VSEPR Theory Questions in English

(C) $CO_2$ has a linear structure with a bond angle of $180^{\circ}$.
$HgCl_2$ also has a linear structure with a bond angle of $180^{\circ}$.
$SnCl_2$ and $SO_2$ are bent (angular) molecules due to the presence of a lone pair on the central atom.
Therefore,$CO_2$ is isostructural with $HgCl_2$.

(A) To be isoelectronic,species must have the same number of electrons.
For $NO_3^-$: $7 + (3 \times 8) + 1 = 32$ electrons.
For $CO_3^{2-}$: $6 + (3 \times 8) + 2 = 32$ electrons.
Both $NO_3^-$ and $CO_3^{2-}$ have $32$ electrons,so they are isoelectronic.
Both species have a central atom bonded to $3$ oxygen atoms with no lone pairs on the central atom,resulting in a trigonal planar geometry. Thus,they are isostructural.
Therefore,the correct pair is $NO_3^-$ and $CO_3^{2-}$.

(D) To determine the number of lone pairs on the central atom,we calculate the valence electrons and subtract the electrons involved in bonding:
$1$. In $[ClO_3]^-$,the central $Cl$ atom has $7$ valence electrons + $1$ (negative charge) = $8$. It forms $3$ double bonds with $O$ atoms ($6$ electrons used). Lone pairs = $(8-6)/2 = 1$.
$2$. In $XeF_4$,the central $Xe$ atom has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms ($4$ electrons used). Lone pairs = $(8-4)/2 = 2$.
$3$. In $SF_4$,the central $S$ atom has $6$ valence electrons. It forms $4$ single bonds with $F$ atoms ($4$ electrons used). Lone pairs = $(6-4)/2 = 1$.
$4$. In $[I_3]^-$,the central $I$ atom has $7$ valence electrons + $1$ (negative charge) = $8$. It forms $2$ single bonds with $I$ atoms ($2$ electrons used). Lone pairs = $(8-2)/2 = 3$.
Thus,$[I_3]^-$ has the maximum number of lone pairs $(3)$. The correct option is $(D)$.

(B) $BF_3$ has $sp^2$ hybridization and a trigonal planar geometry.
$ClF_3$ has $sp^3d$ hybridization with two lone pairs,resulting in a $T$-shaped (planar) geometry.
$NH_3$ has $sp^3$ hybridization with one lone pair,resulting in a trigonal pyramidal (non-planar) geometry.
Therefore,$NH_3$ is the non-planar molecule.

(C) The Lewis structure of $SO_2$ shows that the sulfur atom is bonded to two oxygen atoms by double bonds.
Each double bond consists of one $\sigma$ bond and one $\pi$ bond.
Therefore,there are $2$ $\sigma$ bonds and $2$ $\pi$ bonds.
Additionally,the sulfur atom has one lone pair of electrons.
Thus,the correct description is $2$ $\sigma$ bonds,$2$ $\pi$ bonds,and $1$ lone pair.

(D) $H_2O$ is not linear because the oxygen atom in $H_2O$ is $sp^3$ hybridized and possesses two lone pairs of electrons,resulting in a bent molecular geometry.

(D) In a water molecule $(H_2O)$,the central oxygen atom undergoes $sp^3$ hybridization.
According to the $VSEPR$ theory,the presence of two lone pairs on the oxygen atom causes repulsion,which compresses the bond angle from the ideal tetrahedral angle of $109^o 28'$ to approximately $104.5^o$ (often represented as $104^o 30'$).
Therefore,the correct option is $(d)$.

(C) $CO_2$ has $sp$ hybridization and a linear geometry with a bond angle of $180^{\circ}$. The central carbon atom is bonded to two oxygen atoms with no lone pairs,resulting in a linear shape.

(D) To determine the geometry,we calculate the number of electron pairs around the central atom using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $BF_3$:
Central atom $B$ has $3$ valence electrons.
Number of monovalent atoms $(F)$ = $3$.
$H = \frac{1}{2}(3 + 3) = 3$.
Since there are $3$ bond pairs and $0$ lone pairs,the hybridization is $sp^2$ and the geometry is trigonal planar.
For $IF_3$: $H = \frac{1}{2}(7 + 3) = 5$ ($sp^3d$,$T$-shaped).
For $PCl_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$,trigonal pyramidal).
For $NH_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$,trigonal pyramidal).
Therefore,the correct option is $D$.

