VSEPR Theory Questions in English

(C) In the $OF_2$ molecule,the oxygen atom is bonded to two fluorine atoms by two single covalent bonds.
Thus,there are $2$ bond pairs.
Each fluorine atom has $3$ lone pairs ($6$ electrons) and the oxygen atom has $2$ lone pairs ($4$ electrons).
Total number of lone pair electrons = $(3 \times 2) + 4 = 6 + 4 = 10$ electrons.

(A) In $NH_3$,the nitrogen atom undergoes $sp^3$ hybridization. Due to the presence of one lone pair of electrons on the nitrogen atom,the molecular geometry is pyramidal.

(A) In $SF_2Cl_2$,the central sulfur atom has $5$ electron pairs ($4$ bond pairs and $1$ lone pair),resulting in $sp^3d$ hybridization. According to $VSEPR$ theory,the presence of one lone pair in a trigonal bipyramidal geometry leads to a seesaw molecular shape.

(C) The bond angles for the given molecules are as follows:
$NH_3$: $107^o$ (due to one lone pair)
$BeF_2$: $180^o$ (linear geometry)
$H_2O$: $104.5^o$ (due to two lone pairs)
$CH_4$: $109.5^o$ (tetrahedral geometry)
Comparing these values,$H_2O$ has the smallest bond angle of $104.5^o$.

(B) The structure of $O_2F_2$ is similar to that of $H_2O_2$. Both molecules exhibit a non-planar,open-book structure due to the presence of lone pairs on the oxygen atoms,resulting in a dihedral angle.

(D) All given molecules involve $sp^3$ hybridization. $NH_3$ and $PH_3$ have one lone pair,while $H_2O$ and $H_2S$ have two lone pairs.
As the number of lone pairs increases,the bond angle decreases due to increased lone pair-bond pair repulsion.
Comparing $H_2O$ and $H_2S$,the electronegativity of $O$ is higher than that of $S$.
In $H_2S$,the bond pair electrons are further from the central atom,leading to less bond pair-bond pair repulsion,resulting in a smaller bond angle of approximately $92^o$ compared to $104.5^o$ in $H_2O$.

(D) $NO_2^+$ has a linear geometry with a bond angle of $180^\circ$.
$NO_3^-$ has a trigonal planar geometry with a bond angle of $120^\circ$.
$NO_2$ is bent with a bond angle of approximately $134^\circ$.
$NO_2^-$ is bent with a bond angle of approximately $115^\circ$.
Therefore,$NO_2^+$ has the maximum bond angle.

(D) In the $BrF_3$ molecule,the central $Br$ atom has $2$ lone pairs and $3$ bond pairs,resulting in a $T$-shaped geometry.
According to $VSEPR$ theory,lone pairs occupy equatorial positions to minimize the repulsion between them and with the bond pairs.
The equatorial lone pairs are positioned to minimize both $lp-lp$ (lone pair-lone pair) and $lp-bp$ (lone pair-bond pair) repulsions.

(B) To determine if species are isostructural,we check their hybridization and geometry:
$1$. $PF_6^-$ ($sp^3d^2$,octahedral) and $SF_6$ ($sp^3d^2$,octahedral) are isostructural.
$2$. $SiF_4$ ($sp^3$,tetrahedral) and $SF_4$ ($sp^3d$,see-saw) are $NOT$ isostructural.
$3$. $IO_3^-$ ($sp^3$,pyramidal) and $XeO_3$ ($sp^3$,pyramidal) are isostructural.
$4$. $BH_4^-$ ($sp^3$,tetrahedral) and $NH_4^+$ ($sp^3$,tetrahedral) are isostructural.
Thus,the pair $SiF_4$ and $SF_4$ is not isostructural.

(A) $XeF_2$ has $sp^3d$ hybridization with $3$ lone pairs on the central atom,resulting in a linear geometry.
$IF_2^-$ also has $sp^3d$ hybridization with $3$ lone pairs on the central atom,resulting in a linear geometry.
Therefore,both $XeF_2$ and $IF_2^-$ have the same linear structure.

