Hybridisation Questions in English

(C) In $sp^{2}$-hybridisation,one $s$ and two $p$ orbitals mix to form three equivalent $sp^{2}$ hybrid orbitals.
These three $sp^{2}$ hybrid orbitals form $3$ $\sigma$-bonds.
The remaining unhybridised $p$ orbital forms $1$ $\pi$-bond.
Therefore,a carbon atom with $sp^{2}$-hybridisation forms $3$ $\sigma$-bonds and $1$ $\pi$-bond.

(D) In $H_3O^+$,the oxygen atom is bonded to three hydrogen atoms and has one lone pair of electrons.
Using the formula for steric number: $Steric \ Number = \frac{1}{2} [V + M - C + A] = \frac{1}{2} [6 + 3 - 1 + 0] = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridization.
The geometry is pyramidal with a bond angle of approximately $107^\circ$ due to the presence of one lone pair,which causes repulsion.
Therefore,the orbitals used by oxygen are $sp^3$ hybrid orbitals.

(D) In the allene molecule $(CH_2=C=CH_2)$:
$1$. The central carbon atom is bonded to two other carbon atoms by double bonds,making it $sp$ hybridized.
$2$. The two terminal carbon atoms are each bonded to two hydrogen atoms and one central carbon atom by a double bond,making them $sp^2$ hybridized.
$3$. To form two separate $\pi-$ bonds,the $p-$ orbitals on the central carbon must be perpendicular to each other. Consequently,the two $CH_2$ groups at the ends must lie in mutually perpendicular planes.
Therefore,all the given statements are correct.

(B) The given order is $sp < sp^2 < sp^3$.
$(i)$ Electronegativity: It is directly proportional to $s$-character. Since $sp$ $(50\% \ s)$ $> sp^2$ $(33.3\% \ s)$ $> sp^3$ $(25\% \ s)$,the order is $sp > sp^2 > sp^3$.
$(ii)$ Bond angle: $sp$ $(180^{\circ})$ $> sp^2$ $(120^{\circ})$ $> sp^3$ $(109.5^{\circ})$.
$(iii)$ Size: Higher $s$-character leads to more compact orbitals,hence smaller size. Thus,$sp < sp^2 < sp^3$ is the correct order.
$(iv)$ Energy level: $s$-orbitals are lower in energy than $p$-orbitals. Higher $p$-character leads to higher energy. Thus,$sp < sp^2 < sp^3$ is the correct order.
Therefore,$(iii)$ and $(iv)$ follow the given order.

(A) The $s$-character in hybrid orbitals is related to the hybridization as follows: $sp$ $(50\% \ s)$,$sp^2$ $(33.3\% \ s)$,and $sp^3$ $(25\% \ s)$.
$(I)$ As $s$-character decreases,the bond angle decreases (e.g.,$180^{\circ}$ $\rightarrow 120^{\circ}$ $\rightarrow 109.5^{\circ}$),so $(I)$ is correct.
$(II)$ Bond strength is directly proportional to $s$-character. As $s$-character decreases,bond strength decreases,so $(II)$ is incorrect.
$(III)$ Bond length is inversely proportional to $s$-character. As $s$-character decreases,bond length increases,so $(III)$ is correct.
$(IV)$ Orbitals with higher $p$-character (lower $s$-character) are larger in size because $p$-orbitals are more diffuse than $s$-orbitals. Thus,as $s$-character decreases,the size of the orbital increases,so $(IV)$ is correct.
Therefore,statements $(I)$,$(III)$,and $(IV)$ are correct.

(C) For $sp^3d^3$ hybridization,the steric number is $7$,which requires $7$ orbitals $(s + 3p + 3d)$.
Option $C$ lists only $6$ orbitals $(s + P_x + P_y + P_z + d_{x^2-y^2} + d_{z^2})$,which corresponds to $sp^3d^2$ hybridization,not $sp^3d^3$.
Therefore,option $C$ is incorrectly matched.