(C) By looking at the Lewis structure of an ammonia $(NH_3)$ molecule,the shape is described as trigonal pyramidal.
The $N$ atom in ammonia is $sp^3$ hybridized,having one lone pair and three bond pairs of electrons.
This results in a tetrahedral electron pair geometry and a trigonal pyramidal molecular geometry.

(A) In a methane molecule $(CH_4)$,the central carbon atom is $sp^{3}$ hybridized.
According to the $VSEPR$ theory,the four bonding electron pairs arrange themselves in space to minimize repulsion,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.

(B) $H_2S$ has a bond angle of approximately $92^{\circ}$. This is because the central atom $S$ uses almost pure $p$-orbitals for bonding with $H$ atoms,as the electronegativity difference between $S$ and $H$ is very small,leading to minimal hybridization. In contrast,$NH_3$ $(107^{\circ})$,$H_2O$ $(104.5^{\circ})$,and $CH_4$ $(109.5^{\circ})$ exhibit $sp^3$ hybridization with significant bond angles due to the repulsion between electron pairs.

(B) $PF_5$ molecule has $5$ bond pairs and $0$ lone pairs around the central phosphorus atom.
This corresponds to $sp^3d$ hybridization.
Therefore,the geometry of the $PF_5$ molecule is trigonal bipyramidal.

(B) In $SO_3$,the central sulphur atom is $sp^2$ hybridized.
There are no lone pairs on the sulphur atom.
According to $VSEPR$ theory,the geometry of a molecule with $sp^2$ hybridization and no lone pairs is trigonal planar.

(D) $N_2O$ has $8 + 7 + 7 = 22$ electrons and a linear structure.
$CO_2$ has $6 + 8 + 8 = 22$ electrons and a linear structure.
Both are isoelectronic and isostructural.

(C) The structure of nitric acid $(HNO_3)$ is planar because the central nitrogen atom is $sp^2$ hybridized,resulting in a trigonal planar geometry around the nitrogen atom.
$H_2SO_4$ has a tetrahedral geometry around the sulfur atom.
$H_2O$ has a bent ($V$-shaped) geometry.
$CCl_4$ has a tetrahedral geometry.
Therefore,$HNO_3$ is the compound with planar symmetry.

(A) The geometry of a molecule is determined by the number of bonding pairs and lone pairs of electrons around the central atom.
$SnCl_2$ has a central $Sn$ atom with $2$ bond pairs and $1$ lone pair,resulting in a bent or $V$-shaped geometry.
$HCl$ is a diatomic molecule and is linear.
$CO_2$ has a central $C$ atom with $2$ double bonds and no lone pairs,resulting in a linear geometry $(O=C=O)$.
$HgCl_2$ has a central $Hg$ atom with $2$ bond pairs and no lone pairs,resulting in a linear geometry $(Cl-Hg-Cl)$.
Therefore,$SnCl_2$ is the only non-linear compound among the options.

(C) The hybridization of the central atom in $CCl_4$,$SiCl_4$,and $CF_4$ is $sp^3$,which results in a tetrahedral geometry.
In $SF_4$,the sulfur atom has $6$ valence electrons. It forms $4$ bonds with fluorine atoms and retains $1$ lone pair.
According to $VSEPR$ theory,the presence of $4$ bond pairs and $1$ lone pair results in $sp^3d$ hybridization and a 'see-saw' shape,not a tetrahedral shape.

(C) To determine the shape,we look at the hybridisation and the number of lone pairs on the central atom:
$1$. $NO_3^-$: The central $N$ atom has $sp^2$ hybridisation with $0$ lone pairs,resulting in a trigonal planar shape.
$2$. $H_2O$: The central $O$ atom has $sp^3$ hybridisation with $2$ lone pairs,resulting in a bent or $V$-shape.
$3$. $H_3O^{+}$: The central $O$ atom has $sp^3$ hybridisation with $1$ lone pair,resulting in a pyramidal shape.
$4$. $NH_4^+$: The central $N$ atom has $sp^3$ hybridisation with $0$ lone pairs,resulting in a tetrahedral shape.
Therefore,the correct answer is $H_3O^{+}$.

