each species:
$(a)$ In $OCN^{-}$,the structure is $[:\ddot{O}-C \equiv N:]^{-}$. It has $4$ bond pairs and $4$ lone pairs.
$(b)$ In $H_2O$,the structure is $H-\ddot{O}-H$. It has $2$ bond pairs and $2$ lone pairs.
$(c)$ In $C_2H_2Cl_2$,the structure is $H_2C=C(Cl)_2$. It has $6$ bond pairs and $6$ lone pairs.
$(d)$ In $O_3$,the structure is $:\ddot{O}=\ddot{O}-\ddot{O}:$. It has $3$ bond pairs and $6$ lone pairs.
Thus,$O_3$ does not have the same number of bond pairs and lone pairs.

(B) The bond angles for the given compounds are as follows:
$OH_2$ (water): $104.5^{\circ}$
$SH_2$ (hydrogen sulfide): $92.5^{\circ}$
$NH_3$ (ammonia): $106.5^{\circ}$
$SO_2$ (sulfur dioxide): $119.5^{\circ}$
Comparing these values,$SH_2$ has the smallest bond angle because the central atom $S$ is larger than $O$ and $N$,leading to less repulsion between the bonding pairs and a bond angle closer to $90^{\circ}$.

(B) In the hydrides of group $15$ elements $(NH_3, PH_3, AsH_3, SbH_3, BiH_3)$,the central atom undergoes $sp^3$ hybridization.
As we move down the group,the electronegativity of the central atom decreases.
According to Drago's rule,as the electronegativity of the central atom decreases,the bond pair-bond pair repulsion decreases,and the bond angle approaches $90^{\circ}$.
Therefore,the bond angle decreases as we move down the group: $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ}) > BiH_3 (90^{\circ})$.
Thus,the correct increasing order is $BiH_3 < SbH_3 < AsH_3 < PH_3 < NH_3$.

(C) In $BF_3$,the central atom $B$ is $sp^2$ hybridized with a trigonal planar geometry,resulting in a bond angle of $120^{\circ}$.
In $PF_3$,the central atom $P$ is $sp^3$ hybridized with a trigonal pyramidal geometry. Due to the presence of one lone pair,the bond angle is compressed to approximately $96^{\circ}$.
In $ClF_3$,the central atom $Cl$ is $sp^3d$ hybridized with a $T$-shaped geometry. Due to the presence of two lone pairs in the equatorial positions,the bond angle is further compressed to approximately $87^{\circ}$.
Thus,the increasing order of bond angles is $ClF_3 (87^{\circ}) < PF_3 (96^{\circ}) < BF_3 (120^{\circ})$.

(D) $O_3$ molecule involves $sp^2$ hybridization with one lone pair,resulting in a bond angle of approximately $117^{\circ}$.
$(B)$ $I_3^-$ involves $sp^3d$ hybridization with three lone pairs on the central atom,resulting in a linear geometry with a bond angle of $180^{\circ}$.
$(C)$ $NO_2^-$ involves $sp^2$ hybridization with one lone pair,resulting in a bond angle of approximately $115^{\circ}$.
$(D)$ $PH_3$ involves $sp^3$ hybridization but exhibits significant $s-p$ mixing,leading to a bond angle close to $93^{\circ}$ due to the presence of a lone pair and the lack of hybridization of the $P$ orbitals.
$\therefore$ Among the given species,$PH_3$ has the least bond angle.

(B) According to $VSEPR$ theory,the bond angle decreases as the number of lone pairs on the central atom increases due to increased lone pair-bond pair repulsion.
$NH_4^+$ has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109.5^\circ$.
$NH_3$ has $3$ bond pairs and $1$ lone pair,which exerts repulsion on the bond pairs,reducing the bond angle to approximately $107^\circ$.
$NH_2^-$ has $2$ bond pairs and $2$ lone pairs,which exert even greater repulsion,further reducing the bond angle to approximately $104.5^\circ$.
Therefore,the order of bond angles is $NH_4^+ > NH_3 > NH_2^-$.

