Hybridisation Questions in English

(A) For $BF_3$: The central atom $B$ has $3$ valence electrons and forms $3$ bonds with $F$ atoms. Steric number $= 3 + 0 = 3$,which corresponds to $sp^2$ hybridisation.
For $SnCl_2$: The central atom $Sn$ has $4$ valence electrons. It forms $2$ bonds with $Cl$ atoms and has $1$ lone pair. Steric number $= 2 + 1 = 3$,which corresponds to $sp^2$ hybridisation.
For $HgCl_2$: The central atom $Hg$ has $2$ valence electrons and forms $2$ bonds with $Cl$ atoms. Steric number $= 2 + 0 = 2$,which corresponds to $sp$ hybridisation.
Therefore,the correct sequence is $sp^2, sp^2, sp$.

(D) Let us analyze the hybridisation of the central atoms in each set:
$A$: $H_2O (sp^3), NH_3 (sp^3), CH_4 (sp^3)$. All are $sp^3$.
$B$: $ClF_3 (sp^3d), XeF_2 (sp^3d), PCl_5 (sp^3d)$. All are $sp^3d$.
$C$: $CO_2 (sp), CO (sp), BeF_2 (sp)$. All are $sp$.
$D$: $SF_4 (sp^3d), XeF_4 (sp^3d^2), CH_4 (sp^3)$. The hybridisations are different.

(A) For $BF_3$: Number of bond pairs around $B = 3$,number of lone pairs around $B = 0$. $\Rightarrow$ Hybridization state $= sp^2$.
For $BeF_2$: Number of bond pairs around $Be = 2$,number of lone pairs around $Be = 0$. $\Rightarrow$ Hybridization state $= sp$.
For $BrF_3$: Number of bond pairs around $Br = 3$,number of lone pairs around $Br = 2$. $\Rightarrow$ Hybridization state $= sp^3d$.

(D) Let us analyze each molecule:
$1$. $ClF_3$: Central atom $Cl$ has $sp^3d$ hybridisation (uses $1$ $d$-orbital) and has $2$ lone pairs.
$2$. $PCl_5$: Central atom $P$ has $sp^3d$ hybridisation (uses $1$ $d$-orbital) and has $0$ lone pairs.
$3$. $BrF_5$: Central atom $Br$ has $sp^3d^2$ hybridisation (uses $2$ $d$-orbitals) and has $1$ lone pair.
$4$. $SF_4$: Central atom $S$ has $sp^3d$ hybridisation (uses $1$ $d$-orbital) and has $1$ lone pair.
In $SF_4$,the number of lone pairs on the central atom is $1$ and the number of $d$-orbitals involved in hybridisation is $1$. Thus,they are the same.

(D) $1$. In $C_2H_4$,$C$ is $sp^2$ hybridized,while in $C_2H_6$,$C$ is $sp^3$ hybridized.
$2$. In $PCl_5$,$P$ is $sp^3d$ hybridized,while in $PCl_6^{-}$,$P$ is $sp^3d^2$ hybridized.
$3$. In $BF_3$,$B$ is $sp^2$ hybridized,while in $BF_4^{-}$,$B$ is $sp^3$ hybridized.
$4$. In $NH_3$,$N$ is $sp^3$ hybridized (with one lone pair),and in $NH_4^{+}$,$N$ is also $sp^3$ hybridized (with four bond pairs and no lone pair). Thus,there is no change in the hybridisation of the central atom in this case.

(B) To determine the hybridisation of the central $I$ atom in $I_3^{-}$,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.
For $I_3^{-}$,the central $I$ atom has $V = 7$ valence electrons,$M = 2$ (two other $I$ atoms),$C = 0$,and $A = 1$.
$H = \frac{1}{2} (7 + 2 + 1) = \frac{10}{2} = 5$.
$A$ value of $5$ corresponds to $sp^3d$ hybridisation.
In $sp^3d$ hybridisation,the electron geometry is trigonal bipyramidal. The central $I$ atom has $3$ lone pairs and $2$ bond pairs. According to $VSEPR$ theory,the lone pairs occupy the equatorial positions to minimize repulsion,resulting in a linear shape for the $I_3^{-}$ ion.

