VSEPR Theory Questions in English

(A) In $H_2S$,the central sulfur atom is $sp^3$ hybridized and possesses two lone pairs of electrons,resulting in a bent (angular) molecular geometry.
In contrast,$C_2H_2$ ($sp$ hybridization),$BeH_2$ ($sp$ hybridization),and $CO_2$ ($sp$ hybridization) all possess a linear arrangement of atoms due to their $sp$ hybridization and lack of lone pairs on the central atom.

(A) According to $VSEPR$ theory,the shape of a molecule is determined by the number of bond pairs and lone pairs around the central atom.
$BCl_3$: The central Boron atom has $3$ bond pairs and $0$ lone pairs,resulting in $sp^2$ hybridization and a trigonal planar geometry.
$NCl_3$: The central Nitrogen atom has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization and a pyramidal geometry due to the repulsion between the lone pair and bond pairs.

(B) According to $VSEPR$ theory,when the central atom has $4$ electron pairs in its valence shell,it adopts a $Tetrahedral$ geometry to minimize the repulsion between electron pairs.
Therefore,the correct shape is $Tetrahedral$.

(D) To determine the molecular shapes and lone pairs,we calculate the number of lone pairs on the central atom for each molecule:
$1$. For $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $(6 - 4) / 2 = 1$ lone pair. The shape is see-saw.
$2$. For $CF_4$: The central atom $C$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $(4 - 4) / 2 = 0$ lone pairs. The shape is tetrahedral.
$3$. For $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $(8 - 4) / 2 = 2$ lone pairs. The shape is square planar.
Since the shapes and the number of lone pairs are different for all three,option $D$ is correct.

(A) To determine the geometry,we calculate the hybridization of the central atom:
$1$. For $CO_3^{2-}$: The central carbon atom is $sp^2$ hybridized with no lone pairs,resulting in a trigonal planar geometry.
$2$. For $NH_2^-$: The nitrogen atom has $2$ bond pairs and $2$ lone pairs,resulting in $sp^3$ hybridization and a bent ($V$-shaped) geometry.
$3$. For $PCl_3$: The phosphorus atom has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization and a trigonal pyramidal geometry.
Therefore,only $CO_3^{2-}$ is planar.

(B) In $H_2S$,the central atom $S$ is in the $3rd$ period and has a larger atomic size compared to $O$ in $H_2O$.
Due to the larger size of $S$,the $S-H$ bonds are formed by almost pure $p$-orbitals with very little hybridization.
Consequently,the bond angle in $H_2S$ is approximately $92^\circ$,which is very close to $90^\circ$.

(A) In $H_2O$,the oxygen atom undergoes $sp^3$ hybridization,which would ideally result in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.
However,the presence of two lone pairs on the oxygen atom exerts strong repulsion on the two $O-H$ bond pairs.
According to the $VSEPR$ theory,the order of repulsion is $lp-lp > lp-bp > bp-bp$.
Due to the strong $lp-lp$ repulsion,the bond angle is compressed from the ideal $109.5^{\circ}$ to $104.5^{\circ}$.

(C) In $NH_3$,the central nitrogen atom has $5$ valence electrons. It forms $3$ covalent bonds with $3$ hydrogen atoms (bond pairs) and retains $1$ lone pair of electrons.
Thus,the nitrogen atom in $NH_3$ has $3$ bond pairs and $1$ lone pair.

(A) The bond angle depends on the electronegativity of the central atom and the size of the surrounding atoms.
As we move down the group $16$ from $O$ to $Te$,the electronegativity of the central atom decreases.
Consequently,the bond pairs move further away from the central atom,and the bond angle decreases.
Thus,the correct order is $H_2O (104.5^{\circ}) > H_2S (92.1^{\circ}) > H_2Se (91^{\circ}) > H_2Te (90^{\circ})$.

(C) In $NH_3$,$PH_3$,and $PCl_3$,the central atom has one lone pair of electrons.
In $BF_3$,the central boron atom has $3$ valence electrons,all of which are involved in bonding with $3$ fluorine atoms.
Therefore,$BF_3$ has no lone pair on the central atom.