(C) According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,the geometry of a molecule is determined by the repulsion between electron pairs in the valence shell of the central atom.
$1.$ In $CH_4$,there are $4$ bonding pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.
$2.$ In $H_2O$,the oxygen atom has $2$ bonding pairs and $2$ lone pairs.
$3.$ According to $VSEPR$ theory,the order of repulsion is: $\text{lone pair-lone pair} > \text{lone pair-bonding pair} > \text{bonding pair-bonding pair}$.
$4.$ The strong repulsion between the two lone pairs on the oxygen atom pushes the bonding pairs closer together,causing the $H-O-H$ bond angle to decrease from the ideal tetrahedral angle of $109.5^{\circ}$ to approximately $104.5^{\circ}$.

(A) In the group $16$ hydrides $(H_2O, H_2S, H_2Se, H_2Te)$,the central atom is bonded to two hydrogen atoms and has two lone pairs.
As we move down the group,the electronegativity of the central atom decreases.
This leads to an increase in the bond length and a decrease in the bond pair-bond pair repulsion.
Consequently,the bond angle decreases as the size of the central atom increases.
Therefore,$H_2O$ has the largest bond angle of $104.5^{\circ}$.

(D) tetrahedral molecule $MX_4$ has a central atom $M$ bonded to four $X$ atoms at the corners of a regular tetrahedron.
In a tetrahedron,the number of angles formed by the bonds is given by the combination formula $^nC_2$,where $n$ is the number of bonds.
Here,$n = 4$,so the number of $\angle XMX$ angles is $^4C_2 = \frac{4 \times 3}{2} = 6$.

(B) $N_2O$ has the structure $\overset{-}{N}=\overset{+}{N}=O$,so it does not contain an $O-N-O$ bond angle.
Comparing the bond angles of the other species:
$NO_2^+$ is linear with a bond angle of $180^\circ$.
$NO_2^-$ is bent with a bond angle of approximately $115^\circ$ due to the presence of a lone pair on the nitrogen atom.
$NO_3^-$ is trigonal planar with a bond angle of $120^\circ$.
Therefore,$NO_2^+$ has the maximum $O-N-O$ bond angle.

(C) The correct order of bond angle is $H_3Te^{+} < H_3Se^{+} < H_3S^{+} < H_3O^{+}$.
Explanation:
$(a)$ In $NH_3$ group hydrides,as the electronegativity of the central atom decreases,the bond angle decreases: $NH_3 > PH_3 > AsH_3 > SbH_3$.
$(b)$ In $OF_2$ and $H_2O$,the bond angle is affected by the electronegativity of the surrounding atoms. The correct order is $OF_2 < H_2O < Cl_2O$.
$(c)$ In $H_3X^{+}$ species,as the electronegativity of the central atom $X$ increases,the bond angle increases. Thus,$H_3Te^{+} < H_3Se^{+} < H_3S^{+} < H_3O^{+}$ is correct.
$(d)$ In $BX_3$ molecules,the bond angle is $120^{\circ}$ for all due to $sp^2$ hybridization,so $BF_3 = BCl_3 = BBr_3 = BI_3$.

(B) $1$. For $XeF^{-}_5$: The structure is pentagonal planar with two lone pairs above and below the plane. The central $Xe$ atom and the five $F$ atoms lie in the same plane. Thus,the number of atoms in the $XY$ plane is $6$.
$2$. For $SF_6$: The structure is octahedral. The central $S$ atom and four $F$ atoms lie in the equatorial plane. Thus,the number of atoms in the $XY$ plane is $5$.
$3$. For $IF_7$: The structure is pentagonal bipyramidal. The central $I$ atom and five $F$ atoms lie in the equatorial plane. Thus,the number of atoms in the $XY$ plane is $6$.
$4$. Comparing the values: $SF_6$ has the minimum number of atoms $(5)$ in the $XY$ plane.

(B) The molecule $ML_x$ has $7$ electron pairs in the valence shell of the central atom $M$.
For the molecule to be planar with $7$ electron pairs,the geometry must be pentagonal bipyramidal,where $5$ bond pairs are in the equatorial plane and $2$ lone pairs occupy the axial positions (above and below the plane).
This arrangement minimizes electronic repulsion.
Since there are $5$ bond pairs,the number of ligands $x$ is $5$.