(B) The involvement of $d$-orbitals in hybridization depends on the geometry of the molecule:
$(a)$ Pentagonal planar geometry involves $sp^3d^3$ hybridization,which utilizes the $d_{z^2}$ orbital.
$(b)$ Trigonal planar geometry involves $sp^2$ hybridization,which uses only $s$ and two $p$ orbitals $(p_x, p_y)$. Thus,the $d_{z^2}$ orbital is not involved.
$(c)$ Linear geometry can involve $sp$ $(s+p_z)$,$sp^3d$ $(s+p_x+p_y+p_z+d_{z^2})$,or $sp^3d^2$ hybridization depending on the central atom's coordination.
$(d)$ Square planar geometry typically involves $dsp^2$ or $sp^3d^2$ hybridization,where the $d_{x^2-y^2}$ or $d_{z^2}$ orbitals are involved.
Therefore,the $d_{z^2}$ orbital is not required for the $sp^2$ hybridization of a trigonal planar molecule.

(C) $I$. $SF_4$ has $sp^3d$ hybridisation and $XeF_4$ has $sp^3d^2$ hybridisation.
$II$. $I_3^-$ has $sp^3d$ hybridisation and $XeF_2$ has $sp^3d$ hybridisation.
$III$. $ICl_4^+$ has $sp^3d$ hybridisation and $SiCl_4$ has $sp^3$ hybridisation.
$IV$. $ClO_3^-$ has $sp^3$ hybridisation and $PO_4^{3-}$ has $sp^3$ hybridisation.
Thus,pairs $II$ and $IV$ have the same hybridisation.

(C) The bond length of a $C-C$ single bond depends on the hybridization of the carbon atoms involved. The bond length decreases as the $s$-character of the hybrid orbitals increases.
$sp^3-sp^3$ bond length $\approx 1.54 \ \mathring{A}$
$sp^3-sp^2$ bond length $\approx 1.50 \ \mathring{A}$
$sp^3-sp$ bond length $\approx 1.46 \ \mathring{A}$
$sp^2-sp^2$ bond length $\approx 1.48 \ \mathring{A}$
$sp^2-sp$ bond length $\approx 1.43 \ \mathring{A}$
$sp-sp$ bond length $\approx 1.37 \ \mathring{A}$
Analyzing the options:
$(a)$ $CH_2=CH-C \equiv CH$: $sp^2-sp$ bond.
$(b)$ $HC \equiv C-C \equiv CH$: $sp-sp$ bond.
$(c)$ $CH_3-CH=CH_2$: $sp^3-sp^2$ bond.
$(d)$ $CH_2=CH-CH=CH_2$: $sp^2-sp^2$ bond.
Comparing the $s$-character:
$(c)$ involves $sp^3-sp^2$ ($25$% $s$ and $33.3$% $s$),which has the lowest total $s$-character among the options,resulting in the longest bond length.

(C) If we assume no hybridization,carbon in its excited state has one $2s$ orbital and three $2p$ orbitals $(2p_x, 2p_y, 2p_z)$.
Three $C-H$ bonds are formed by the overlap of three $2p$ orbitals of carbon with the $1s$ orbitals of three hydrogen atoms. These three bonds are at $90^{\circ}$ to each other.
The fourth $C-H$ bond is formed by the overlap of the $2s$ orbital of carbon with the $1s$ orbital of the fourth hydrogen atom.
Since the $2s$ orbital is spherical (non-directional),this bond will have different characteristics compared to the other three bonds.
Because the bond angles are $90^{\circ}$ and the bond types are different,the molecule cannot have a tetrahedral geometry.
Therefore,the statement that the shape of the molecule will be tetrahedral is false.

(D) The structure of $C_2H_2$ (acetylene) is linear,with a bond angle of $180^{\circ}$ and $sp$ hybridization of carbon atoms.
To be isostructural,the molecule must have the same geometry and hybridization.
$CO_2$ has the structure $O=C=O$,which is linear with $sp$ hybridization at the central carbon atom.
Therefore,$C_2H_2$ and $CO_2$ are isostructural.

(B) The bond angle of $120^o$ corresponds to $sp^2$ hybridization.
In $sp^2$ hybridization,one $s$ orbital and two $p$ orbitals are mixed.
The total number of orbitals involved is $3$.
The fraction of $s-$ character is $\frac{1}{3}$.
The percentage of $s-$ character is $\frac{1}{3} \times 100 \approx 33.33 \%$,which is nearly $33 \%$.

