Hybridisation Questions in English

(B) The total number of valence electrons in $NH_3$ is $5 + (3 \times 1) = 8$.
Using the formula for steric number $X = \frac{1}{2} (V + M - C + A)$,where $V = 5$ (valence electrons of $N$),$M = 3$ (monovalent atoms),$C = 0$,and $A = 0$.
$X = \frac{8}{2} = 4$.
Since the steric number is $4$,the hybridization is $sp^3$.

(C) In $BF_4^-$,the central boron atom undergoes $sp^3$ hybridization.
Since there are no lone pairs of electrons on the central atom,the molecule adopts a regular tetrahedral geometry.

(B) The $sp^3d^2$ hybridization corresponds to an octahedral geometry where all bond angles between adjacent bond pairs are $90^\circ$. In $dsp^2$ (square planar),the bond angles are also $90^\circ$,but $sp^3d^2$ is the standard octahedral case where this geometry is defined by the presence of $90^\circ$ angles in three dimensions.

(C) For an $MX_3$ molecule to have a dipole moment of zero $(\mu = 0)$,the molecule must be symmetric with a trigonal planar geometry.
This geometry is achieved when the central atom $M$ undergoes $sp^2$ hybridization.

(D) To determine the hybridization of the $C$ atom,we count the number of sigma bonds and lone pairs attached to it.
$1$. In $HCOOH$ (formic acid),the $C$ atom is bonded to $H$,$OH$,and double-bonded to $O$. It is $sp^2$ hybridized.
$2$. In $(H_2N)_2CO$ (urea),the $C$ atom is bonded to two $NH_2$ groups and double-bonded to $O$. It is $sp^2$ hybridized.
$3$. In $HCHO$ (formaldehyde),the $C$ atom is bonded to two $H$ atoms and double-bonded to $O$. It is $sp^2$ hybridized.
$4$. In $CH_3CHO$ (acetaldehyde),the methyl carbon $(CH_3)$ is bonded to three $H$ atoms and one $C$ atom. Since it has four sigma bonds,it is $sp^3$ hybridized.

(A) To determine the hybridization of the central atom $Cl$ in $ClO_2^-$,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 $(Cl = 7)$,
$M$ is the number of monovalent atoms attached ($O$ is divalent,so $M = 0$),
$C$ is the cationic charge $(0)$,
$A$ is the anionic charge $(1)$.
$X = \frac{1}{2} [7 + 0 - 0 + 1] = \frac{8}{2} = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridization.
Thus,the chlorine atom in $ClO_2^-$ undergoes $sp^3$ hybridization.

(A) In $H_2CO_3$,the central carbon atom is $sp^2$ hybridized and the molecule is polar due to the presence of different electronegative atoms ($O$ and $OH$ groups) attached to the central carbon.
In $BF_3$,the central boron atom is $sp^2$ hybridized,but it has a trigonal planar geometry,making it non-polar.
$SiF_4$ has $sp^3$ hybridization.
$HClO_2$ has $sp^3$ hybridization.
Therefore,$H_2CO_3$ is the correct answer.

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

(B) To determine the hybridization,we calculate the steric number $(X)$ using the formula: $X = \frac{1}{2}(V + M - C + A)$,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 $NO_2^+$: $V = 5$,$M = 0$,$C = 1$,$A = 0$. $X = \frac{1}{2}(5 + 0 - 1 + 0) = 2$. Hybridization is $sp$.
$2$. For $NO_3^-$: $V = 5$,$M = 0$,$C = 0$,$A = 1$. $X = \frac{1}{2}(5 + 0 - 0 + 1) = 3$. Hybridization is $sp^2$.
$3$. For $NH_4^+$: $V = 5$,$M = 4$,$C = 1$,$A = 0$. $X = \frac{1}{2}(5 + 4 - 1 + 0) = 4$. Hybridization is $sp^3$.
Thus,the hybridization states are $sp, sp^2, sp^3$ respectively.

(A) In the molecule $H_2C = CH - C \equiv N$:
$1$. The first carbon atom $(CH_2)$ is bonded to two hydrogens and one double bond,so it is $sp^2$ hybridized.
$2$. The second carbon atom $(CH)$ is bonded to one hydrogen,one double bond,and one single bond,so it is $sp^2$ hybridized.
$3$. The third carbon atom $(C)$ is bonded to one single bond and one triple bond,so it is $sp$ hybridized.
$4$. The nitrogen atom $(N)$ is bonded to one triple bond and has one lone pair,so it is $sp$ hybridized.
Thus,the hybridization sequence is $sp^2 - sp^2 - sp - sp$.