(D) For $I^{+}_3$: The central $I$ atom has $2$ bond pairs and $2$ lone pairs. Hybridization is $sp^3$. The geometry is bent,making it a polar species $(\mu \neq 0)$. It is a planar species.
For $I^{-}_3$: The central $I$ atom has $2$ bond pairs and $3$ lone pairs. Hybridization is $sp^3d$. The geometry is linear,making it a non-polar species $(\mu = 0)$. It is a planar species.
Comparing both: Both $I^{+}_3$ and $I^{-}_3$ are planar species. Thus,option $D$ is correct.

(C) The central atom $X$ has two $\sigma$-bonds and two $\pi$-bonds.
This implies that $X$ is involved in two double bonds (or one single and one triple bond,but given $XY_2$ and the presence of one lone pair,the structure is $Y=X=Y$).
The total number of electron domains around $X$ is determined by the number of $\sigma$-bonds and lone pairs.
Here,there are $2$ $\sigma$-bonds and $1$ lone pair,totaling $3$ electron domains.
According to $VSEPR$ theory,$3$ electron domains correspond to $sp^2$ hybridization with a trigonal planar geometry.
Since one of these domains is a lone pair,the molecular shape is bent,but the arrangement of the electron pairs (lone pair and bond pairs) is trigonal planar.

(C) To determine the geometry and lone pairs,we use the formula for steric number: $SN = \frac{1}{2}(V + M - C + A)$,where $V$ is valence electrons,$M$ is monovalent atoms,$C$ is cationic charge,and $A$ is anionic charge.
$I. SF_4$: $SN = \frac{1}{2}(6 + 4) = 5$ ($sp^3d$ hybridization). It has $1$ lone pair and a see-saw shape.
$II. XeO_4$: $SN = \frac{1}{2}(8 + 0) = 4$ ($sp^3$ hybridization). It has $0$ lone pairs and a tetrahedral shape.
$III. XeF_4$: $SN = \frac{1}{2}(8 + 4) = 6$ ($sp^3d^2$ hybridization). It has $2$ lone pairs and a square planar shape.
$IV. ICl_4^-$: $SN = \frac{1}{2}(7 + 4 + 1) = 6$ ($sp^3d^2$ hybridization). It has $2$ lone pairs and a square planar shape.
Thus,$III$ and $IV$ possess two lone pairs and are square planar.

(D) The correct set is $BF_3, BCl_3, BBr_3$.
In all these molecules,the central Boron atom undergoes $sp^2$ hybridization.
Since all these molecules have a trigonal planar geometry with no lone pairs on the central atom,they all exhibit a bond angle of $120^{\circ}$.
In contrast,$NF_3$ and $NH_3$ have lone pairs on the central atom,which cause distortion and result in bond angles different from $120^{\circ}$.

(D) For series $(A)$,as the electronegativity of the halogen decreases,the electron density around the central atom increases,leading to greater repulsion between the bonding pairs and the lone pair,which decreases the bond angle. Thus,$OSF_2$ has the smallest bond angle.
For series $(B)$,as the electronegativity of the halogen decreases,the bond angle increases. Thus,$SbCl_3$ has the smallest bond angle.
For series $(C)$,as we move down the group,the central atom size increases and the electronegativity decreases,leading to a decrease in the bond angle. Thus,$SbI_3$ has the smallest bond angle.
Therefore,the correct sequence is $OSF_2, SbCl_3$ and $SbI_3$.

(C) The nitrite ion $(NO_2^-)$ has a central nitrogen atom bonded to two oxygen atoms with one lone pair on the nitrogen atom. According to $VSEPR$ theory,the presence of the lone pair causes the shape to be angular (or bent).
In the context of chemical nomenclature,'nitrile' refers to the functional group $-CN$. However,if the question refers to the nitrite ion $(NO_2^-)$ and the nitro group $(-NO_2)$,both species involve a central nitrogen atom with a lone pair or specific bonding geometry that results in an angular shape.
Specifically,the nitrite ion $(NO_2^-)$ is angular due to $sp^2$ hybridization of the nitrogen atom ($1$ lone pair,$2$ bond pairs).
Therefore,both species are angular.