(B) In the dimer of $BH_3$,which is diborane $(B_2H_6)$,each Boron atom is bonded to four hydrogen atoms (two terminal and two bridging). Thus,the Boron atom undergoes $sp^3$ hybridisation.
In the dimer of $BeH_2$ $(Be_2H_4)$,each Beryllium atom is bonded to three hydrogen atoms (one terminal and two bridging). Thus,the Beryllium atom undergoes $sp^2$ hybridisation.
Therefore,the hybridisation states are $sp^3$ for $B$ in $B_2H_6$ and $sp^2$ for $Be$ in $Be_2H_4$.

(B) $NO_2$ involves $sp^2$ hybridization.
To determine the hybridization of the $N$ atom in $NO_2$,we look at the Lewis structure.
The $N$ atom is bonded to two $O$ atoms via one sigma bond each and possesses one odd electron (radical) on the nitrogen atom.
The total number of electron domains around the $N$ atom is calculated as: $(\text{Number of sigma bonds}) + (\text{Number of lone pairs}) + (\text{Number of odd electrons}) = 2 + 0 + 1 = 3$.
Since the sum is $3$,the hybridization of the $N$ atom is $sp^2$.

(C) The central atom $M$ is Tellurium ($Te$,atomic number $52$).
In $MX_4$ (e.g.,$TeCl_4$),the hybridization of $Te$ is $sp^3d$,which involves the $d_{z^2}$ orbital. The structure is a see-saw shape with one lone pair.
In $MX_4X'_2$ (e.g.,$TeCl_4F_2$),the hybridization of $Te$ is $sp^3d^2$,which involves $d_{x^2-y^2}$ and $d_{z^2}$ orbitals.
The axial positions in the octahedral geometry of $MX_4X'_2$ are formed by hybrid orbitals $(sp^3d^2)$.
Since $X'$ is more electronegative,it occupies the axial positions to minimize repulsion.
In $sp^3d^2$ hybridization,the $d-$ orbitals used are $d_{x^2-y^2}$ and $d_{z^2}$. Both are axial $d-$ orbitals. Therefore,the central atom does not use any non-axial $d-$ orbitals (like $d_{xy}, d_{yz}, d_{zx}$) in the hybridization of the final product.
Thus,statement $C$ is correct.

(C) The $d$-orbital having $xz$ and $yz$ as nodal planes is the $d_{xy}$ orbital.
$1$. $IO_2F_2^-$: Hybridization is $sp^3d$,involving $d_{z^2}$ orbital.
$2$. $ClF_4^-$: Hybridization is $sp^3d^2$,involving $d_{x^2-y^2}$ and $d_{z^2}$ orbitals.
$3$. $IF_7$: Hybridization is $sp^3d^3$,involving $d_{x^2-y^2}$,$d_{xy}$,and $d_{z^2}$ orbitals.
Since $IF_7$ involves the $d_{xy}$ orbital in its hybridization,it is the correct species.

(D) According to Bent's Rule,more electronegative substituents prefer to be attached to a hybrid orbital that has more $p-$character.
As the number of fluorine atoms increases on the carbon atoms,the $p-$character in the $C-C$ bond increases.
Greater $p-$character in the hybrid orbitals forming the $C-C$ bond leads to a longer bond length.
In $CH_3-CF_3$,the $C-C$ bond is formed by $sp^3-sp^3$ hybridization.
In $CF_3-CF_3$,the $C-C$ bond has the highest $p-$character due to the presence of the most electronegative fluorine atoms,resulting in the longest $C-C$ bond length.

(B) Back bonding occurs when a lone pair from one atom is donated into a vacant orbital of an adjacent atom. This often leads to a change in the hybridisation of the central atom to maximize stability.
$(i)$ $CCl_3^-$: The carbon atom has a lone pair,but chlorine does not have a vacant orbital available for back bonding. Thus,hybridisation remains $sp^3$.
$(ii)$ $CCl_2$: This is a carbene where carbon has a lone pair and a vacant $p$-orbital. Back bonding from $Cl$ to $C$ occurs,but the hybridisation of $C$ remains $sp^2$.
$(iii)$ $(SiH_3)_2O$: Oxygen has lone pairs and silicon has vacant $d$-orbitals. Back bonding occurs,changing the geometry,but the hybridisation of $O$ is often considered $sp^3$ in the context of bond angle changes.
$(iv)$ $N(SiH_3)_3$: Nitrogen has a lone pair and silicon has vacant $d$-orbitals. Back bonding occurs,changing the hybridisation of $N$ from $sp^3$ to $sp^2$.
Therefore,for $(i)$ and $(iii)$,the hybridisation does not change significantly or is not driven by the back bonding mechanism in the same way as $(iv)$.