(C) In the $C_2H_4$ molecule,each carbon atom is $sp^2$ hybridized.
Each carbon atom forms three $sp^2$ sigma bonds (two with hydrogen atoms and one with the other carbon atom) and one $\pi$ bond.
The $sp^2$ hybridization results in a trigonal planar geometry around each carbon atom,making the entire molecule planar with a bond angle of approximately $120^{\circ}$.

(C) The given electronic configuration shows two $sp$ hybrid orbitals and two unhybridized $p$ orbitals ($2p_y$ and $2p_z$).
Each of these four orbitals contains one unpaired electron.
The two $sp$ hybrid orbitals will form two sigma $(\sigma)$ bonds.
The two unhybridized $p$ orbitals ($2p_y$ and $2p_z$) will form two pi $(\pi)$ bonds.
Therefore,the configuration can form two sigma $(\sigma)$ bonds and two pi $(\pi)$ bonds.

(C) The bond length of a $C-C$ single bond depends on the hybridization of the carbon atoms involved.
As the percentage of $s$-character increases,the electronegativity of the carbon atom increases,and the bond length decreases.
Hybridization states for the $C-C$ single bond in each compound:
$(I) CH_2 = CH - C \equiv CH$: The $C-C$ single bond is between $sp^2$ and $sp$ hybridized carbons.
$(II) CH \equiv C - C \equiv CH$: The $C-C$ single bond is between $sp$ and $sp$ hybridized carbons.
$(III) CH_3 - CH = CH_2$: The $C-C$ single bond is between $sp^3$ and $sp^2$ hybridized carbons.
$(IV) CH_2 = CH - CH = CH_2$: The $C-C$ single bond is between $sp^2$ and $sp^2$ hybridized carbons.
Comparing the $s$-character:
$sp^3-sp^2$ (lowest $s$-character) > $sp^2-sp^2$ > $sp^2-sp$ > $sp-sp$ (highest $s$-character).
Therefore,the order of bond length is: $(III) > (IV) > (I) > (II)$.

(A) In $Acetone$ $(CH_3COCH_3)$,the carbonyl carbon is $sp^2$ hybridized.
In $Acetic$ $acid$ $(CH_3COOH)$,the carboxyl carbon is $sp^2$ hybridized.
In $Acetamide$ $(CH_3CONH_2)$,the carbonyl carbon is $sp^2$ hybridized.
In $Acetonitrile$ $(CH_3CN)$,the carbon in the $CN$ group is $sp$ hybridized.
Since options $A$,$B$,and $D$ all contain $sp^2$ hybridized carbons,this question is ambiguous. However,typically in such multiple-choice questions,if one must choose,all three carbonyl-containing compounds listed ($Acetone$,$Acetic$ $acid$,$Acetamide$) possess $sp^2$ hybridized carbons.

(C) In $CH_2 = CH - CH = CH_2$ ($1$,$3$-butadiene),all four carbon atoms are $sp^2$ hybridized.
$sp^2$ hybridized carbon atoms have a trigonal planar geometry.
Since all carbon atoms are connected in a conjugated system,the entire molecule lies in a single plane.
Therefore,the shape of the molecule is planar.

(A) The bond length depends on the hybridization of the carbon atoms involved in the bond.
$1$. $C_2H_6$ (Ethane): $sp^3-sp^3$ hybridization, bond length $\approx 154 \ pm$.
$2$. $C_6H_6$ (Benzene): $sp^2-sp^2$ hybridization with resonance, bond length $\approx 139 \ pm$.
$3$. $C_2H_4$ (Ethene): $sp^2-sp^2$ hybridization, bond length $\approx 134 \ pm$.
$4$. $C_2H_2$ (Ethyne): $sp-sp$ hybridization, bond length $\approx 120 \ pm$.
As the $s$-character increases, the bond length decreases.
Therefore, the order of bond length is $C_2H_6 (154 \ pm) > C_6H_6 (139 \ pm) > C_2H_4 (134 \ pm) > C_2H_2 (120 \ pm)$.