(B) The hybridization of the central atom in $NO_2^+$ is $sp$,which results in a linear geometry with a bond angle of $180^{\circ}$.
$SO_2$ has $sp^2$ hybridization and a bent shape.
$NO_2^-$ has $sp^2$ hybridization and a bent shape.
$SCl_2$ has $sp^3$ hybridization and a bent shape.

(A) . In group $15$ hydrides,the bond angle decreases down the group as the electronegativity of the central atom decreases. The bond angle order is $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ}) > BiH_3 (90^{\circ})$. Therefore,the bond angle in $PH_3$ is much less than in $NH_3$.

(B) The hybridization of the central atom in $NH_4^+$ is $sp^3$,which corresponds to a tetrahedral geometry.
$CO_3^{2-}$ has a trigonal planar geometry due to $sp^2$ hybridization.
$K_4[Fe(CN)_6]$ is an octahedral complex.
Therefore,$NH_4^+$ is the correct answer.

(B) The central boron atom in $BF_3$ undergoes $sp^2$ hybridization.
It has three bond pairs and zero lone pairs around the central atom.
According to $VSEPR$ theory,this arrangement results in a trigonal planar geometry with bond angles of $120^{\circ}$.

(A) The bond angle in hydrides of group $16$ elements decreases as we move down the group due to the increase in the size of the central atom and the decrease in the electronegativity of the central atom.
The bond angles are: $H_2O$ $(104.5^\circ)$,$H_2S$ $(92.1^\circ)$,$H_2Se$ $(91^\circ)$,and $H_2Te$ $(90^\circ)$.
Therefore,the minimum bond angle is in $H_2Te$.

(D) The hybridization and structure of the given species are as follows:
$1$. $NH_4^+$: Central atom $N$ has $4$ bond pairs and $0$ lone pairs. It is $sp^3$-hybridized and has a tetrahedral structure.
$2$. $SO_4^{2-}$: Central atom $S$ has $4$ bond pairs and $0$ lone pairs. It is $sp^3$-hybridized and has a tetrahedral structure.
Since both species have the same hybridization and geometry,they have the same structure. Therefore,the correct option is $(D)$.

(A) $IF_7$ exhibits $sp^3d^3$ hybridization with a pentagonal bipyramidal geometry.
In this structure,the equatorial bond angles are $72^o$,which is the smallest among the given molecules.
$CH_4$ has a bond angle of $109.5^o$,$BeF_2$ has $180^o$,and $BF_3$ has $120^o$.

(B) To determine the geometry,we look at the hybridization and lone pairs on the central atom:
$1$. $CO_2$: The central $C$ atom is $sp$ hybridized with $0$ lone pairs,resulting in a linear shape.
$2$. $ClO_2$: The central $Cl$ atom has $7$ valence electrons. It forms $2$ bonds with $O$ atoms and has $1$ lone pair and $1$ unpaired electron ($sp^3$ hybridization),resulting in a bent or $V$-shaped geometry.
$3$. $I_3^-$: The central $I$ atom has $3$ lone pairs and $2$ bond pairs ($sp^3d$ hybridization),resulting in a linear geometry.
Therefore,$ClO_2$ is not linear.

(A) To determine the geometry,we calculate the hybridization and number of lone pairs for each species:
$1$. $SCl_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $Cl$ and has $1$ lone pair. The steric number is $5$ ($sp^3d$ hybridization),resulting in a see-saw geometry.
$2$. $SO_4^{2-}$: Sulfur has $6$ valence electrons,plus $2$ from the charge. It forms $4$ bonds with $O$ atoms. The steric number is $4$ ($sp^3$ hybridization),resulting in a tetrahedral geometry.
$3$. $Ni(CO)_4$: Nickel is in the $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry.
$4$. $NiCl_4^{2-}$: Nickel is in the $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,so no pairing occurs. The hybridization is $sp^3$,resulting in a tetrahedral geometry.
Therefore,$SCl_4$ is the only species that is not tetrahedral.

(B) The central iodine atom in the $IF_7$ molecule undergoes $sp^3d^3$ hybridization.
There are $7$ bond pairs and $0$ lone pairs around the central iodine atom.
According to $VSEPR$ theory,this arrangement results in a pentagonal bipyramidal geometry.