(C) $NCO^-$,$CS_2$,and $NO_2^+$ exhibit $sp$ hybridization,resulting in a linear geometry.
In the solid state,$BeH_2$ exists as a polymeric chain structure where each $Be$ atom is $sp^3$ hybridized,making it non-linear.

(D) $S_2Cl_2$ has an open book-like structure.
Whereas:
$SOCl_2$ has $sp^3$ hybridization and a pyramidal shape.
$CO_2$ has $sp$ hybridization and a linear shape.
$H_2S$ has $sp^3$ hybridization and a bent shape.
$H_2O_2$ has an open book-like structure,hence it is analogous to $S_2Cl_2$.
Therefore,option $D$ is the correct answer.

(A) The $AsF_3Cl_2$ molecule adopts a trigonal bipyramidal geometry.
According to Bent's rule,more electronegative atoms prefer to occupy the axial positions to minimize repulsion.
Therefore,the three $F$ atoms occupy two axial positions and one equatorial position,while the two $Cl$ atoms occupy the remaining two equatorial positions.
The bond angles involving the axial $F$ atoms and the equatorial $F$ atom are $90^{\circ}$,and the bond angle between the two axial $F$ atoms is $180^{\circ}$.

(C) The bond angle depends on the repulsion between lone pairs and bond pairs,as well as the electronegativity of the central and terminal atoms.
$1$. For $H_2O$,the bond angle is $104.5^{\circ}$.
$2$. For $F_2O$,the bond angle is $102^{\circ}$. The electronegative $F$ atoms pull electron density away from the oxygen,reducing the repulsion between bond pairs.
$3$. For $Cl_2O$,the bond angle is $109.5^{\circ}$. The large size of the $Cl$ atoms and their lower electronegativity compared to $F$ leads to significant lone pair-bond pair and bond pair-bond pair repulsions,resulting in a larger bond angle.
$4$. For $H_2S$,the bond angle is approximately $92^{\circ}$ due to the larger size of the $S$ atom and the lack of hybridization (Drago's rule).
Comparing these values,$Cl_2O$ has the largest bond angle. Therefore,option $(C)$ is correct.

(C) $SeOCl_2$ has a trigonal pyramidal geometry with $sp^3$ hybridization of the central $Se$ atom.
$1$. The molecule has a plane of symmetry passing through the $Se=O$ bond and bisecting the $Cl-Se-Cl$ angle.
$2$. Due to the presence of a lone pair and the double bond $(Se=O)$,the $Cl-Se-Cl$ bond angle is compressed to approximately $106^\circ$,while the $Cl-Se-O$ bond angle is approximately $108^\circ$. Thus,$Cl-Se-Cl < Cl-Se-O$.
$3$. According to Bent's rule,the lone pair occupies an orbital with higher $s-$ character. In $sp^3$ hybridization,the $s-$ character is $25\%$. The lone pair in $SeOCl_2$ has more than $25\% \, s-$ character (often cited as $> 33.3\%$ in specific theoretical models for such molecules).
$4$. $Se$ does not use $d-$ orbitals for hybridization in this molecule; it is $sp^3$ hybridized.

(B) molecule has a regular geometry if all bond angles are equal and all bond lengths are identical,which typically occurs when the central atom has no lone pairs and all surrounding atoms are the same.
$CH_4$ has a tetrahedral geometry with $sp^3$ hybridization and no lone pairs,resulting in a regular geometry.
$CHCl_3$ has different atoms ($H$ and $Cl$) attached to the central carbon atom,leading to unequal bond angles and bond lengths,thus it lacks regular geometry.
$SF_4$ has $sp^3d$ hybridization with one lone pair on the sulfur atom,which causes distortion in the bond angles (seesaw shape),resulting in an irregular geometry.
Therefore,$CHCl_3$ and $SF_4$ do not have regular geometry.