(C) The central atom in $SO_2$ is sulphur $(S)$.
Sulphur has $6$ valence electrons.
In $SO_2$,sulphur forms two double bonds with two oxygen atoms.
Number of electron pairs around sulphur = $2$ (bonding pairs) + $1$ (lone pair) = $3$.
Since the steric number is $3$,the hybridization is $sp^2$.

(B) To determine the hybridization,we use the formula: $\text{Hybridization} = \frac{1}{2} [V + M - C + A]$,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 $NO_2^+$: $\text{Hybridization} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$,which corresponds to $sp$ hybridization.
$2$. For $NO_3^-$: $\text{Hybridization} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$,which corresponds to $sp^2$ hybridization.
$3$. For $NH_4^+$: $\text{Hybridization} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^3$ hybridization.
Thus,the correct order is $sp$,$sp^2$,and $sp^3$ respectively.

(C) The correct answer is $C$ ($sp$ hybridization).

Hybridization Type	Bond Angle
$sp^3$	$109.5^\circ$
$sp^2$	$120^\circ$
$sp^3d$	$90^\circ$ and $120^\circ$
$sp$	$180^\circ$

As shown in the table,$sp$ hybridization corresponds to a linear geometry with a bond angle of $180^\circ$,which is the highest among the given options.

(C) To determine the species with a bond angle of $120^{\circ}$,we analyze the hybridization and geometry of each molecule:
$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The hybridization is $sp^3d$,and the geometry is $T$-shaped with bond angles less than $90^{\circ}$.
$2$. $NCl_3$: The central atom $N$ has $5$ valence electrons. It forms $3$ bonds with $Cl$ atoms and has $1$ lone pair. The hybridization is $sp^3$,and the geometry is pyramidal with bond angles approximately $107^{\circ}$.
$3$. $BCl_3$: The central atom $B$ has $3$ valence electrons. It forms $3$ bonds with $Cl$ atoms and has no lone pairs. The hybridization is $sp^2$,and the geometry is trigonal planar with bond angles of $120^{\circ}$.
$4$. $PH_3$: The central atom $P$ has $5$ valence electrons. It forms $3$ bonds with $H$ atoms and has $1$ lone pair. The geometry is pyramidal with bond angles close to $93^{\circ}$.
Therefore,$BCl_3$ is the correct species.

(C) To determine the hybridization of the central nitrogen atom,we use the formula: $\text{Number of hybrid orbitals} = \frac{1}{2} [V + M - C + A]$,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 $NO_2^+$: $V = 5$,$M = 0$,$C = 1$,$A = 0$. $\text{Hybrid orbitals} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. Hence,$sp$ hybridization.
$2)$ For $NO_3^-$: $V = 5$,$M = 0$,$C = 0$,$A = 1$. $\text{Hybrid orbitals} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Hence,$sp^2$ hybridization.
$3)$ For $NH_4^+$: $V = 5$,$M = 4$,$C = 1$,$A = 0$. $\text{Hybrid orbitals} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$. Hence,$sp^3$ hybridization.
Therefore,the hybridizations are $sp$,$sp^2$,and $sp^3$ respectively.

(D) In $XeF_4$,the central atom $Xe$ has $8$ valence electrons.
It forms $4$ single bonds with $F$ atoms and has $2$ lone pairs of electrons.
Total electron pairs = $4 \text{ (bond pairs)} + 2 \text{ (lone pairs)} = 6$.
This corresponds to $sp^3d^2$ hybridization.
Due to the presence of $2$ lone pairs,the geometry is square planar.

(C) The chemical formula of ethene is $CH_2=CH_2$.
In this molecule,there are $4$ $C-H$ sigma $(\sigma)$ bonds and $1$ $C-C$ sigma $(\sigma)$ bond,making a total of $5$ sigma $(\sigma)$ bonds.
Additionally,there is $1$ $C-C$ pi $(\pi)$ bond.
Therefore,the total number of bonds is $5$ sigma $(\sigma)$ and $1$ pi $(\pi)$ bonds.

(D) In $BF_4^-$,the central atom $B$ has $4$ bond pairs and $0$ lone pairs,resulting in $sp^3$ hybridization and a tetrahedral shape.
In $NH_4^+$,the central atom $N$ has $4$ bond pairs and $0$ lone pairs,resulting in $sp^3$ hybridization and a tetrahedral shape.
Since both species have the same hybridization and shape,they are isostructural.