(C) The correct decreasing order of bond angles is $NO_2^+ > NO_2 > NO_2^-$.
$1$. $NO_2^+$ is a linear molecule with $sp$ hybridization and a bond angle of $180^{\circ}$. It has no lone pair on the central nitrogen atom.
$2$. $NO_2$ has one unpaired electron on the nitrogen atom,which exerts less repulsion than a lone pair,resulting in a bond angle of approximately $134^{\circ}$.
$3$. $NO_2^-$ has one lone pair on the nitrogen atom,which exerts strong repulsion on the bonding pairs,reducing the bond angle to approximately $115^{\circ}$.

(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-bond pair repulsion decreases.
Consequently,the bond angle decreases as we move down the group.
Thus,the correct order of decreasing bond angles is $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ})$.

(D) The $Xe$ atom has $8$ valence electrons in its outermost shell.
In $XeF_2$,$2$ electrons are used for bonding with $F$ atoms,leaving $6$ electrons which form $3$ lone pairs.
In $XeF_4$,$4$ electrons are used for bonding,leaving $4$ electrons which form $2$ lone pairs.
In $XeF_6$,$6$ electrons are used for bonding,leaving $2$ electrons which form $1$ lone pair.
Therefore,the number of lone pairs in $XeF_2$,$XeF_4$,and $XeF_6$ are $3, 2, 1$ respectively.

(C) To determine the bond angles,we analyze the hybridization and the number of lone pairs on the central atom:

Compound	Hybridization	Lone Pairs	Bond Angle
$H_2S$	$sp^3$	$2$	$\approx 92.2^o$
$NH_3$	$sp^3$	$1$	$\approx 107^o$
$SiH_4$	$sp^3$	$0$	$109.5^o$
$BF_3$	$sp^2$	$0$	$120^o$

As the number of lone pairs increases,the bond angle decreases due to increased lone pair-bond pair repulsion.
Therefore,the correct order is $H_2S < NH_3 < SiH_4 < BF_3$.

(A) In $PCl_3$,the central phosphorus atom undergoes $sp^3$ hybridization.
It has $3$ bond pairs and $1$ lone pair of electrons.
According to $VSEPR$ theory,the presence of one lone pair results in a pyramidal geometry.

(D) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$A) [ClO_3]^-$: Central atom $Cl$ $(V=7)$. $O$ is divalent $(M=0)$. Anionic charge $A=1$. $\text{Lone pairs} = \frac{1}{2} (7 - 0 - 0 + 1) = 4$ electrons,which corresponds to $1$ lone pair.
$B) XeF_4$: Central atom $Xe$ $(V=8)$. $F$ is monovalent $(M=4)$. $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 2$ lone pairs.
$C) SF_4$: Central atom $S$ $(V=6)$. $F$ is monovalent $(M=4)$. $\text{Lone pairs} = \frac{1}{2} (6 - 4) = 1$ lone pair.
$D) [I_3]^-$: Central atom $I$ $(V=7)$. Two $I$ atoms are attached $(M=2)$. Anionic charge $A=1$. $\text{Lone pairs} = \frac{1}{2} (7 - 2 + 1) = 3$ lone pairs.
Comparing the values,$[I_3]^-$ has the maximum number of lone pairs $(3)$.

(B) In ethane $(CH_3-CH_3)$,each carbon atom is $sp^3$ hybridized.
According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,an $sp^3$ hybridized carbon atom forms four sigma bonds directed towards the corners of a regular tetrahedron.
Therefore,the geometry around each carbon atom in ethane is tetrahedral.

(C) $1$. $CH_3^+$ has $sp^2$ hybridization with $0$ lone pairs,resulting in a trigonal planar shape.
$2$. $BF_3$ has $sp^2$ hybridization with $0$ lone pairs,resulting in a trigonal planar shape.
$3$. $NH_3$ has $sp^3$ hybridization with $1$ lone pair on the nitrogen atom,which causes a distortion in the geometry,resulting in a pyramidal shape.
$4$. $CH_3^-$ has $sp^3$ hybridization with $1$ lone pair on the carbon atom,resulting in a pyramidal shape.
Note: In many standard chemistry contexts,$NH_3$ is the classic example of a pyramidal molecule.