(A) When $KI$ and $I_2$ react,they form $KI_3$ as follows:
$KI + I_2 \rightarrow KI_3$
The species $KI_3$ dissociates into $K^+$ and $I_3^-$ ions.
The central iodine atom in the $I_3^-$ ion has $3$ lone pairs and $2$ bond pairs.
According to $VSEPR$ theory,the hybridization is $sp^3d$,which corresponds to a trigonal bipyramidal electron geometry.
Since there are $3$ lone pairs occupying the equatorial positions,the remaining $2$ bond pairs are at the axial positions,resulting in a linear shape.
Therefore,the shape of the $I_3^-$ ion is linear.

(D) To determine if a species is planar,we analyze its molecular geometry based on $VSEPR$ theory:
$1$. $I_{2}Cl_{6}$: This molecule exists as a dimer with a planar structure ($sp^{3}d^{2}$ hybridization at $I$ atoms).
$2$. $XeF_{2}$: It has $sp^{3}d$ hybridization with $3$ lone pairs in the equatorial plane,resulting in a linear (planar) geometry.
$3$. $BrF_{4}^{-}$: It has $sp^{3}d^{2}$ hybridization with $2$ lone pairs in axial positions,resulting in a square planar geometry.
$4$. $XeF_{5}^{-}$: It has $sp^{3}d^{3}$ hybridization with $2$ lone pairs in axial positions,resulting in a pentagonal planar geometry.
Since all species in option $D$ are planar,it is the correct answer.

(C) $I. XeF_5^-$: Central atom $Xe$ has $8$ valence electrons. It forms $5$ bonds with $F$ and has $1$ negative charge,total $9$ electrons. $5$ are used in bonding,leaving $4$ electrons,which equals $2$ lone pairs.
$II. BrF_3$: Central atom $Br$ has $7$ valence electrons. $3$ are used in bonding with $F$,leaving $4$ electrons,which equals $2$ lone pairs.
$III. XeF_2$: Central atom $Xe$ has $8$ valence electrons. $2$ are used in bonding,leaving $6$ electrons,which equals $3$ lone pairs.
$IV. H_3S^{+}$: Central atom $S$ has $6$ valence electrons. $3$ are used in bonding with $H$,and $1$ is lost due to positive charge,leaving $2$ electrons,which equals $1$ lone pair.
$V. CH_2$: Central atom $C$ has $4$ valence electrons. $2$ are used in bonding with $H$,leaving $2$ non-bonding electrons (radical),which equals $0$ lone pairs.
Thus,$XeF_5^-$ and $BrF_3$ both have $2$ lone pairs.

(D) $1$. $B_3N_3H_6$ (Borazine) is planar,similar to benzene.
$2$. $C_3N_3(NH_2)_3$ (Melamine) is planar due to the resonance of the triazine ring.
$3$. $SO_3$ is planar with a trigonal planar geometry ($sp^2$ hybridization).
$4$. $C_3N_3(N_3)_3$ (Cyanuric triazide) is not perfectly flat because the azide groups $(-N_3)$ attached to the triazine ring are not coplanar with the ring due to steric hindrance and the geometry of the azide group.

(C) $1$. $Na_3B_3O_6$ contains a $B_3O_6^{3-}$ ring where boron atoms are $sp^2$ hybridized,making the ring planar.
$2$. $I_2Cl_6$ is a planar molecule where iodine atoms are $sp^3d^2$ hybridized.
$3$. Sheet silicates consist of $SiO_4$ tetrahedra linked in a two-dimensional sheet. While the individual sheets are extended,the overall three-dimensional structure of silicate minerals is non-planar due to the tetrahedral geometry of silicon.
$4$. Inorganic graphite (boron nitride) consists of layers where both $B$ and $N$ are $sp^2$ hybridized,making each individual layer planar.
$5$. Among the given options,the sheet silicate structure is considered non-planar in its three-dimensional arrangement.