(A) To determine the hybridization,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.
$1. NO_2^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.
$2. NO_3^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.
$3. NH_2^-: SN = \frac{1}{2} (5 + 2 - 0 + 1) = 4 \rightarrow sp^3$ hybridization.
$4. NH_4^+: SN = \frac{1}{2} (5 + 4 - 1 + 0) = 4 \rightarrow sp^3$ hybridization.
$5. SCN^-: SN = \frac{1}{2} (4 + 0 - 0 + 1) = 2 \rightarrow sp$ hybridization.
Comparing the results,$NO_2^-$ and $NO_3^-$ both exhibit $sp^2$ hybridization.

(B) For $sp^{2}$ hybridisation,the steric number (sum of $\sigma$-bonds and lone pairs) must be $3$.
$(I)$ $NO_{2}^{-}$: Central $N$ atom has $2$ $\sigma$-bonds and $1$ lone pair,so steric number $= 3$,which corresponds to $sp^{2}$ hybridisation.
$(II)$ $NH_{3}$: Central $N$ atom has $3$ $\sigma$-bonds and $1$ lone pair,so steric number $= 4$,which corresponds to $sp^{3}$ hybridisation.
$(III)$ $BF_{3}$: Central $B$ atom has $3$ $\sigma$-bonds and $0$ lone pairs,so steric number $= 3$,which corresponds to $sp^{2}$ hybridisation.
$(IV)$ $NH_{2}^{-}$: Central $N$ atom has $2$ $\sigma$-bonds and $2$ lone pairs,so steric number $= 4$,which corresponds to $sp^{3}$ hybridisation.
$(V)$ $H_{2}O$: Central $O$ atom has $2$ $\sigma$-bonds and $2$ lone pairs,so steric number $= 4$,which corresponds to $sp^{3}$ hybridisation.
Thus,the pair $BF_{3}$ and $NO_{2}^{-}$ both exhibit $sp^{2}$ hybridisation.

(C) The hybridization of the central atom can be determined using the formula $H = \frac{1}{2} [V + X - C + A]$,where $V$ is the number of valence electrons,$X$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1. SF_4: H = \frac{1}{2} [6 + 4] = 5 \rightarrow sp^3d$
$2. I_3^-: H = \frac{1}{2} [7 + 2 + 1] = 5 \rightarrow sp^3d$
$3. PCl_5: H = \frac{1}{2} [5 + 5] = 5 \rightarrow sp^3d$
$4. SbCl_5^{2-}: H = \frac{1}{2} [5 + 5 + 2] = 6 \rightarrow sp^3d^2$
Thus,$SbCl_5^{2-}$ has a different hybridization $(sp^3d^2)$ compared to the others $(sp^3d)$.

(B) The number of hybrid orbitals $(H)$ 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 total positive charge,and $A$ is the negative charge.
For $CH_4$: $H = \frac{1}{2} [4 + 4 - 0 + 0] = 4$,which corresponds to $sp^3$ hybridization.
For $SF_4$: $H = \frac{1}{2} [6 + 4 - 0 + 0] = 5$,which corresponds to $sp^3d$ hybridization.
For $BF_4^-$: $H = \frac{1}{2} [3 + 4 - 0 + 1] = 4$,which corresponds to $sp^3$ hybridization.
For $NH_4^+$: $H = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^3$ hybridization.
Thus,the central atom in $SF_4$ does not have $sp^3$ hybridization.

(C) For $sp^2$ hybridization,the steric number should be $3$.
In $BF_3$ molecule,the central atom $B$ is bonded to $3$ atoms and has $0$ lone pairs,so steric number $= 3 + 0 = 3$. Hence,it is $sp^2$ hybridized.
In $NO_2^-$ ion,the central atom $N$ is bonded to $2$ oxygen atoms and has $1$ lone pair,so steric number $= 2 + 1 = 3$. Hence,it is $sp^2$ hybridized.
In $NH_2^-$ ion,the central atom $N$ is bonded to $2$ hydrogen atoms and has $2$ lone pairs,so steric number $= 2 + 2 = 4$. Hence,it is $sp^3$ hybridized.
In $H_2O$ molecule,the central atom $O$ is bonded to $2$ hydrogen atoms and has $2$ lone pairs,so steric number $= 2 + 2 = 4$. Hence,it is $sp^3$ hybridized.
Thus,in $BF_3$ and $NO_2^-$,the central atom is $sp^2$ hybridised.