(A) The central atom $S$ in $H_2S$ has $6$ valence electrons. It forms $2$ bond pairs with $H$ atoms and has $2$ lone pairs of electrons.
According to $VSEPR$ theory,the presence of $2$ lone pairs causes the molecule to adopt a bent or angular geometry.
Due to the difference in electronegativity between $S$ and $H$ and the angular shape,the bond dipoles do not cancel each other out.
Therefore,$H_2S$ has a non-zero dipole moment.

(D) . $H_3O^{+}$ has $sp^3$ hybridization. The oxygen atom has one lone pair and three bond pairs,resulting in a triangular pyramidal shape due to the presence of the lone pair.

(D) The bond angles for the given species are as follows:
$NO_2^+$: The nitrogen atom is $sp$ hybridized,resulting in a linear geometry with a bond angle of $180^\circ$.
$NO_2$: The nitrogen atom is $sp^2$ hybridized with one unpaired electron,resulting in a bent geometry with a bond angle of approximately $134^\circ$.
$NO_2^-$: The nitrogen atom is $sp^2$ hybridized with one lone pair,resulting in a bent geometry with a bond angle of approximately $115^\circ$.
$NO_3^-$: The nitrogen atom is $sp^2$ hybridized,resulting in a trigonal planar geometry with a bond angle of $120^\circ$.
Comparing these values,the $O-N-O$ bond angle is maximum in $NO_2^+$.
Therefore,the correct option is $(D)$.

(D) To be isoelectronic,the total number of valence electrons must be the same.
For $IBr_{2}^{-}$: $7 (I) + 2 \times 7 (Br) + 1 (\text{charge}) = 22$ valence electrons.
For $XeF_{2}$: $8 (Xe) + 2 \times 7 (F) = 22$ valence electrons.
Both species have $22$ valence electrons,so they are isoelectronic.
Both $IBr_{2}^{-}$ and $XeF_{2}$ have $3$ lone pairs and $2$ bond pairs around the central atom,resulting in a linear geometry ($sp^3d$ hybridization),making them isostructural.

(D) According to $VSEPR$ theory,the bond angles are influenced by the number of lone pairs on the central atom.
$CH_4$ has $0$ lone pairs,$NH_3$ has $1$ lone pair,and $H_2O$ has $2$ lone pairs.
As the number of lone pairs increases,the repulsion between lone pairs and bond pairs increases,which decreases the bond angle.
The bond angles are:
$CH_4$: $109.5^{\circ}$ $(109^{\circ} 28')$
$NH_3$: $107^{\circ}$
$H_2O$: $104.5^{\circ}$
Comparing these values:
$104.5^{\circ} < 107^{\circ} < 109.5^{\circ}$.
Statement $A$ is true $(104.5^{\circ} < 107^{\circ})$.
Statement $B$ is true $(109.5^{\circ} > 107^{\circ})$.
Statement $C$ is true (all are $> 90^{\circ}$).
Statement $D$ is false because $104.5^{\circ}$ is not larger than $109.5^{\circ}$.

(C) According to the $VSEPR$ theory,the magnitude of repulsion between electron pairs follows the order: $lone \ pair - lone \ pair > lone \ pair - bond \ pair > bond \ pair - bond \ pair$.
This is because a lone pair is held by only one nucleus,whereas a bond pair is shared between two nuclei. Consequently,lone pairs occupy more space around the central atom,leading to greater repulsion.