(D) To determine the geometry of the given species,we use the $VSEPR$ theory:
$1$. $(CN)_2$ (Cyanogen): The structure is $N \equiv C-C \equiv N$. The carbon atoms are $sp$ hybridized,resulting in a linear geometry.
$2$. $OCN^{-}$ (Cyanate ion): The structure is $[O=C=N]^{-}$. The central carbon atom is $sp$ hybridized with no lone pairs,resulting in a linear geometry.
$3$. $XeF_2$: The central $Xe$ atom has $2$ bond pairs and $3$ lone pairs. According to $VSEPR$ theory,the lone pairs occupy the equatorial positions of a trigonal bipyramid to minimize repulsion,resulting in a linear molecular geometry.
$4$. $S_3^{2-}$ (Trisulfide ion): The central sulfur atom has $2$ bond pairs and $2$ lone pairs (total $4$ electron pairs,$sp^3$ hybridization). Due to the presence of $2$ lone pairs,the geometry is bent (angular),similar to $H_2O$.
Therefore,$S_3^{2-}$ is not linear.

(B) The correct answer is $(B)$.
In a trigonal bipyramidal geometry,the electron pairs are distributed in equatorial and axial positions.
According to $VSEPR$ theory,lone pairs occupy equatorial positions to minimize repulsion.
Therefore,a trigonal bipyramidal geometry cannot result in a triangular planar molecular shape,as that would require placing lone pairs in axial positions,which is energetically unfavorable due to increased repulsion.

(NONE) The correct matches are:
$A$. $SF_4$ has $sp^3d$ hybridization with $1$ lone pair,resulting in a see-saw shape.
$B$. $BrF_3$ has $sp^3d$ hybridization with $2$ lone pairs,resulting in a $T$-shaped geometry.
$C$. $XeF_4$ has $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar geometry.
$D$. $ICl_2^-$ has $sp^3d$ hybridization with $3$ lone pairs,resulting in a linear geometry.
Since all given options are correct matches,there is no incorrect match in the provided list.

(A) To determine which molecule has all equal $X-F$ bond lengths,we analyze the geometry and hybridization of each:
$1$. $SOCl_2F_2$: The molecule has a trigonal bipyramidal geometry with $S$ as the central atom. The $S=O$ bond and the two $Cl$ atoms occupy positions such that the two $S-F$ bonds are equivalent due to symmetry,resulting in equal $S-F$ bond lengths.
$2$. $SeF_4$: This molecule has a see-saw geometry ($sp^3d$ hybridization) with one lone pair. The axial and equatorial $Se-F$ bonds have different lengths.
$3$. $PBr_2F_3$: This molecule has a trigonal bipyramidal geometry. The $F$ atoms occupy both axial and equatorial positions,which have different bond lengths.
$4$. $IF_7$: This molecule has a pentagonal bipyramidal geometry. The axial and equatorial $I-F$ bonds have different lengths.
Therefore,$SOCl_2F_2$ is the correct molecule where all $S-F$ bonds are equal.

(C) From the provided image,the bond lengths are:
$a = 1.577 \ \mathring{A}, b = 1.534 \ \mathring{A}, c = 1.612 \ \mathring{A}, d = 1.543 \ \mathring{A}, e = 1.643 \ \mathring{A}, f = 1.553 \ \mathring{A}, g = 2.19 \ \mathring{A}, h = 2.04 \ \mathring{A}$.
Comparing the values:
$g (2.19) > h (2.04) > e (1.643) > c (1.612) > a (1.577) > f (1.553) > d (1.543) > b (1.534)$.
Evaluating the options:
$A$. $g (2.19) > a (1.577) > d (1.543) > b (1.534)$ (Correct order)
$B$. $g (2.19) > e (1.643) > f (1.553) > b (1.534)$ (Correct order)
$C$. $f (1.553) > d (1.543) > a (1.577) > b (1.534)$ (Incorrect order,as $a > f$)
$D$. $c (1.612) > f (1.553) > d (1.543) > b (1.534)$ (Correct order)
Thus,option $C$ is the incorrect order.