(A) In $BrF_3$,the central atom $Br$ has $7$ valence electrons. It forms $3$ $Br-F$ bonds and has $2$ lone pairs,resulting in a total of $5$ electron pairs ($sp^3d$ hybridization).
According to $VSEPR$ theory,lone pairs prefer equatorial positions in a trigonal bipyramidal geometry to minimize lone pair-lone pair repulsion and lone pair-bond pair repulsion.
By placing both lone pairs in equatorial positions,the $90^{\circ}$ lone pair-lone pair interactions are avoided,which significantly reduces the overall repulsion in the molecule.

(D) The central atom in $NH_3$ is nitrogen $(N)$,which has $5$ valence electrons.
It forms $3$ covalent bonds with $3$ hydrogen atoms and retains $1$ lone pair of electrons.
According to $VSEPR$ theory,the presence of $1$ lone pair and $3$ bond pairs results in $sp^3$ hybridization with a distorted tetrahedral geometry.
Due to the lone pair-bond pair repulsion,the bond angle is reduced from $109.5^{\circ}$ to $107^{\circ}$,resulting in a pyramidal shape.

(B) The bond angle in hydrides of group $15$ elements decreases as the electronegativity of the central atom decreases down the group $(NH_3 > PH_3 > AsH_3)$.
For $H_2O$,the bond angle is approximately $104.5^\circ$ (or $105^\circ$),which is smaller than the bond angle in $NH_3$ $(107^\circ)$.
Therefore,$NH_3$ has the largest bond angle among the given options.

(C) In $NH_3$,$sp^3$ hybridization is present.
According to $VSEPR$ theory,the presence of one lone pair on the nitrogen atom causes $lp-bp$ repulsion.
This repulsion is greater than $bp-bp$ repulsion,which compresses the bond angle from the ideal tetrahedral angle of $109^o 28'$ to $106^o 45'$.

(B) $NH_3$ has a pyramidal structure because the nitrogen atom is $sp^3$ hybridized and contains one lone pair of electrons.
According to $VSEPR$ theory,the presence of one lone pair and three bond pairs results in a trigonal pyramidal geometry.

(C) $SiF_4$ has a symmetrical tetrahedral shape due to $sp^3$ hybridization of the central silicon atom.
$SF_4$ has a distorted tetrahedral or see-saw geometry due to $sp^3d$ hybridization of the central sulfur atom and the presence of one lone pair of electrons in an equatorial hybrid orbital.
Since $SiF_4$ is tetrahedral and $SF_4$ is see-saw,they are not isostructural.

(D) The bond angle depends on the electronegativity of the central atom and the size of the surrounding atoms.
In $NH_3$ and $PH_3$,the bond angles are $107^\circ$ and $93^\circ$ respectively.
In $H_2O$ and $H_2Se$,the bond angles are $104.5^\circ$ and $91^\circ$ respectively.
As the electronegativity of the central atom decreases down the group,the bond angle decreases.
Comparing all,$H_2Se$ has the smallest bond angle of $91^\circ$.

(C) To identify isostructural pairs,we determine the hybridization and geometry of each species:
$1$. $NF_3$: $N$ is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
$2$. $H_3O^+$: $O$ is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
$3$. $NO_3^-$: $N$ is $sp^2$ hybridized with no lone pairs,resulting in a trigonal planar geometry.
$4$. $BF_3$: $B$ is $sp^2$ hybridized with no lone pairs,resulting in a trigonal planar geometry.
$5$. $HN_3$: This is hydrazoic acid,which has a linear/bent structure depending on resonance,but it is not isostructural with the others.
Thus,the isostructural pairs are $[NF_3, H_3O^+]$ (both pyramidal) and $[NO_3^-, BF_3]$ (both trigonal planar).
Therefore,option $C$ is correct.

(D) $SF_4$ has a see-saw shape due to $1$ lone pair on the $S$ atom.
$CF_4$ has a tetrahedral geometry with $0$ lone pairs on the $C$ atom.
$XeF_4$ has a square planar structure with $2$ lone pairs on the $Xe$ atom.
Therefore,the molecular shapes are different,with $1, 0$ and $2$ lone pairs of electrons respectively.