(C) Two species are isostructural if they have the same hybridization and same geometry.
$A$. Diamond $(C)$ and Silicon carbide $(SiC)$ both have a tetrahedral structure ($sp^3$ hybridization).
$B$. $NH_3$ and $PH_3$ both have a trigonal pyramidal structure ($sp^3$ hybridization with one lone pair).
$C$. $XeF_4$ has $sp^3d^2$ hybridization and a square planar geometry,whereas $XeO_4$ has $sp^3$ hybridization and a tetrahedral geometry. Thus,they are not isostructural.
$D$. $SiCl_4$ and $PCl_{4}^{+}$ both have $sp^3$ hybridization and a tetrahedral geometry.

(B) Isoelectronic species have the same number of electrons,and isostructural species have the same geometry.
$1$. Calculate the number of electrons:
$SO_3^{2-}: 16 + (3 \times 8) + 2 = 42 \ e^-$
$ClO_3^-: 17 + (3 \times 8) + 1 = 42 \ e^-$
$CO_3^{2-}: 6 + (3 \times 8) + 2 = 32 \ e^-$
$NO_3^-: 7 + (3 \times 8) + 1 = 32 \ e^-$
$2$. Determine the structure:
$SO_3^{2-}$ and $ClO_3^-$ both have $sp^3$ hybridization with one lone pair on the central atom,resulting in a pyramidal geometry.
$CO_3^{2-}$ and $NO_3^-$ both have $sp^2$ hybridization with no lone pair on the central atom,resulting in a trigonal planar geometry.
Therefore,$ClO_3^-$ and $SO_3^{2-}$ are both isoelectronic $(42 \ e^-)$ and isostructural (pyramidal).

(A) The bond angles are determined by the hybridization and the presence of lone pairs or unpaired electrons on the central nitrogen atom:
$1.$ $NO_{2}^{+}$: Nitrogen is $sp$ hybridized,resulting in a linear geometry with a bond angle of $180^{\circ}$.
$2.$ $NO_{3}^{-}$: Nitrogen is $sp^{2}$ hybridized,resulting in a trigonal planar geometry with a bond angle of $120^{\circ}$.
$3.$ $NO_{2}$: Nitrogen is $sp^{2}$ hybridized with one unpaired electron,resulting in a bent shape with a bond angle of approximately $134^{\circ}$.
$4.$ $NO_{2}^{-}$: Nitrogen is $sp^{2}$ hybridized with one lone pair,resulting in a bent shape with a bond angle of approximately $115^{\circ}$.
Therefore,the maximum bond angle is in $NO_{2}^{+}$.

(B) Species with $sp^2$ hybridization exhibit a plane triangular shape.
In $NO_3^-$,the central $N$ atom is $sp^2$ hybridized and has no lone pair of electrons,resulting in a trigonal planar (plane triangular) geometry.
$N_3^-$ (azide ion),$NO_2^-$ (nitrite ion),and $CO_2$ (carbon dioxide) are $sp$ hybridized and possess a linear shape.

(D) To determine the isostructural species,we calculate the hybridization and geometry of the molecules:
$1$. $XeF_2$: The central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms and has $3$ lone pairs. The steric number is $2 + 3 = 5$,corresponding to $sp^3d$ hybridization. Due to $3$ lone pairs in the equatorial positions,the geometry is linear.
$2$. $ICl_2^-$: The central atom $I$ has $7$ valence electrons. Including the negative charge,it has $8$ electrons. It forms $2$ bonds with $Cl$ atoms and has $3$ lone pairs. The steric number is $2 + 3 = 5$,corresponding to $sp^3d$ hybridization. Like $XeF_2$,it has a linear geometry.
$3$. $SbCl_3$: The central atom $Sb$ has $5$ valence electrons. It forms $3$ bonds with $Cl$ atoms and has $1$ lone pair. The steric number is $3 + 1 = 4$,corresponding to $sp^3$ hybridization and a pyramidal geometry.
$4$. $TeF_2$: The central atom $Te$ has $6$ valence electrons. It forms $2$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $2 + 2 = 4$,corresponding to $sp^3$ hybridization and a $V$-shaped (bent) geometry.
Therefore,$XeF_2$ and $ICl_2^-$ are isostructural.