(D) $1$. The molecule $BF_2NH_2$ exhibits back bonding from $N$ to $B$,which gives the $B-N$ bond partial double bond character.
$2$. Both boron and nitrogen atoms are $sp^2$ hybridized to maintain planarity and facilitate back bonding.
$3$. The $FBF$ bond angle is less than $120^{\circ}$ due to the presence of the $B=N$ double bond character,which exerts more repulsion than single bonds.
$4$. The $HNH$ bond angle is greater than $109^{\circ}28'$ because the nitrogen atom is $sp^2$ hybridized (ideal angle $120^{\circ}$).
$5$. Since the nitrogen atom is $sp^2$ hybridized,option $D$ is incorrect.
$6$. The molecule contains $N-H$ bonds,which allows for intermolecular hydrogen bonding.

(D) Let us analyze the hybridization of the central atom in each reaction:
$(i)$ $PCl_3$ $(sp^3)$ $\to$ $H_3PO_3$ $(sp^3)$. The hybridization remains unchanged.
$(ii)$ $SF_4$ $(sp^3d)$ $\to$ $H_2SO_3$ $(sp^3)$. The hybridization changes from $sp^3d$ to $sp^3$.
$(iii)$ $BCl_3$ $(sp^2)$ $\to$ $H_3BO_3$ $(sp^2)$. The hybridization remains unchanged.
$(iv)$ $XeF_6$ $(sp^3d^3)$ $\to$ $XeO_3$ $(sp^3)$. The hybridization changes from $sp^3d^3$ to $sp^3$.
Therefore,the statement in option $(D)$ is incorrect because the hybridization of $Xe$ changes from $sp^3d^3$ to $sp^3$.

(B) The reaction is $Cl_2O_6 + HF \to [ClO_2]^+[ClO_4]^- + HF \to [ClO_2]^+ + [ClO_4]^-$.
Given that $H$ of $HF$ attaches to $Q$,$Q$ must be the $[ClO_4]^-$ ion,making $P$ the $[ClO_2]^+$ ion.
For $P$ $([ClO_2]^+)$: The $Cl$ atom is $sp^2$ hybridized with one lone pair,resulting in a bent shape with $\angle OClO < 120^\circ$.
For $Q$ $([ClO_4]^-)$: The $Cl$ atom is $sp^3$ hybridized with a tetrahedral geometry,resulting in $\angle OClO = 109^\circ 28'$.
Comparing with the given options,$B$ is the correct statement regarding $Q$.

(C) In $SO_2$,the sulfur atom is bonded to two oxygen atoms and has one lone pair of electrons.
Using the formula for hybridization: $H = \frac{1}{2} (V + M - C + A)$,where $V = 6$ (valence electrons of $S$),$M = 0$ (monovalent atoms),$C = 0$,and $A = 0$.
$H = \frac{1}{2} (6 + 0) = 3$.
$A$ value of $3$ corresponds to $sp^2$ hybridization.

(C) In $SF_3Cl_3$,the central sulfur atom undergoes $sp^3d^2$ hybridization.
Since there are no lone pairs on the central atom,the molecule adopts an octahedral geometry.

(B) The central atom $S$ in $SF_4$ has $6$ valence electrons.
It forms $4$ bond pairs with $F$ atoms and has $1$ lone pair.
Total electron pairs = $4 + 1 = 5$.
Therefore,the hybridization is $sp^3d$.

(A) In $SO_3$,the central sulfur atom is bonded to three oxygen atoms with no lone pairs.
Using the formula: $\text{Hybridization} = \frac{1}{2} (V + M - C + A)$,where $V = 6$ (valence electrons of $S$),$M = 0$ (monovalent atoms),$C = 0$,and $A = 0$.
$\text{Hybridization} = \frac{1}{2} (6 + 0) = 3$,which corresponds to $sp^2$ hybridization.