(A) To determine the bond angle,we look at the hybridization and the number of lone pairs on the central atom:
$1$. $CO_2$: $sp$ hybridization,linear geometry,bond angle = $180^{\circ}$.
$2$. $CH_4$: $sp^3$ hybridization,tetrahedral geometry,bond angle = $109.5^{\circ}$.
$3$. $NH_3$: $sp^3$ hybridization,trigonal pyramidal geometry,one lone pair,bond angle = $107^{\circ}$.
$4$. $H_2O$: $sp^3$ hybridization,bent geometry,two lone pairs,bond angle = $104.5^{\circ}$.
Due to the presence of two lone pairs on the oxygen atom in $H_2O$,the lone pair-lone pair repulsion is greater than the lone pair-bond pair repulsion,which compresses the $H-O-H$ bond angle to $104.5^{\circ}$,the smallest among the given options.

(D) The $H-O-H$ bond angle in a water molecule is approximately $104.5^\circ$ (often rounded to $105^\circ$).
This deviation from the ideal tetrahedral angle of $109.5^\circ$ is due to the presence of two lone pairs of electrons on the oxygen atom,which exert greater repulsion on the bonding pairs according to the $VSEPR$ theory.

(A) Carbon suboxide $(C_3O_2)$ has a linear structure.
In this molecule, the central carbon atom is $sp$ hybridized, and the terminal carbon atoms are also $sp$ hybridized.
The structure is $O=C=C=C=O$.
The $C-C$ bond length is $130 \text{ pm}$ and the $C-O$ bond length is $120 \text{ pm}$.

(B) The correct order is $BCl_3 > PCl_3 > AsCl_3 > BiCl_3$.
$BCl_3$ has a trigonal planar geometry with a bond angle of $120^{\circ}$ due to $sp^2$ hybridization.
In the case of $PCl_3$,$AsCl_3$,and $BiCl_3$,the central atom is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
As we move down the group from $P$ to $Bi$,the electronegativity of the central atom decreases,causing the bond pairs to move further away from the central atom.
Consequently,the bond angle decreases as we move down the group.

(D) The correct option is $(D)$. As we move down the group $16$,the electronegativity of the central atom decreases. Consequently,the bond pairs are further away from the central atom,leading to reduced repulsion between them,which results in a decrease in the bond angle.
All these molecules are hydrides of group $16$ elements. Although the central atoms undergo $sp^3$ hybridization (theoretically expecting a bond angle of $109.5^{\circ}$),the actual bond angles are significantly smaller due to the repulsion between lone pairs and bond pairs. As the size of the central atom increases from $O$ to $Te$,the bond length increases and the repulsion between the bond pairs decreases,leading to a minimum bond angle in $H_2Te$.

$H_2O$	$104.5^{\circ}$
$H_2S$	$92.1^{\circ}$
$H_2Se$	$91^{\circ}$
$H_2Te$	$90^{\circ}$

(B) Both $O_2F_2$ and $H_2O_2$ have a non-planar,open-book structure.
In both molecules,the central oxygen atoms are bonded to each other with a single bond,and each oxygen atom is further bonded to a terminal atom ($F$ in $O_2F_2$ and $H$ in $H_2O_2$).
The dihedral angle in $O_2F_2$ is $87.5^{\circ}$,which is similar to the dihedral angle in $H_2O_2$ ($111.5^{\circ}$ in gas phase).
Therefore,the shape of $O_2F_2$ is similar to that of $H_2O_2$.

(D) The central atom $Xe$ has $8$ valence electrons.
In $XeF_2$,$Xe$ forms $2$ sigma bonds with $F$ atoms and has $3$ lone pairs.
The steric number is $2 + 3 = 5$,which corresponds to $sp^3d$ hybridization.
According to $VSEPR$ theory,the $3$ lone pairs occupy the equatorial positions of the trigonal bipyramidal geometry to minimize repulsion.
Therefore,the molecule adopts a linear shape.