(C) The electronegativity of hybrid orbitals depends on the $s$-character present in the orbital.
As the $s$-character increases,the electron-attracting tendency (electronegativity) increases.
The $s$-character in different hybrid orbitals is as follows:
$sp$: $50\%$
$sp^2$: $33.3\%$
$sp^3$: $25\%$
Therefore,the correct order of electronegativity is $sp > sp^2 > sp^3$.

(B) The geometry of a molecule depends on the hybridization of its carbon atoms.
$sp^{3}$ hybridized carbon atoms have a tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.
$sp^{2}$ hybridized carbon atoms have a trigonal planar geometry with a bond angle of $120^{\circ}$.
$sp$ hybridized carbon atoms have a linear geometry with a bond angle of $180^{\circ}$.
In $CH_3 - C \equiv C - CH_3$ (but$-2-$yne),the two central carbon atoms are $sp$ hybridized,making the central part of the molecule linear.

(C) $XeF_6$ has $sp^3d^3$ hybridization and distorted octahedral shape due to $6$ bond pairs and $1$ lone pair.
$(B)$ $XeO_3$ has $sp^3$ hybridization and pyramidal shape due to $3$ bond pairs and $1$ lone pair.
$(C)$ $XeOF_4$ has $sp^3d^2$ hybridization and square pyramidal shape due to $5$ bond pairs and $1$ lone pair.
$(D)$ $XeF_4$ has $sp^3d^2$ hybridization and square planar shape due to $4$ bond pairs and $2$ lone pairs.
Therefore,the correct matching is $A-(i), B-(iii), C-(iv), D-(ii)$.

(B) The hybridization of the central atom can be 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 $NO_{3}^{-}$: $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$,which corresponds to $sp^{2}$ hybridization.
For $NO_{2}^{+}$: $H = \frac{1}{2} [5 + 0 - 1 + 0] = 2$,which corresponds to $sp$ hybridization.
For $NH_{4}^{+}$: $H = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^{3}$ hybridization.
Thus,the hybridization states are $sp^{2}, sp, sp^{3}$ respectively.

(C) To determine the hybridization of the $N$ atom,we use the formula: $\text{Steric Number} = \frac{1}{2} [V + M - C + A]$,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 $NO_{2}^{+}$: $\text{Steric Number} = \frac{1}{2} [5 + 0 - 1 + 0] = \frac{4}{2} = 2$. $A$ steric number of $2$ corresponds to $sp$ hybridization.
For $NO_{3}^{-}$: $\text{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$ ($sp^2$ hybridization).
For $NO_{2}$: The $N$ atom has one unpaired electron and is $sp^2$ hybridized.
For $NO_{2}^{-}$: $\text{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$ ($sp^2$ hybridization).
Thus,the $N$ atom is $sp$ hybridized in $NO_{2}^{+}$.
Therefore,the correct option is $C$.

(A) $H_2O$: Central atom $O$ is $sp^3$ hybridized with $2$ lone pairs,resulting in a bent ($V$-shape) geometry.
$NH_3$: Central atom $N$ is $sp^3$ hybridized with $1$ lone pair,resulting in a trigonal pyramidal geometry.
Since both $H_2O$ and $NH_3$ are $sp^3$ hybridized but have different shapes (bent vs pyramidal),option $A$ is correct.
$Ni(CO)_4$ is $sp^3$ (tetrahedral) and $[Ni(CN)_4]^{4-}$ is $dsp^2$ (square planar).
$XeF_4$ is $sp^3d^2$ (square planar) and $[Fe(CO)_4]^{2-}$ is $sp^3$ (tetrahedral).
$SF_4$ is $sp^3d$ (see-saw) and $CF_4$ is $sp^3$ (tetrahedral).