(C) According to Bent's rule,the bond angle is inversely proportional to the $p$-character of the hybrid orbitals forming the bonds.
$(I) \ SF_4$: The lone pair occupies an equatorial position in the trigonal bipyramidal geometry. The lone pair-bond pair repulsion compresses the equatorial $FSF$ bond angle to approximately $101^{\circ}$.
$(II) \ OSF_4$: The highly electronegative oxygen atom is double-bonded to the sulfur atom. According to Bent's rule,the double bond prefers an orbital with more $s$-character,leaving more $p$-character for the $S-F$ bonds. This results in a smaller bond angle compared to $SF_4$.
$(III) \ H_2CSF_4$: The $CH_2$ group is attached via a double bond. The presence of the $CH_2$ group further influences the hybridization,leading to an equatorial $FSF$ bond angle of approximately $97^{\circ}$.
Comparing the effects,the bond angle decreases as the $p$-character of the $S-F$ bonds increases due to the presence of more electronegative or double-bonded substituents. The correct order is $(I) > (II) > (III)$.

(C) To determine the bond angle,we consider hybridization,lone pairs,and electronegativity of the central and surrounding atoms.
$1$. For option $A$: $OCl_2$ $(111^{\circ})$,$SF_2$ $(103^{\circ})$,$AsH_3$ $(92^{\circ})$,$H_2Se$ $(91^{\circ})$. The order is correct.
$2$. For option $B$: $NH_3$ $(107^{\circ})$,$PF_3$ $(96^{\circ})$,$PH_3$ $(93^{\circ})$,$H_2S$ $(92^{\circ})$. The order is correct.
$3$. For option $C$: All these species ($XeO_4$,$ClO^{-}_4$,$SO^{2-}_4$,$CF_4$) have $sp^3$ hybridization and a tetrahedral geometry. In an ideal tetrahedral geometry,the bond angle is $109^{\circ} 28'$. Since all have the same geometry,their bond angles are equal. Therefore,the order $XeO_4 > ClO^{-}_4 > SO^{2-}_4 > CF_4$ is incorrect because they are all equal.
Thus,option $C$ represents the incorrect order.

(D) $1$. In $SSF_2$,the sulfur atom bonded to fluorine has a lone pair,resulting in a bond angle $\angle FSF < 109^{\circ}28'$.
$2$. In $SF_6$,the geometry is octahedral,and the bond angle $\angle FSF$ is $90^{\circ}$.
$3$. In $SF_2$,the sulfur atom has two lone pairs,resulting in a bond angle $\angle FSF < 109^{\circ}28'$.
$4$. In $F_3SSF$,the sulfur atom bonded to three fluorine atoms and one sulfur atom has a lone pair,causing significant repulsion,leading to a bond angle $\angle FSF < 90^{\circ}$.
$5$. Comparing all,the minimum $\angle FSF$ bond angle is in $F_3SSF$.

(B) The bond angles for the given molecules are as follows:
$OF_2$: $103^\circ$
$H_2O$: $104.5^\circ$
$Cl_2O$: $111^\circ$
$ClO_2$: $117.5^\circ$
In $OF_2$,$H_2O$,and $Cl_2O$,the central oxygen atom is $sp^3$ hybridized with $2$ lone pairs. The bond angle decreases as the electronegativity of the surrounding atom increases because the bonding electron pairs are pulled closer to the central atom,increasing lone pair-lone pair repulsion.
$Cl_2O$ has a larger bond angle than $H_2O$ and $OF_2$ due to the steric repulsion between the bulky chlorine atoms.
$ClO_2$ has a different electronic structure (it is a radical with an odd electron on the central atom),which results in a larger bond angle compared to the others.
Thus,the increasing order of bond angles is: $OF_2 < H_2O < Cl_2O < ClO_2$.

(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 central $N$ atom is $sp$ hybridized with $0$ lone pairs. The geometry is linear, so the bond angle is $180^{\circ}$.
$2$. $NO_{2}$: The central $N$ atom is $sp^{2}$ hybridized with $1$ odd electron (radical). Due to the presence of the odd electron, the bond angle is slightly less than $120^{\circ}$ (approximately $134^{\circ}$ in some contexts, but generally treated as $sp^{2}$ with repulsion).
$3$. $NO^{-}_{2}$: The central $N$ atom is $sp^{2}$ hybridized with $1$ lone pair. The lone pair-bond pair repulsion decreases the bond angle to approximately $115^{\circ}$.
Comparing these, the order is $NO^{+}_{2} (180^{\circ}) > NO_{2} (> 120^{\circ}) > NO^{-}_{2} (\,115^{\circ})$.
Thus, the correct order is $NO^{+}_{2} > NO_{2} > NO^{-}_{2}$.