(A) To determine the hybridization of the central atom,we calculate the number of electron pairs (steric number) around it:
$(A)$ In $H_2CO_3$,the central carbon atom is bonded to three oxygen atoms (one double bond and two single bonds). Steric number = $3$ (sigma bonds) + $0$ (lone pairs) = $3$,which corresponds to $sp^2$ hybridization.
$(B)$ In $SiF_4$,the central silicon atom is bonded to four fluorine atoms. Steric number = $4$ (sigma bonds) + $0$ (lone pairs) = $4$,which corresponds to $sp^3$ hybridization.
$(C)$ In $BF_3$,the central boron atom is bonded to three fluorine atoms. Steric number = $3$ (sigma bonds) + $0$ (lone pairs) = $3$,which corresponds to $sp^2$ hybridization.
$(D)$ In $HClO_2$,the central chlorine atom is bonded to two oxygen atoms and one hydrogen atom,with two lone pairs. Steric number = $3$ (sigma bonds) + $2$ (lone pairs) = $5$,which corresponds to $sp^3d$ hybridization.
Thus,only $H_2CO_3$ $(A)$ and $BF_3$ $(C)$ have $sp^2$ hybridized central atoms.

(D) In the molecule $IF_7$,the central atom $I$ (Iodine) has $7$ valence electrons.
All $7$ valence electrons are involved in forming $7$ sigma bonds with $7$ fluorine atoms.
The steric number is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 7 + 0 = 7$.
$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridization.
The geometry and shape of $IF_7$ is pentagonal bipyramidal,where $5$ $F$ atoms are in the equatorial plane and $2$ $F$ atoms are in the axial positions.

(A) In the $CO_2$ molecule $(O=C=O)$,the carbon atom is bonded to two oxygen atoms via two double bonds.
Carbon has $2 \sigma$ bonds and $2 \pi$ bonds,so its steric number is $2$,which corresponds to $sp$ hybridization.
Each oxygen atom is bonded to the carbon atom via one double bond. An oxygen atom in $CO_2$ has $1 \sigma$ bond,$1 \pi$ bond,and $2$ lone pairs.
The steric number for oxygen is $1 (\sigma) + 2 (\text{lone pairs}) = 3$,which corresponds to $sp^2$ hybridization.

(A) To determine the hybridisation of carbon atoms,we count the number of sigma $(\sigma)$ bonds and lone pairs (if any) attached to the carbon atom.
$1$. $C_1$ is involved in a triple bond with the terminal carbon and a single bond with $C_2$. It has $2$ sigma bonds and $2$ pi bonds. Thus,it is $sp$ hybridised.
$2$. $C_2$ is bonded to $C_1$,$H$,and $C_3$. It has $3$ sigma bonds and $1$ pi bond. Thus,it is $sp^2$ hybridised.
$3$. $C_3$ is bonded to $C_2$,$H$,and $CH_2$. It has $3$ sigma bonds and $1$ pi bond. Thus,it is $sp^2$ hybridised.
Therefore,the hybridisations are $C_1: sp$,$C_2: sp^2$,$C_3: sp^2$.

(B) To determine the hybridisation,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 $BCl_3$: $V=3, M=3$. $H = \frac{1}{2}(3+3) = 3$,which corresponds to $sp^2$ hybridisation.
$2$. For $PCl_5$: $V=5, M=5$. $H = \frac{1}{2}(5+5) = 5$,which corresponds to $sp^3d$ hybridisation.
$3$. For $NH_3$: $V=5, M=3$. $H = \frac{1}{2}(5+3) = 4$,which corresponds to $sp^3$ hybridisation.
$4$. For $SF_6$: $V=6, M=6$. $H = \frac{1}{2}(6+6) = 6$,which corresponds to $sp^3d^2$ hybridisation.
Thus,the correct hybridisation is $(BCl_3: sp^2); (PCl_5: sp^3d); (NH_3: sp^3)$ and $(SF_6: sp^3d^2)$.