(A) $1$. Analyze the hybridization of $A$ $(PCl_4^+)$: The central atom $P$ has $5 + 4 - 1 = 8$ valence electrons,forming $4$ bonds. Hybridization is $sp^3$ ($0$ $d$-orbitals).
$2$. Analyze the hybridization of $B$ $(PCl_6^-)$: The central atom $P$ has $5 + 6 + 1 = 12$ valence electrons,forming $6$ bonds. Hybridization is $sp^3d^2$ ($2$ $d$-orbitals).
$3$. Evaluate $(I)$: From $B$ $(sp^3d^2)$ to $A$ $(sp^3)$,the number of $d$-orbitals involved decreases from $2$ to $0$. Thus,statement $(I)$ is $INCORRECT$.
$4$. Evaluate $(II)$: In $A$ $(sp^3)$,$p$-character is $75\%$. In $B$ $(sp^3d^2)$,$p$-character is $3/6 = 50\%$. Thus,from $A$ to $B$,$p$-character decreases. Statement $(II)$ is $INCORRECT$.
$5$. Evaluate $(III)$: $A$ has $4$ covalent bonds; $B$ has $6$ covalent bonds. They are not the same. Statement $(III)$ is $INCORRECT$.
$6$. Evaluate $(IV)$: $PCl_6^-$ is formed by $PCl_5 + Cl^- \rightarrow PCl_6^-$. $PCl_5$ acts as a Lewis acid. This is a super-octet molecule. Statement $(IV)$ is $CORRECT$.
$7$. Since $(I), (II),$ and $(III)$ are incorrect,the answer is $A$.

(A) In $BF_3$,the boron atom is $sp^2$ hybridized with a bond angle of $120^\circ$. In $BF_4^-$,the boron atom is $sp^3$ hybridized with a bond angle of $109^\circ 28'$. Thus,this process involves a change in both hybridization and bond angle.
In $NH_3$,the nitrogen atom is $sp^3$ hybridized with a bond angle of $107^\circ$. In $NH_4^+$,the nitrogen atom is $sp^3$ hybridized with a bond angle of $109^\circ 28'$. While the bond angle changes,the hybridization remains $sp^3$.
In $BF_3 \cdot NH_3$,boron changes from $sp^2$ to $sp^3$ and nitrogen remains $sp^3$ (but its geometry changes from pyramidal to tetrahedral). However,$BF_3 + F^- \to BF_4^-$ is the most direct example of a change in hybridization from $sp^2$ to $sp^3$ accompanied by a change in bond angle.

(C) To determine the hybridisation and shape,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 $ICl_4^-$:
$V = 7$ (for $I$),$M = 4$ (for $Cl$),$A = 1$.
$SN = \frac{1}{2}(7 + 4 + 1) = 6$.
$A$ steric number of $6$ corresponds to $sp^3d^2$ hybridisation.
With $4$ bond pairs and $2$ lone pairs,the geometry is octahedral and the shape is square planar.

Chemical species	Hybridisation
$SF_4$	$sp^3d$
$BF_4^-$	$sp^3$
$ICl_4^-$	$sp^3d^2$
$XeO_2F_2$	$sp^3d$

(B) In the formation of $PF_5$,phosphorus undergoes $sp^3d$ hybridization.
During hybridization,the orbitals involved ($3s$,$3p$,and $3d$) undergo mixing to form five equivalent $sp^3d$ hybrid orbitals.
The $3d$ orbital,which is initially very diffuse (large radial distance of $2.4 \ \mathring{A}$),contracts significantly to participate in bonding with the electronegative fluorine atoms.
The $3s$ and $3p$ orbitals remain relatively unchanged in their radial distribution compared to the $3d$ orbital.
Therefore,the radial distances become approximately $0.40 \ \mathring{A}$,$0.48 \ \mathring{A}$,and $0.85 \ \mathring{A}$ respectively,reflecting the contraction of the $3d$ orbital.

(A) Step $1$: $2KI + MnO_2 + 2H_2SO_4 \to K_2SO_4 + MnSO_4 + 2H_2O + I_2(g)$. Thus,$(B) = I_2$.
Step $2$: $I_2 + K_2S_2O_8 \to K_2SO_4 + I_2(SO_4)$ (or similar oxidation reaction). Here $(D)$ is $I_2$.
Step $3$: $I_2 + KI \to KI_3$. The brown complex $(E)$ is $KI_3$,which contains the triiodide ion $I_3^-$.
Step $4$: In $I_3^-$,the central iodine atom has $3$ lone pairs and $2$ bond pairs,resulting in $5$ electron pairs. The hybridisation is $sp^3d$.

(D) In $XeF_2$,the central atom $Xe$ undergoes $sp^3d$ hybridisation.
The $d-$orbital involved in $sp^3d$ hybridisation is the $d_{z^2}$ orbital.
The $d_{z^2}$ orbital has $0$ nodal planes because it consists of a lobe along the $z-$axis and a ring of electron density in the $xy-$plane,meaning it does not have planar nodes passing through the nucleus like other $d-$orbitals.
Therefore,the number of nodal planes is $0$.