(D) To determine the number of bond pairs and lone pairs,we analyze the Lewis structure of each species:
$1$. $H_2O$: The central oxygen atom has $2$ bond pairs (with $H$) and $2$ lone pairs.
$2$. $BF_3$: The central boron atom has $3$ bond pairs (with $F$) and $0$ lone pairs.
$3$. $NH_2^-$: The central nitrogen atom has $2$ bond pairs (with $H$) and $2$ lone pairs.
$4$. $PCl_3$: The central phosphorus atom has $3$ bond pairs (with $Cl$) and $1$ lone pair.
Thus,$PCl_3$ is the species that contains three bond pairs and one lone pair around the central atom.

(D) As the number of lone pairs of electrons increases,the bond angle decreases.
$NO_2^+$ is isoelectronic with $CO_2$. It is a linear ion with $sp$-hybridization at the central $N$ atom,resulting in a bond angle of $180^{\circ}$.
In $NO_2^-$,the $N$ atom undergoes $sp^2$-hybridization. Due to the presence of one lone pair,the bond angle is reduced from the ideal $120^{\circ}$ to approximately $115^{\circ}$.
In $NO_2$,the $N$ atom has one unpaired electron in an $sp^2$-hybrid orbital. The bond angle is approximately $134^{\circ}$ (often cited as $132^{\circ}-134^{\circ}$),which is greater than $120^{\circ}$ due to the repulsion of the unpaired electron.
Therefore,the increasing order of bond angles is $NO_2^- < NO_2 < NO_2^+$ $(115^{\circ} < 134^{\circ} < 180^{\circ})$.

(C) To be isostructural,the species must have the same hybridization and the same geometry.
$(a)$ $SO_{3}^{2-}$ has $sp^{3}$ hybridization (pyramidal) and $NO_{3}^{-}$ has $sp^{2}$ hybridization (trigonal planar).
$(b)$ $BF_{3}$ has $sp^{2}$ hybridization (trigonal planar) and $NF_{3}$ has $sp^{3}$ hybridization (pyramidal).
$(c)$ $BrO_{3}^{-}$ has $sp^{3}$ hybridization with one lone pair (pyramidal) and $XeO_{3}$ has $sp^{3}$ hybridization with one lone pair (pyramidal). Thus,they are isostructural.
$(d)$ $SF_{4}$ has $sp^{3}d$ hybridization (see-saw) and $XeF_{4}$ has $sp^{3}d^{2}$ hybridization (square planar).

(D) Generally,$AB_5$ molecules have a trigonal bipyramidal structure due to $sp^3d$ hybridization.
For example,$PCl_5$ has a trigonal bipyramidal geometry.
Therefore,the statement that every $AB_5$ molecule has a square pyramidal structure is incorrect.

(D) $NO_2^+$ has a linear shape due to $sp$ hybridization of the $N$ atom.
$O_3$,$NO_2^-$,and $SO_2$ all have an angular (bent) shape due to the presence of a lone pair on the central atom.

(B) $SiCl_4$ has $sp^3$ hybridization and a tetrahedral geometry.
$NH_4^+$,$SO_4^{2-}$,and $PO_4^{3-}$ also exhibit $sp^3$ hybridization and a tetrahedral geometry,making them isostructural with $SiCl_4$.
$SCl_4$ has $sp^3d$ hybridization with one lone pair on the $S$ atom,resulting in a see-saw geometry.
Therefore,$SCl_4$ is not isostructural with $SiCl_4$.

(D) To determine the bond angle,we analyze the hybridization and the number of lone pairs on the central atom:
$1$. $ClO_2^-$: The central $Cl$ atom has $2$ lone pairs and $2$ bond pairs. The bond angle is approximately $111^\circ$.
$2$. $Cl_2O$: The central $O$ atom has $2$ lone pairs and $2$ bond pairs. Due to the lower electronegativity of $Cl$ compared to $O$,the bond pairs are further from the central atom,resulting in a smaller bond angle of approximately $110.9^\circ$.
$3$. $ClO_2$: The central $Cl$ atom has $1$ lone pair and $2$ bond pairs. With fewer lone pairs,there is less repulsion,leading to a larger bond angle of approximately $117.5^\circ$.
Thus,the correct order of increasing bond angles is $ClO_2^- < Cl_2O < ClO_2$.