(B) is the correct answer.
$SO_2$ molecule undergoes $sp^2$ hybridization and possesses a bent (angular) structure due to the presence of one lone pair on the sulfur atom.
$CO_2$,$BeCl_2$,and $C_2H_2$ all exhibit linear geometry.

(A) In $PCl_3$,the central $P$ 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 causes the geometry to be pyramidal,similar to $NH_3$.

(C) In a regular octahedral molecule,the central atom $M$ is surrounded by six $X$ atoms located at the corners of an octahedron.
These six positions consist of three pairs of atoms that are diametrically opposite to each other.
Each such pair forms a linear $X-M-X$ bond with a bond angle of $180^{\circ}$.
Therefore,there are $3$ such $X-M-X$ bonds at $180^{\circ}$.

(C) To determine the bond angles,we analyze the hybridization and molecular geometry of each molecule:
$1$. $H_2S$: The central atom $S$ has two bond pairs and two lone pairs. Due to the presence of lone pairs,the bond angle is significantly reduced from the tetrahedral angle,approximately $92.6^o$.
$2$. $NH_3$: The central atom $N$ has three bond pairs and one lone pair. The lone pair-bond pair repulsion reduces the bond angle to approximately $107^o$.
$3$. $SiH_4$: The central atom $Si$ is $sp^3$ hybridized with four bond pairs and no lone pairs,resulting in a perfect tetrahedral geometry with a bond angle of $109^o 28'$.
$4$. $BF_3$: The central atom $B$ is $sp^2$ hybridized with three bond pairs and no lone pairs,resulting in a trigonal planar geometry with a bond angle of $120^o$.
Thus,the correct order of bond angles (smallest first) is $H_2S < NH_3 < SiH_4 < BF_3$ $(92.6^o < 107^o < 109^o 28' < 120^o)$.

(A) The correct answer is $(A)$.
In the $PCl_3$ molecule,the central phosphorus atom is $sp^3$ hybridized.
It has three bond pairs and one lone pair of electrons.
Due to the presence of this lone pair,the molecule adopts a pyramidal geometry according to $VSEPR$ theory.
In contrast,$SO_3$,$CO_3^{2-}$,and $NO_3^-$ all have a trigonal planar geometry.

(A) Except $CO_3^{2-}$,the other choices $CO_2$,$CS_2$,and $BeCl_2$ have $sp$ hybridization and exhibit a linear structure.
$CO_3^{2-}$ has $sp^2$ hybridization and exhibits a trigonal planar structure,which is non-linear.

(A) The structures are as follows:
$NO_2^+$: $[O=N=O]^+$
$NCO^-$: $[N \equiv C-O]^-$
$CS_2$: $S=C=S$
In $CS_2$,the central carbon atom is $sp$-hybridised,resulting in a linear geometry.
$NCO^-$ and $NO_2^+$ are isoelectronic with $CS_2$ (each having $16$ valence electrons),and they also exhibit a linear geometry.

(C) The geometry of a molecule is determined by the number of bond pairs and lone pairs around the central atom.
$CH_4$ has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry.
$CCl_4$ has $4$ bond pairs and $0$ lone pairs,also resulting in a tetrahedral geometry.
Therefore,the pair having similar geometry is $CH_4$ and $CCl_4$.

(B) In $NH_3$,the central nitrogen atom has $5$ valence electrons.
It forms $3$ covalent bonds with $3$ hydrogen atoms,leaving one lone pair (non-bonding electrons) on the nitrogen atom.
$CH_4$ has $0$ lone pairs,$H_2O$ has $2$ lone pairs,and $HF$ has $3$ lone pairs.
Therefore,the correct option is $(B)$.

(A) In $PH_3$,the phosphorus atom is $sp^3$ hybridized,but the three $P-H$ bonds are formed by the overlap of $p$-orbitals of phosphorus with $s$-orbitals of hydrogen.
Due to the presence of a lone pair on the phosphorus atom and the lack of significant hybridization,the bond angle is very close to $90^o$.
The actual bond angle $H-P-H$ in $PH_3$ is approximately $93^o$.

(B) In the $BF_3$ molecule,the central Boron atom is $sp^2$ hybridized,resulting in a trigonal planar geometry.
Therefore,all atoms in $BF_3$ lie in the same plane.