(C) To determine the geometry and hybridisation of $SF_4$,we first calculate the number of valence electrons: $S$ $(6)$ + $4 \times F$ $(7)$ = $34$ electrons.
Using the steric number formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V = 6$ (valence electrons of $S$),$M = 4$ (monovalent atoms),$C = 0$,and $A = 0$.
$SN = \frac{1}{2} (6 + 4) = 5$.
$A$ steric number of $5$ corresponds to $sp^3d$ hybridisation.
The molecule has $4$ bonding pairs and $1$ lone pair on the sulfur atom.
According to $VSEPR$ theory,the presence of one lone pair in a trigonal bipyramidal electron geometry results in a $See-Saw$ molecular geometry.

(A) In the given structure (dehydrobenzene derivative),there are $4$ carbon atoms involved in double bonds,which are $sp^{2}$ hybridized.
There are $2$ carbon atoms involved in a triple bond,which are $sp$ hybridized.
Therefore,the number of $sp^{2}$ and $sp$ hybridized carbon atoms are $4$ and $2$ respectively.
The correct option is $A$.

(C) In $C(CN)_4$,the central carbon atom is $sp^3$ hybridized,meaning it has four $sp^3$ hybrid orbitals. Each $sp^3$ orbital has $25\%$ $s$ character and $75\%$ $P$ character.
Each of the four cyano group carbon atoms is $sp$ hybridized,meaning each has two $sp$ hybrid orbitals. Each $sp$ orbital has $50\%$ $s$ character and $50\%$ $P$ character.
$33\%$ $P$ character corresponds to $sp^2$ hybridization (where $P$ character is $2/3 \approx 66.6\%$ is incorrect,actually $sp^2$ has $66.6\%$ $P$ character,$sp^3$ has $75\%$ $P$ character,and $sp$ has $50\%$ $P$ character).
Since there are no $sp^2$ hybridized carbon atoms in $C(CN)_4$,the number of hybrid orbitals with $33\%$ $P$ character is $0$.

(C) In $CH_4$,if we use pure unhybridised orbitals,the $3$ $p$-orbitals of carbon $(p_x, p_y, p_z)$ would overlap with $1s$ orbitals of $3$ hydrogen atoms at $90^o$ angles.
The $4^{th}$ hydrogen atom would overlap with the $2s$ orbital of carbon.
Since $p$-orbitals are directional and $s$-orbitals are spherical,the bond lengths and bond angles would not be equivalent to the observed tetrahedral geometry.
Specifically,the geometry would not be tetrahedral,as the $3$ $p$-bonds would be at $90^o$ to each other,and the $4^{th}$ bond would be different in nature.
Therefore,the statement that the geometry will be tetrahedral is incorrect.

(D) In $PCl_5$,the phosphorus atom is $sp^3d$ hybridized,and the molecular geometry is trigonal bipyramidal.
In $PCl_6^-$,the phosphorus atom is $sp^3d^2$ hybridized,and the molecular geometry is octahedral.
Since both the hybridization and the geometry change,the shape of the molecule also changes.
Therefore,all the given statements are correct.

(A) According to Bent's rule,more electronegative elements prefer to occupy orbitals with less $s$-character (i.e.,more $p$-character).
Conversely,less electronegative elements prefer to occupy orbitals with more $s$-character.
As the $s$-character increases,the bond angle increases.
Therefore,the bond angle is directly proportional to the $s$-character of the hybrid orbital involved in the bond.

(C) In $SO_3$,the sulphur atom undergoes $sp^2$ hybridisation. The orbitals involved in $sp^2$ hybridisation are $s$,$p_x$,and $p_y$.
Since the molecule is planar (lying in the $xy$-plane),the $p_z$ orbital remains unhybridised and is involved in $p\pi-p\pi$ bonding.
To form $p\pi-d\pi$ bonds,the $d$ orbitals must have the correct symmetry to overlap with the remaining $p$ orbitals of oxygen.
Given the $sp^2$ hybridisation uses $p_x$ and $p_y$,the $d$ orbitals that can participate in $\pi$-bonding with the $p_z$ orbital of oxygen are $d_{xz}$ and $d_{yz}$.