(B) $BF_3$ has a trigonal planar structure with $sp^2$ hybridization,where all $B-F$ bonds are equivalent.
$AlF_3$ is an ionic compound with a crystal lattice structure where all $Al-F$ bonds are equivalent.
$NF_3$ has a trigonal pyramidal geometry with $sp^3$ hybridization,where all $N-F$ bonds are equivalent.
$ClF_3$ has a $T$-shaped geometry due to $sp^3d$ hybridization with two lone pairs in the equatorial positions. This results in two different bond lengths (axial and equatorial),meaning all bonds are not equal.
Therefore,the correct option is $B$.

(D) In $XeF_4$,the geometry is square planar with all $Xe-F$ bonds being equal due to symmetry.
In $BF_4^-$ and $SiF_4$,the geometry is tetrahedral,where all four bonds are equivalent.
In $SF_4$,the sulfur atom undergoes $sp^3d$ hybridization with one lone pair,resulting in a see-saw geometry.
Due to the presence of the lone pair,the axial and equatorial $S-F$ bonds have different bond lengths and bond angles,making them unequal.

(C) To determine if species are isostructural,we check their hybridization and geometry:
$1$. $CO_3^{2-}$ and $NO_3^-$: Both have $sp^2$ hybridization and are triangular planar.
$2$. $PCl_4^+$ and $SiCl_4$: Both have $sp^3$ hybridization and are tetrahedral.
$3$. $PF_5$: $sp^3d$ hybridization,trigonal bipyramidal geometry.
$BrF_5$: $sp^3d^2$ hybridization,square pyramidal geometry.
Since their geometries differ,they are not isostructural.
$4$. $AlF_6^{3-}$ and $SF_6$: Both have $sp^3d^2$ hybridization and are octahedral.
Therefore,the correct option is $C$.

(C) In the series $NCl_3, PCl_3, AsCl_3$,and $SbCl_3$,all central atoms belong to Group $15$ and possess one lone pair of electrons.
As we move down the group from $N$ to $Sb$,the electronegativity of the central atom decreases,and its atomic size increases.
According to $VSEPR$ theory,as the size of the central atom increases,the bond pairs are located further away from the central atom.
This leads to a reduction in the repulsion between the bond pairs,which results in a decrease in the bond angle.
Therefore,$SbCl_3$ has the smallest bond angle among the given molecules.

(A) The central atom is $Xe$ (Xenon),which has $8$ valence electrons.
In $XeO_2F_2$,$Xe$ forms $2$ double bonds with $O$ atoms ($4$ electrons used) and $2$ single bonds with $F$ atoms ($2$ electrons used).
Total electrons used in bonding = $4 + 2 = 6$.
Number of valence electrons remaining = $8 - 6 = 2$.
These $2$ electrons form $1$ lone pair on the $Xe$ atom.
Thus,the total number of lone pair$(s)$ on the central atom is $1$.

(D) The bond angle in $H_2O$ is approximately $104.5^o$ due to two lone pairs on the oxygen atom causing repulsion.
The bond angle in $O_3$ (ozone) is approximately $117^o$ due to the presence of one lone pair on the central oxygen atom.
The bond angle in $CO_2$ is $180^o$ because it has a linear geometry with $sp$ hybridization.
Therefore,the correct sequence is $H_2O$,$O_3$ and $CO_2$.