(D) In $PF_3$ and $PCl_3$,both have a central $P$ atom with one lone pair and three bond pairs.
According to $VSEPR$ theory,the bond angle depends on the electronegativity of the surrounding atoms.
As the electronegativity of the surrounding atom decreases $(Cl < F)$,the bond pairs move further away from the central atom,reducing the repulsion between them.
However,in $PCl_3$,the larger size of $Cl$ atoms causes more steric repulsion,which increases the bond angle compared to $PF_3$.
Thus,the bond angle order is $PCl_3 > PF_3$.

(C) According to the $VSEPR$ theory,the bond angle depends on the number of lone pairs $(n)$ and the geometry of the molecule.
For $n=0$ ($AX_2$,linear),the angle is $180^{\circ}$.
For $n=1$ ($AX_2L$,bent),the angle is $< 120^{\circ}$.
For $n=2$ ($AX_2L_2$,bent),the angle is $< 109.5^{\circ}$.
For $n=3$ ($AX_2L_3$,linear),the lone pairs occupy the equatorial positions of a trigonal bipyramidal geometry,resulting in an $X-A-X$ bond angle of $180^{\circ}$.
Thus,the bond angle is maximum for $n=3$ in this specific series.

(B) $1$. In $N(SiH_3)_3$,the lone pair on $N$ is donated into the empty $d$-orbitals of $Si$ ($p\pi-d\pi$ back bonding). This makes the $N$ atom $sp^2$ hybridized,resulting in a planar geometry.
$2$. In $Me_3N$ (trimethylamine),there is no back bonding. The $N$ atom is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
$3$. In $(SiH_3)_3P$,the central atom $P$ is larger than $N$. The energy gap between the $P$ lone pair and the empty $d$-orbitals of $Si$ is large,making back bonding insignificant. Thus,it retains a pyramidal geometry.

(D) In trisilyamine $(SiH_3)_3N$,the lone pair of electrons on the nitrogen atom is donated into the empty $d$-orbitals of the silicon atoms through $p\pi-d\pi$ back-bonding.
This delocalization of the lone pair makes the nitrogen atom $sp^2$ hybridized,resulting in a trigonal planar geometry.
Due to this back-bonding,the lone pair is not available for donation,making the molecule extremely weakly basic.

(D) The perxenate ion $(XeO_6^{4-})$ has a central $Xe$ atom surrounded by $6$ oxygen atoms.
Its hybridization is $sp^3d^2$,resulting in an octahedral geometry.
In an octahedral geometry,all $Xe-O$ bond lengths are equal,and the bond angles are $90^{\circ}$ and $180^{\circ}$.
Since the geometry is perfectly symmetrical (octahedral),the dipole moment is zero,making it a non-polar species.
For the $Xe-O$ bond order: The total number of bonds is $12$ (each $Xe-O$ is a single bond,plus $6$ additional pi-bonds distributed via resonance) across $6$ positions,giving a bond order of $12/6 = 2$ is incorrect; rather,considering the resonance of $6$ double bonds across $6$ positions,the bond order is $1.5$ or $2$ depending on the model,but the statement regarding bond angles is the most accurate description of its symmetry.
However,in the context of standard chemistry problems,the octahedral structure implies bond angles of $90^{\circ}$ and $180^{\circ}$,and the symmetry makes it non-polar. Option $D$ is the most chemically accurate statement regarding its high symmetry.

(A) In $BCl_3$,the boron atom is $sp^2$ hybridized with no lone pair,resulting in a trigonal planar geometry.
In $NCl_3$,the nitrogen atom is $sp^3$ hybridized with one lone pair of electrons,which causes a repulsion leading to a pyramidal geometry.