(B) In $SO_2$,the sulfur atom is bonded to two oxygen atoms and has one lone pair of electrons. The steric number is $2 + 1 = 3$,which corresponds to $sp^2$ hybridization.

(D) The central iodine atom in $ICl_2^-$ has $7$ valence electrons.
It forms $2$ single bonds with chlorine atoms and has $3$ lone pairs of electrons.
Total electron pairs = $2 \text{ (bond pairs)} + 3 \text{ (lone pairs)} = 5$.
For $5$ electron pairs,the hybridization is $sp^3d$.

(C) The central atom $Br$ in $BrF_5$ has $7$ valence electrons.
It forms $5$ bonds with $F$ atoms and has $1$ lone pair.
Total electron pairs = $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization.
Due to the presence of one lone pair,the geometry is octahedral,but the molecular shape is square pyramidal.

(B) The hybridization of the central atom is calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $I_3^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.
For $ICl_2^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.
For $ICl_4^-$: $H = \frac{1}{2}(7 + 4 + 1) = 6$,which corresponds to $sp^3d^2$ hybridization.
Thus,the correct sequence is $sp^3d, sp^3d, sp^3d^2$.

(C) The hybridization of the central atom is calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $ClO_4^-$: $H = \frac{1}{2}(7 + 0 - 0 + 1) = 4$ ($sp^3$ hybridization).
For $ClO_3^-$: $H = \frac{1}{2}(7 + 0 - 0 + 1) = 4$ ($sp^3$ hybridization).
For $ClO_2^-$: $H = \frac{1}{2}(7 + 0 - 0 + 1) = 4$ ($sp^3$ hybridization).
Thus,the chlorine atom in all these ions is $sp^3$ hybridized.

(A) In $ICl_7$,the central iodine atom is bonded to $7$ chlorine atoms.
Using the formula for steric number: $Steric \ Number = \frac{1}{2} [V + M - C + A]$,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 $ICl_7$: $V = 7$ (iodine),$M = 7$ (chlorine atoms),$C = 0$,$A = 0$.
$Steric \ Number = \frac{1}{2} [7 + 7] = 7$.
$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridization,which results in a pentagonal bipyramidal geometry.

(A) In the chlorite ion $(ClO_2^-)$,the central chlorine atom is bonded to two oxygen atoms and has two lone pairs of electrons.
Using the formula for steric number: $Steric \ Number = \frac{1}{2} \times (V + M - C + A)$,where $V = 7$ (valence electrons of $Cl$),$M = 0$,$C = 0$,$A = 1$ (charge on ion).
$Steric \ Number = \frac{1}{2} \times (7 + 0 - 0 + 1) = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridization.

(C) The hybridization of $Xe$ in $XeF_2$ is calculated 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 $XeF_2$,$V = 8$ (valence electrons of $Xe$),$M = 2$ (two $F$ atoms).
$H = \frac{1}{2} (8 + 2) = 5$.
$A$ value of $5$ corresponds to $sp^3d$ hybridization.

(C) In $XeF_4$,the central atom $Xe$ has $8$ valence electrons.
It forms $4$ bonds with $F$ atoms and has $2$ lone pairs.
Total electron pairs = $4 \text{ (bond pairs)} + 2 \text{ (lone pairs)} = 6$.
Therefore,the hybridization is $sp^3d^2$.
According to $VSEPR$ theory,with $4$ bond pairs and $2$ lone pairs,the geometry is octahedral,but the shape is square planar.

(A) In $XeO_3$,the central atom $Xe$ has $8$ valence electrons. It forms $3$ double bonds with $O$ atoms and has $1$ lone pair. The steric number is $3 + 1 = 4$,which corresponds to $sp^3$ hybridization and a pyramidal geometry.
$BCl_3$ and $BBr_3$ have $sp^2$ hybridization.
$XeF_4$ has $sp^3d^2$ hybridization.

(A) The given structure is a six-membered ring with two double bonds and one triple bond (benzyne-like system).
In the ring,the two carbon atoms involved in the triple bond are $sp$ hybridized.
The remaining four carbon atoms in the ring are $sp^2$ hybridized.
Thus,there are $4$ $sp^2$ hybridized carbon atoms and $2$ $sp$ hybridized carbon atoms.
Therefore,the correct option is $A$.