(A)

Ionic species	Steric Number/Hybridisation/Geometry
$BH_4^{-}$	$4, sp^3$,Tetrahedral
$NH_2^{-}$	$4, sp^3$,$V$-shaped
$CO_3^{2-}$	$3, sp^2$,Trigonal Planar
$H_3O^{+}$	$4, sp^3$,Pyramidal

The central atom in $BH_4^{-}$ is $B$ (Boron). The steric number is $4$ ($4$ bond pairs,$0$ lone pairs),which corresponds to $sp^3$ hybridisation and a Tetrahedral geometry. Thus,$BH_4^{-}$ is the correct answer.

(A) To determine the hybridisation,we use the formula: $\text{Steric Number} = \text{Number of bond pairs} + \text{Number of lone pairs}$.
For $NO_2^{+}$:
$\text{Steric Number} = 2 + 0 = 2$. Hybridisation is $sp$.
For $NO_3^{-}$:
$\text{Steric Number} = 3 + 0 = 3$. Hybridisation is $sp^2$.
For $NH_4^{+}$:
$\text{Steric Number} = 4 + 0 = 4$. Hybridisation is $sp^3$.
Thus,the hybridisation states are $sp, sp^2, sp^3$ respectively.

(C) Ethyne $(C_2H_2)$: The carbon atom is $sp$ hybridized with $2$ $\sigma$-bonds and $0$ lone pairs,resulting in a linear geometry.
Methane $(CH_4)$: The carbon atom is $sp^3$ hybridized with $4$ $\sigma$-bonds and $0$ lone pairs,resulting in a tetrahedral geometry.
Therefore,the shapes are linear and tetrahedral.

(D) In $XeF_6$,the central atom $Xe$ has $8$ valence electrons. It forms $6$ bonds with $F$ atoms,leaving $2$ electrons as one lone pair.
Steric number = (Number of sigma bonds) + (Number of lone pairs) = $6 + 1 = 7$.
$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridisation.
Thus,$XeF_6$ has $sp^3d^3$ hybridisation and $1$ lone pair of electrons.

(A) In $BeCl_2$,the central atom $Be$ undergoes $sp$-hybridization,resulting in a linear geometry with a bond angle of $180^{\circ}$.
$H_2O$ has a bent (angular) shape due to two lone pairs on oxygen.
$SO_2$ has a bent (angular) shape due to one lone pair on sulfur.
$CH_4$ has a tetrahedral geometry.
Therefore,$BeCl_2$ is the linear molecule.

(D) The ground state electronic configuration of $Cl$ $(Z=17)$ is $[Ne] 3s^2 3p_x^2 3p_y^2 3p_z^1$.
In the first excited state,one electron from $3p$ moves to $3d$,giving $3$ unpaired electrons.
In the second excited state,one electron from $3s$ moves to $3d$,giving $5$ unpaired electrons.
In the third excited state,one more electron from $3p$ moves to $3d$,resulting in $7$ unpaired electrons $(3s^1, 3p_x^1, 3p_y^1, 3p_z^1, 3d^3)$.
These $7$ unpaired electrons react with $7$ fluorine atoms to form $ClF_7$.
The hybridization is $sp^3d^3$,which corresponds to a pentagonal bipyramidal geometry.

(B) When $SO_3$ is dissolved in heavy water $(D_2O)$,it forms deuterated sulphuric acid $(D_2SO_4)$ as the compound $X$.
The chemical reaction is: $SO_3 + D_2O \longrightarrow D_2SO_4$ $(X)$.
In $D_2SO_4$,the central sulphur atom is bonded to two $OD$ groups and two oxygen atoms via double bonds.
The steric number of sulphur is calculated as: $\text{Number of sigma bonds} + \text{Number of lone pairs} = 4 + 0 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(A) In $H_2O$,the oxygen atom is $sp^3$ hybridized with two lone pairs and two bond pairs,resulting in an angular (bent) shape.
$NH_4^+$ is $sp^3$ hybridized and tetrahedral.
$CH_4$ is $sp^3$ hybridized and tetrahedral.
Therefore,the correct set is $H_2O, sp^3$,angular.