(D) The correct answer is $(D)$.
$XeF_4$ has $sp^3d^2$ hybridization with two lone pairs and four bond pairs.
According to $VSEPR$ theory,the two lone pairs occupy the axial positions to minimize repulsion,while the four fluorine atoms occupy the equatorial positions,resulting in a square planar geometry.
$XeF_2$ is linear,$XeO_3F$ is tetrahedral,and $XeO_2F_2$ has a see-saw geometry.

(C) In $CCl_4$,the carbon atom is $sp^3$ hybridized.
According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,the four $C-Cl$ bonds are arranged in a tetrahedral geometry to minimize electron repulsion.
Therefore,the four chlorine atoms lie at the four corners of a regular tetrahedron.

(A) In the $NH_3$ molecule,the nitrogen atom is at the apex of a trigonal pyramid,and the three hydrogen atoms form the base.
Rotation of the molecule by $120^o$ $(360^o / 3)$ about the $C_3$ axis passing through the nitrogen atom and the center of the hydrogen triangle results in an identical configuration.
Therefore,$NH_3$ possesses a three-fold axis of symmetry ($C_3$ axis).

(A) Trimethylamine,$(CH_3)_3N$,has a pyramidal geometry.
In this molecule,the central nitrogen atom is $sp^3$ hybridized.
It contains three bond pairs ($N$-$C$ bonds) and one lone pair of electrons,which results in a trigonal pyramidal shape due to $VSEPR$ theory.

(D) $1$. $BF_{3}$ is a trigonal planar molecule with a dipole moment of $0$. This statement is correct.
$2$. $B_{2}H_{6}$ (diborane) is a classic example of an electron-deficient molecule due to $3c-2e^{-}$ bonds. This statement is correct.
$3$. Hydroboration-oxidation involves the syn-addition of $H$ and $OH$ across the double bond following anti-Markovnikov regioselectivity. This statement is correct.
$4$. $BF_{3}$ has a trigonal planar shape ($sp^{2}$ hybridization),whereas $BrF_{3}$ has a $T$-shaped geometry ($sp^{3}d$ hybridization with two lone pairs). Therefore,they have different shapes. The statement claiming they have the same shape is incorrect.

(D) In a water molecule $(H_2O)$,the oxygen atom is $sp^3$ hybridized. Due to the presence of two lone pairs of electrons on the oxygen atom,the bond angle is compressed from the ideal tetrahedral angle of $109.5^\circ$ to approximately $104.5^\circ$ (often cited as $104.45^\circ$ in precise measurements) due to lone pair-bond pair repulsion.

(A) In the $NH_3$ molecule,the principal axis of rotation is $120^o$.
Therefore,this axis is known as a three-fold axis of symmetry or a triad axis.

(A) The structure of carbon dioxide $(CO_2)$ is linear,with $sp$ hybridization at the central carbon atom.
$HgCl_2$ also exhibits a linear geometry with $sp$ hybridization at the mercury atom.
Therefore,$HgCl_2$ is isostructural with $CO_2$.

(B) The bond angles are as follows:
$CH_4$ has a bond angle of $109.5^o$.
$NH_3$ has a bond angle of $107^o$.
$H_2O$ has a bond angle of $104.5^o$.
$H_2S$ has a bond angle of approximately $92^o$.
As the electronegativity of the central atom decreases,the bond pair-bond pair repulsion decreases,leading to a smaller bond angle. Since $S$ is less electronegative than $O$,the bond angle in $H_2S$ is the smallest.

(B) In $NH_3$,the nitrogen atom undergoes $sp^3$ hybridization and possesses one lone pair of electrons.
Due to the presence of one lone pair and three bond pairs,the molecule adopts a trigonal pyramidal geometry.

(C) In $SO_4^{2-}$,the central sulfur atom undergoes $sp^3$ hybridization.
Since there are no lone pairs on the sulfur atom,the geometry is perfectly tetrahedral.

(A) In the corresponding compound $NH_3$,the bond angle is $107^o$.
However,in $PH_3$,the bond angle is approximately $90^o$.
This is because,despite having a similar electronic structure,the electronegativity of the central atom $(P)$ is lower than that of $N$. As a result,the bonding electrons are further from the central atom,reducing the $bp-bp$ repulsion and causing the bond angle to decrease towards $90^o$.