(D) To determine the hybridization of the triiodide ion $(I_3^-)$,we use the formula:
$\text{Number of Hybrid Orbitals} = \frac{1}{2} \times (V + M - C + A)$
Where:
$V = \text{Number of valence electrons of the central atom} = 7$
$M = \text{Number of monovalent atoms attached} = 2$
$C = \text{Cationic charge} = 0$
$A = \text{Anionic charge} = 1$
Substituting the values:
$\text{Number of Hybrid Orbitals} = \frac{1}{2} \times (7 + 2 - 0 + 1) = \frac{10}{2} = 5$
$A$ hybridization number of $5$ corresponds to $sp^3d$ hybridization.

(B) To determine the hybridization,we use 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.
$1$. For $I_3^-$: Central atom $I$ has $V=7$,$M=2$,$A=1$. $H = \frac{1}{2}(7 + 2 + 1) = 5$. Hybridization is $sp^3d$.
$2$. For $ICl_4^-$: Central atom $I$ has $V=7$,$M=4$,$A=1$. $H = \frac{1}{2}(7 + 4 + 1) = 6$. Hybridization is $sp^3d^2$.
$3$. For $ICl_2^-$: Central atom $I$ has $V=7$,$M=2$,$A=1$. $H = \frac{1}{2}(7 + 2 + 1) = 5$. Hybridization is $sp^3d$.
Thus,the hybridization states are $sp^3d, sp^3d^2, sp^3d$.

(C) To determine the hybridisation of the central atom $N$,we use the formula: $\text{Hybridisation} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1$. For $NO_3^-$: $N = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. This corresponds to $sp^2$ hybridisation.
$2$. For $NO_2^+$: $N = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. This corresponds to $sp$ hybridisation.
$3$. For $NO_2^-$: $N = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. This corresponds to $sp^2$ hybridisation.
Comparing these,both $NO_3^-$ and $NO_2^-$ have $sp^2$ hybridisation. Therefore,statement $C$ is correct.

(C) $X$ is the methyl radical $(\dot{CH_3})$,which is $sp^2$ hybridized (planar,bond angle $120^{\circ}$). $Y$ is the trifluoromethyl radical $(\dot{CF_3})$,which is $sp^3$ hybridized (pyramidal,bond angle $\approx 110^{\circ}$).
$I$. Dimerization of $X$ gives $CH_3-CH_3$ (ethane),where $C$ is $sp^3$ hybridized (bond angle $109.5^{\circ}$). Since $120^{\circ} > 109.5^{\circ}$,the bond angle decreases. Statement $I$ is correct.
$II$. Incorrect,as shown above.
$III$. In $X-Y$ $(CH_3-CF_3)$,the $C-C$ bond length is shorter than in $Y-Y$ $(CF_3-CF_3)$ due to less steric repulsion and electronegativity effects. Statement $III$ is correct.
$IV$. Bond angle in $X$ $(120^{\circ})$ is greater than in $Y$ $(\approx 110^{\circ})$. Statement $IV$ is correct.
Since statements $I, III,$ and $IV$ are correct,and the closest provided option is $C$ $(I, IV)$,we select $C$.

(C) For an $MX_3$ molecule to have a dipole moment of zero,it must have a trigonal planar geometry.
This geometry corresponds to $sp^2$ hybridization of the central atom $M$.
Since the atomic number of $M$ is less than $21$,it belongs to the $s$ or $p$ block elements (e.g.,$BF_3$ where $B$ has atomic number $5$).
Therefore,the central atom $M$ uses $sp^2$ hybridized orbitals for sigma bonding.

(A) To determine the hybridization,we use the formula: $\text{Steric Number} = \frac{1}{2} [V + M - C + A]$,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.
$1$. For $NO^{+}_2$: $V=5, M=0, C=1, A=0$. $\text{Steric Number} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. Hybridization is $sp$.
$2$. For $NO^{-}_3$: $V=5, M=0, C=0, A=1$. $\text{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Hybridization is $sp^2$.
$3$. For $NH^{+}_4$: $V=5, M=4, C=1, A=0$. $\text{Steric Number} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$. Hybridization is $sp^3$.
Thus,the hybridization is $sp, sp^2, sp^3$.