(B) The conjugate base of $NH_4^{+}$ is obtained by removing a proton $(H^{+})$ from it.
$NH_4^{+} \rightarrow NH_3 + H^{+}$.
The conjugate base is $NH_3$.
In $NH_3$,the central nitrogen atom has $3$ bond pairs and $1$ lone pair.
Total electron pairs = $3 + 1 = 4$.
Therefore,the hybridization state of the nitrogen atom is $sp^3$.

(C) The hybridisation of an atom can be determined by calculating the steric number $(SN)$:
$SN = \frac{1}{2} [V + M - C + A]$
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 $NO_2^{+}$:
$V = 5$ (nitrogen),$M = 0$,$C = 1$,$A = 0$.
$SN = \frac{1}{2} [5 + 0 - 1 + 0] = \frac{4}{2} = 2$.
$A$ steric number of $2$ corresponds to $sp$ hybridisation.
Therefore,the nitrogen atom in $NO_2^{+}$ is $sp$ hybridised.

(A, C) $CO_{2}$ has a linear shape due to $sp$ hybridization and the absence of lone pairs on the central carbon atom.
$HgCl_{2}$ ($sp$ hybridization,linear) and $C_{2}H_{2}$ ($sp$ hybridization,linear) both have a shape similar to $CO_{2}$.
$SnCl_{2}$ and $NO_{2}$ have bent shapes due to the presence of a lone pair on the central atom.

(C) In $POCl_{3}$,the central phosphorus atom is bonded to one oxygen atom via a double bond and three chlorine atoms via single bonds.
Total number of electron pairs around the central atom = $4$ (three $P-Cl$ $\sigma$-bonds and one $P=O$ bond,where the double bond counts as one region of electron density for hybridization).
Since there are $4$ bonding pairs and no lone pairs,the hybridization is $sp^{3}$.
Therefore,the hybridization is $sp^{3}$ and the number of lone pairs is $0$.

(D) The hybridization $sp^3d^2$ involves the mixing of one $s$,three $p$,and two $d$ orbitals to form six equivalent hybrid orbitals.
These six hybrid orbitals are directed towards the corners of an octahedron.
Therefore,a molecule with $sp^3d^2$ hybridization of the central atom exhibits an octahedral geometry.

(A) In $PCl_5$,the central phosphorus atom $(P)$ has $5$ valence electrons. It forms $5$ covalent bonds with $5$ chlorine atoms.
Using the formula for steric number: $\text{Steric Number} = \frac{1}{2} (V + M - C + A)$,where $V = 5$ (valence electrons of $P$),$M = 5$ (monovalent atoms),$C = 0$ (cationic charge),and $A = 0$ (anionic charge).
$\text{Steric Number} = \frac{1}{2} (5 + 5) = 5$.
$A$ steric number of $5$ corresponds to $sp^3d$ hybridisation,resulting in a trigonal bipyramidal geometry.

(A) In $NF_3$,the central nitrogen atom is bonded to three fluorine atoms by three $\sigma$-bonds and has one lone pair of electrons.
To calculate the hybridisation,we use the formula:
$\text{Steric Number} = \text{Number of } \sigma \text{-bonds} + \text{Number of lone pairs}$.
$\text{Steric Number} = 3 + 1 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(B) To determine the $H-C-H$ bond angle,we look at the hybridization and geometry of the central carbon atom in each molecule:
$I$ (Cyclopropylidene): The central carbon atom is $sp^2$ hybridized. Due to the ring strain in the three-membered ring,the $H-C-H$ bond angle is approximately $115^{\circ}$.
$II$ (Ethene): The carbon atom is $sp^2$ hybridized. In a standard alkene,the $H-C-H$ bond angle is approximately $120^{\circ}$.
$III$ (Methane): The carbon atom is $sp^3$ hybridized with a tetrahedral geometry. The $H-C-H$ bond angle is $109^{\circ} 28^{\prime}$ (or $\approx 109.5^{\circ}$).
Comparing these values: $120^{\circ} (II) > 115^{\circ} (I) > 109.5^{\circ} (III)$.
Therefore,the correct decreasing order is $II > I > III$.