(D) $(i)$ According to Drago's rule,as the electronegativity of the central atom decreases,the bond angle decreases. Thus,$NH_3 > PH_3 > AsH_3$. The smallest bond angle is in $AsH_3$.
$(ii)$ In $O_3$,there is a lone pair on the central oxygen atom,whereas in $O_3^+$,there is a single electron. The lone pair-bond pair repulsion is greater than single electron-bond pair repulsion,leading to a smaller bond angle in $O_3$ compared to $O_3^+$. Thus,$O_3$ has the smallest bond angle.
$(iii)$ According to Bent's rule,more electronegative substituents prefer orbitals with more $p$-character. In $NO_2^-$,the $N-O$ bond has more $p$-character than in $O_3$. Consequently,the bond angle in $NO_2^-$ is smaller than in $O_3$.
$(iv)$ In $SOF_2$ and $SOCl_2$,the $S-F$ bond is more polar and has more $p$-character than the $S-Cl$ bond. According to Bent's rule,the bond angle decreases as the $p$-character of the bonding orbitals increases. Therefore,$SOF_2$ has a smaller $X-S-X$ bond angle than $SOCl_2$.

(C) The geometry of $PCl_5$ (phosphorus pentachloride) is trigonal bipyramidal.
In this molecule,the central $P$ atom and the three equatorial $Cl$ atoms lie in the same plane.
Therefore,the maximum number of atoms in the same plane is $4$ ($1$ $P$ atom and $3$ $Cl$ atoms).

(D) To determine the geometry of the molecules,we use the $VSEPR$ theory:
$1$. $CO_2$ $(O=C=O)$ has $2$ bond pairs and $0$ lone pairs on the central atom,resulting in a linear geometry.
$2$. $HCN$ $(H-C \equiv N)$ has $2$ bond pairs and $0$ lone pairs on the central carbon atom,resulting in a linear geometry.
$3$. $C_2H_2$ $(H-C \equiv C-H)$ has a linear geometry due to $sp$ hybridization of carbon atoms.
$4$. $H_2O$ $(H-O-H)$ has $2$ bond pairs and $2$ lone pairs on the central oxygen atom. Due to the presence of lone pairs,the geometry is bent or angular (non-linear).

(C) The presence of a lone pair in $NH_3$ affects both its geometry and bond angle.
According to $VSEPR$ theory,the repulsion between a lone pair and a bond pair is greater than the repulsion between two bond pairs.
This lone pair-bond pair repulsion causes the $H-N-H$ bond angle to decrease from the ideal tetrahedral angle of $109.5^{\circ}$ to $107^{\circ}$.
Consequently,the molecular shape changes from tetrahedral to trigonal pyramidal.

(A) In a regular octahedral geometry,the central atom $M$ is surrounded by $6$ ligands $X$ arranged at the corners of an octahedron.
These $6$ ligands form $3$ linear axes passing through the central atom.
Each axis consists of two $X$ atoms positioned at $180^{\circ}$ to each other with respect to the central atom $M$.
Therefore,there are $3$ such $X-M-X$ bond angles of $180^{\circ}$.

(D) The bond angle in hydrides of group $15$ and $16$ elements decreases down the group due to the decrease in electronegativity of the central atom and the increase in the size of the central atom.
For group $15$ hydrides $(NH_3, PH_3, AsH_3, SbH_3)$,the bond angle decreases as: $NH_3 (107^\circ) > PH_3 (93.6^\circ) > AsH_3 (91.8^\circ) > SbH_3 (91.3^\circ)$.
For group $16$ hydrides $(H_2O, H_2S, H_2Se, H_2Te)$,the bond angle decreases as: $H_2O (104.5^\circ) > H_2S (92.1^\circ)$.
Comparing all given options,$SbH_3$ has the lowest bond angle of approximately $91.3^\circ$ because $Sb$ is the largest and least electronegative atom among the central atoms listed,leading to maximum $p$-orbital character in the bonding and minimum bond angle.

(A) The central atom $Br$ has $7$ valence electrons.
In $BrF_3$,it forms $3$ single bonds with $F$ atoms and has $2$ lone pairs of electrons.
According to the $VSEPR$ theory,the total number of electron pairs is $3 + 2 = 5$,which corresponds to a trigonal bipyramidal electron geometry.
Due to the presence of $2$ lone pairs at the equatorial positions to minimize repulsion,the molecular shape of $BrF_3$ is a bent $T$-shape.