(A) The hybridization of the central atom is calculated 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 $ClO^{-}_2$:
$V = 7$ (for $Cl$),
$M = 0$ (oxygen is divalent),
$C = 0$,
$A = 1$ (for $-1$ charge).
$H = \frac{1}{2} (7 + 0 - 0 + 1) = \frac{8}{2} = 4$.
$A$ value of $4$ corresponds to $sp^3$ hybridization.

(B) To determine the hybridization of the central atom,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $CO_2$: $H = \frac{1}{2}(4 + 0) = 2$ ($sp$ hybridization).
For $SO_2$: The central $S$ atom has $6$ valence electrons. It forms two double bonds with $O$ atoms and has one lone pair. $H = \frac{1}{2}(6 + 0) = 3$ ($sp^2$ hybridization).
For $N_2O$: The central $N$ atom is $sp$ hybridized.
For $CO$: The central $C$ atom is $sp$ hybridized.
Thus,$SO_2$ is the correct answer.

(C) In $PCl_3$,the phosphorus atom has $3$ bond pairs and $1$ lone pair,resulting in a steric number of $4$,which corresponds to $sp^3$ hybridisation.
In $PCl_5$,the phosphorus atom has $5$ bond pairs and $0$ lone pairs,resulting in a steric number of $5$,which corresponds to $sp^3d$ hybridisation.
Therefore,the hybridisation changes from $sp^3$ to $sp^3d$.

(D) To determine the hybridization,we use the formula: $\text{Steric Number} = \frac{1}{2} [V + M - C + A]$,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.
$A) \ BBr_3$: $\text{Steric Number} = \frac{1}{2} [3 + 3] = 3$ ($sp^2$ hybridization).
$B) \ XeF_4$: $\text{Steric Number} = \frac{1}{2} [8 + 4] = 6$ ($sp^3d^2$ hybridization).
$C) \ BCl_3$: $\text{Steric Number} = \frac{1}{2} [3 + 3] = 3$ ($sp^2$ hybridization).
$D) \ XeO_3$: $\text{Steric Number} = \frac{1}{2} [8 + 0] = 4$ ($sp^3$ hybridization). Note: Oxygen is a divalent atom,so it is not included in the $M$ count.
Therefore,$XeO_3$ has $sp^3$ hybridization.

(C) Electronegativity is directly proportional to the $s$-character in the hybrid orbital.
$sp$ hybrid orbital has $50\%$ $s$-character.
$sp^2$ hybrid orbital has $33.3\%$ $s$-character.
$sp^3$ hybrid orbital has $25\%$ $s$-character.
Since the $s$-character increases in the order $sp^3 < sp^2 < sp$,the electronegativity also increases in the same order: $sp^3 < sp^2 < sp$.

(C) The relationship between the bond angle $(\theta)$ and the $s-$character $(s)$ in hybrid orbitals is given by the formula: $\cos \theta = \frac{s}{s-1}$.
Given $\theta = 105^o$,we have $\cos(105^o) = -0.2588$.
Substituting this into the equation: $-0.2588 = \frac{s}{s-1}$.
$-0.2588(s-1) = s$.
$-0.2588s + 0.2588 = s$.
$1.2588s = 0.2588$.
$s = \frac{0.2588}{1.2588} \approx 0.2056$.
Converting to percentage,$s \approx 20.56 \%$.
Among the given options,the value closest to this result is $22 - 23 \%$.

(C) To determine the hybridisation state,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,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 option $C$:
$1$. $NH_4^+$: $SN = \frac{1}{2} (5 + 4 - 1 + 0) = 4$ ($sp^3$ hybridisation). Shape: Tetrahedral.
$2$. $H_3O^+$: $SN = \frac{1}{2} (6 + 3 - 1 + 0) = 4$ ($sp^3$ hybridisation). Shape: Pyramidal.
$3$. $OF_2$: $SN = \frac{1}{2} (6 + 2 - 0 + 0) = 4$ ($sp^3$ hybridisation). Shape: Bent.
All three species have $sp^3$ hybridisation but different shapes due to the presence of different numbers of lone pairs on the central atom.