(A) For all atoms to lie in a single plane,all carbon atoms and the atoms attached to the benzene ring must be $sp^2$ hybridized to maintain planarity through conjugation.
In $4-$Nitrobenzaldehyde $(O_2N-C_6H_4-CHO)$,the benzene ring,the nitro group $(-NO_2)$,and the aldehyde group $(-CHO)$ are all $sp^2$ hybridized.
The conjugation extends throughout the molecule,allowing all atoms (including the hydrogen of the aldehyde group) to lie in the same plane.
In other options,groups like $-OCH_3$ (in $4-$Methoxybenzaldehyde),$-CH_3$ (in $4-$Methylnitrobenzene),and $-COCH_3$ (in $4-$Nitroacetophenone) contain $sp^3$ hybridized carbon atoms,which possess tetrahedral geometry and prevent all atoms from lying in a single plane.

(D) To determine the hybridization and number of lone pairs for $XeOF_2$:
$1$. The central atom is Xenon $(Xe)$,which has $8$ valence electrons.
$2$. In $XeOF_2$,$Xe$ forms two single bonds with two Fluorine $(F)$ atoms and one double bond with one Oxygen $(O)$ atom.
$3$. The number of bonding electron pairs = $2$ (from $F$) + $2$ (from $O$) = $4$ electron pairs involved in bonding.
$4$. The number of lone pairs on $Xe$ = (Total valence electrons - electrons used in bonding) / $2$ = $(8 - 6) / 2 = 1$ lone pair.
$5$. The steric number = (Number of sigma bonds) + (Number of lone pairs) = $3$ (sigma bonds) + $1$ (lone pair) = $4$. Wait,let us re-evaluate: $Xe$ is bonded to $2$ $F$ atoms and $1$ $O$ atom. Total sigma bonds = $3$. Total lone pairs = $2$. Steric number = $3 + 2 = 5$.
$6$. $A$ steric number of $5$ corresponds to $sp^3d$ hybridization.
$7$. Thus,$XeOF_2$ has $2$ lone pairs and $sp^3d$ hybridization.

(B) Lewis acid is an electron-deficient species. $BF_3$ is a $p$-block fluoride where Boron has an incomplete octet ($6$ electrons),making it a Lewis acid. Its hybridization is $sp^2$ due to $3$ bond pairs and $0$ lone pairs.
$YF_3$ acts as a Lewis base because the central atom has a lone pair of electrons available for donation. $NF_3$ is a $p$-block fluoride where Nitrogen has $3$ bond pairs and $1$ lone pair,making it a Lewis base. Its hybridization is $sp^3$ due to $4$ electron domains ($3$ bond pairs + $1$ lone pair).
Therefore,the hybridization of $XF_3$ $(BF_3)$ is $sp^2$ and $YF_3$ $(NF_3)$ is $sp^3$.

(B) The hybridization $H$ is determined by 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.
$sp^3d$ hybridization corresponds to a steric number of $5$.
$1$) $SF_4$: $H = \frac{1}{2}(6 + 4) = 5$ $(sp^3d)$
$2$) $ClF_3$: $H = \frac{1}{2}(7 + 3) = 5$ $(sp^3d)$
$3$) $XeF_2$: $H = \frac{1}{2}(8 + 2) = 5$ $(sp^3d)$
$4$) $ICl_2^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$ $(sp^3d)$
$5$) $XeF_5^-$: $H = \frac{1}{2}(8 + 5 + 1) = 7$ $(sp^3d^3)$
$6$) $ICl_4^-$: $H = \frac{1}{2}(7 + 4 + 1) = 6$ $(sp^3d^2)$
$7$) $BrF_5$: $H = \frac{1}{2}(7 + 5) = 6$ $(sp^3d^2)$
$8$) $XeF_4$: $H = \frac{1}{2}(8 + 4) = 6$ $(sp^3d^2)$
$9$) $BF_4^-$: $H = \frac{1}{2}(3 + 4 + 1) = 4$ $(sp^3)$
$10$) $NH_4^+$: $H = \frac{1}{2}(5 + 4 - 1) = 4$ $(sp^3)$
The species with $sp^3d$ hybridization are $SF_4, ClF_3, XeF_2, ICl_2^-$. Thus,the total count is $4$.