(D) In $XeF_4$,the central atom $Xe$ undergoes $sp^3d^2$ hybridization.
It has $4$ bond pairs and $2$ lone pairs of electrons.
According to $VSEPR$ theory,the presence of $2$ lone pairs results in a square planar geometry.

(C) In $H_3O^+$,the central oxygen atom undergoes $sp^3$ hybridization.
It has three bond pairs and one lone pair of electrons.
Due to the presence of one lone pair,the geometry is distorted from tetrahedral to trigonal pyramidal,similar to $NH_3$.

(C) In a water molecule $(H_2O)$,the oxygen atom undergoes $sp^3$ hybridization.
Due to the presence of two lone pairs on the oxygen atom,the bond angle is compressed from the ideal tetrahedral angle of $109.5^o$ to $104.5^o$ according to the $VSEPR$ theory.

(A) According to $VSEPR$ theory,the shape of a molecule depends on the number of bonding and non-bonding electron pairs around the central atom.
In $BCl_3$,the central $B$ atom has $3$ bonding pairs and $0$ lone pairs,resulting in $sp^2$ hybridization and a trigonal planar geometry.
In $NCl_3$,the central $N$ atom has $3$ bonding pairs and $1$ lone pair,resulting in $sp^3$ hybridization and a pyramidal geometry due to the lone pair-bond pair repulsion.

(B) In $ICl_2^-$,the central iodine atom undergoes $sp^3d$ hybridization.
It has $3$ lone pairs of electrons and $2$ bond pairs.
According to $VSEPR$ theory,the presence of $3$ lone pairs in the equatorial positions results in a linear molecular geometry.

(B) $BF_3$ has $sp^2$ hybridization with a bond angle of $120^o$.
$CH_4$ has $sp^3$ hybridization with a bond angle of $109.5^o$.
$NH_3$ has $sp^3$ hybridization with a bond angle of $107^o$ due to the presence of one lone pair.
$PF_3$ has $sp^3$ hybridization with a bond angle of approximately $96^o$.
Therefore,the bond angle is maximum in $BF_3$.

(C) In $XeF_4$,the central atom $Xe$ undergoes $sp^3d^2$ hybridization. It has $4$ bond pairs and $2$ lone pairs of electrons. According to $VSEPR$ theory,the two lone pairs occupy the axial positions of an octahedral geometry to minimize repulsion,resulting in a square planar molecular shape.

(B) In the $H_2O$ molecule,the central oxygen atom has $2$ bond pairs and $2$ lone pairs of electrons.
According to $VSEPR$ theory,the presence of $2$ lone pairs causes repulsion,which distorts the expected tetrahedral geometry into a bent or $V$-shaped structure,often described as a distorted tetrahedral geometry.

(C) In $SnCl_2$,the central atom $Sn$ undergoes $sp^2$ hybridization. Due to the presence of one lone pair of electrons on the $Sn$ atom,the molecular geometry is bent or angular.

(D) The Lewis structure of the nitrate ion $(NO_3^-)$ consists of one nitrogen atom bonded to three oxygen atoms.
There is one double bond $(N=O)$ and two single bonds $(N-O)$,resulting in a total of $4$ bond pairs.
The oxygen atoms have lone pairs: the double-bonded oxygen has $2$ lone pairs,and each of the two single-bonded oxygen atoms has $3$ lone pairs.
Total lone pairs = $2 + 3 + 3 = 8$.
Thus,there are $4$ bond pairs and $8$ lone pairs of electrons.

(A) In $ClO_3^-$,the central $Cl$ 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 distorts the geometry from tetrahedral to trigonal pyramidal.

(D) The bond angles for the given molecules are as follows:
$CO_2$: $180^\circ$ (linear geometry)
$CH_4$: $109.5^\circ$ (tetrahedral geometry)
$NH_3$: $107^\circ$ (pyramidal geometry,due to one lone pair)
$H_2O$: $104.5^\circ$ (bent geometry,due to two lone pairs)
Therefore,the correct order is $CO_2 > CH_4 > NH_3 > H_2O$.