(C) The bond angles for the given molecules are as follows:
$BeF_2$: $180^o$ ($sp$ hybridization,linear geometry).
$CH_4$: $109^o 28'$ ($sp^3$ hybridization,tetrahedral geometry).
$NH_3$: $107^o$ ($sp^3$ hybridization,one lone pair,pyramidal geometry).
$H_2O$: $104.5^o$ ($sp^3$ hybridization,two lone pairs,bent geometry).
Due to the presence of two lone pairs on the oxygen atom in $H_2O$,the $lp-lp$ repulsion is greater than the $lp-bp$ repulsion in $NH_3$,resulting in the smallest bond angle among the given options. Thus,the correct option is $(C)$.

(B) The central sulfur atom in $SO_4^{2-}$ has $6$ valence electrons.
It forms $4$ bonds with oxygen atoms.
The steric number is calculated as $\frac{1}{2}(V + M - C + A) = \frac{1}{2}(6 + 0 - 0 + 2) = 4$.
Since the steric number is $4$,the hybridization is $sp^3$.
Therefore,the shape of the $SO_4^{2-}$ ion is tetrahedral.

(B) The central atom in $NH_3$ is Nitrogen $(N)$,which has $5$ valence electrons. It forms $3$ covalent bonds with Hydrogen atoms,leaving $1$ lone pair of electrons on the Nitrogen atom.
In $H_2O$,Oxygen has $2$ lone pairs.
In $CH_4$,Carbon has $0$ lone pairs.
In $PCl_5$,Phosphorus has $0$ lone pairs.
Therefore,the correct option is $(B)$.

(C) The bond angle in these molecules depends on the electronegativity of the halogen atoms attached to the central carbon atom.
As the electronegativity of the halogen atom decreases,the bond pair electrons are pulled less towards the halogen,resulting in greater repulsion between the bond pairs around the central atom.
Electronegativity order: $F > Cl > Br$.
Therefore,the bond angle increases as the electronegativity of the halogen decreases.
The order of bond angles is: $CHF_3 < CHCl_3 < CHBr_3$.
Thus,$CHBr_3$ has the maximum bond angle.

(C) To determine the geometry,we calculate the hybridization of the central atom:
$1$. $NH_4^+$: $N$ has $5$ valence electrons. Steric number = $0.5 \times (5 + 4 - 1) = 4$ ($sp^3$ hybridization,tetrahedral).
$2$. $BF_4^-$: $B$ has $3$ valence electrons. Steric number = $0.5 \times (3 + 4 + 1) = 4$ ($sp^3$ hybridization,tetrahedral).
$3$. $XeF_4$: $Xe$ has $8$ valence electrons. Steric number = $0.5 \times (8 + 4) = 6$ ($sp^3d^2$ hybridization). With $4$ bond pairs and $2$ lone pairs,the geometry is square planar.
$4$. $SCl_4$: $S$ has $6$ valence electrons. Steric number = $0.5 \times (6 + 4) = 5$ ($sp^3d$ hybridization). With $4$ bond pairs and $1$ lone pair,the geometry is see-saw.
Thus,$XeF_4$ has a square planar structure.

(A) The bond angles for the given molecules are as follows:
$1$. $CO_2$: The molecule is linear with $sp$ hybridization,resulting in a bond angle of $180^\circ$.
$2$. $CH_4$: The molecule has tetrahedral geometry with $sp^3$ hybridization,resulting in a bond angle of $109.5^\circ$.
$3$. $NH_3$: The molecule has trigonal pyramidal geometry with $sp^3$ hybridization and one lone pair,resulting in a bond angle of $107^\circ$.
$4$. $H_2O$: The molecule has bent geometry with $sp^3$ hybridization and two lone pairs,resulting in a bond angle of $104.5^\circ$.
Therefore,the bond angle is greatest in $CO_2$.

(A) The $XeF_2$ molecule has a central $Xe$ atom with $8$ valence electrons.
It forms $2$ bonds with $F$ atoms and has $3$ lone pairs of electrons.
The total number of electron pairs is $5$,which corresponds to $sp^3d$ hybridization.
According to $VSEPR$ theory,the $3$ lone pairs occupy the equatorial positions to minimize repulsion,resulting in a linear geometry.

(B) The structure of $H_2O$ is bent (angular) due to the presence of two lone pairs on the central oxygen atom,which exert repulsion on the bonding pairs according to the $VSEPR$ theory.
Thus,the correct option is $B$.