Hybridisation Questions in English

(A) $CO_2$ has a linear structure with $sp$ hybridization.
$HgCl_2$ $(Cl-Hg-Cl)$ and $C_2H_2$ $(H-C\equiv C-H)$ also possess a linear structure due to $sp$ hybridization.
Therefore,$CO_2$ resembles $HgCl_2$ and $C_2H_2$.

(D) The $C-H$ bond energy depends on the hybridization of the carbon atom.
In ethane $(C_2H_6)$,carbon is $sp^3$ hybridized ($25\% \ s$-character).
In ethylene $(C_2H_4)$,carbon is $sp^2$ hybridized ($33.3\% \ s$-character).
In acetylene $(C_2H_2)$,carbon is $sp$ hybridized ($50\% \ s$-character).
As the $s$-character increases,the electronegativity of the carbon atom increases,which pulls the bonding electrons closer to the nucleus,resulting in a shorter and stronger $C-H$ bond.
Therefore,the $C-H$ bond energy is greatest in acetylene.

(D) In $sp^2$ hybridization,the central atom forms three hybrid orbitals that are directed towards the corners of an equilateral triangle.
This arrangement results in a bond angle of $120^{\circ}$ between the orbitals.
Therefore,the geometry of a molecule with $sp^2$ hybridization and no lone pairs on the central atom is Trigonal planar.

(A) In $sp^3$ hybridization,one $s$ orbital and three $p$ orbitals mix to form four equivalent $sp^3$ hybrid orbitals.
Since the total number of orbitals involved is $4$,the contribution of the $s$ orbital to each hybrid orbital is $1/4$.
Therefore,each $sp^3$ hybridized orbital contains $1/4$ $s-$character.

(D) The hybridization $sp^3d$ involves the mixing of one $s$,three $p$,and one $d$ orbital.
This results in a geometry where the central atom is surrounded by five electron pairs arranged in a $Trigonal bipyramidal$ structure.

(D) is the correct answer because $dsp^3$ hybridization results in a trigonal bipyramidal geometry.
In this geometry,the bond angles are $120^o$ (equatorial) and $90^o$ (axial),not all at $90^o$.

(A) The structure of $CO_2$ is $O=C=O$.
In this molecule,the carbon atom forms two double bonds with two oxygen atoms.
According to the steric number rule,the steric number is calculated as the number of sigma bonds plus the number of lone pairs on the central atom.
For carbon in $CO_2$,there are $2$ sigma bonds and $0$ lone pairs,resulting in a steric number of $2$.
$A$ steric number of $2$ corresponds to $sp$ hybridization.

(B) The central atom $Xe$ in $XeF_2$ has $8$ valence electrons.
It forms $2$ sigma bonds with $F$ 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 hybridisation is $sp^3d$ and the geometry is trigonal bipyramidal.

(A) The geometry associated with each hybridisation is as follows:
$A. sp$ hybridisation results in a linear geometry.
$B. sp^2$ hybridisation results in a trigonal planar geometry.
$C. sp^3$ hybridisation results in a tetrahedral geometry,which is non-planar.
$D. dsp^2$ hybridisation results in a square planar geometry.
Therefore,$sp^3$ hybridisation is the one that results in non-planar orbitals.

(B) $sp^3d^2$ hybridization involves the intermixing of $1$ $s$,$3$ $p$,and $2$ $d$ orbitals to form $6$ identical $sp^3d^2$ hybrid orbitals.
These $6$ orbitals are directed towards the corners of an octahedron.
They are oriented at an angle of $90^{\circ}$ to one another,resulting in an octahedral geometry.

(C) In the $OF_2$ molecule,the central oxygen atom is bonded to two fluorine atoms and has two lone pairs of electrons.
Using the formula for steric number: $SN = \frac{1}{2} (V + M - C + A)$,where $V = 6$ (valence electrons of $O$),$M = 2$ (monovalent atoms $F$),$C = 0$,and $A = 0$.
$SN = \frac{1}{2} (6 + 2) = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(A) In $sp^3$ hybridization,one $s$ orbital and three $p$ orbitals are mixed to form four equivalent $sp^3$ hybrid orbitals.
The total number of orbitals involved is $1 + 3 = 4$.
The fraction of $s-$character is $\frac{1}{4} = 0.25$.
Therefore,the percentage of $s-$character is $0.25 \times 100 = 25\%$.
The correct option is $A$.

(B) The bond angle is maximum for $sp$ hybridisation.
In $sp$ hybridisation,the two hybrid orbitals are arranged linearly with a bond angle of $180^\circ$.
In $sp^2$ hybridisation,the bond angle is $120^\circ$.
In $sp^3$ hybridisation,the bond angle is approximately $109.5^\circ$.
In $dsp^2$ hybridisation,the bond angle is $90^\circ$.

(C) The $C-H$ bond length depends on the hybridization of the carbon atom.
As the $s$-character in the hybrid orbital increases,the bond length decreases.
$C_2H_2$ ($sp$ hybridization,$50\% \ s$-character),$C_2H_4$ ($sp^2$ hybridization,$33.3\% \ s$-character),and $C_6H_6$ ($sp^2$ hybridization) have shorter $C-H$ bonds.
In $C_2H_4Br_2$ ($1$,$2$-dibromoethane),the carbon atoms are $sp^3$ hybridized ($25\% \ s$-character).
Since $sp^3$ hybridized carbon has the least $s$-character,the $C-H$ bond is the longest.
Therefore,the correct option is $C$.

(C) In $CH_2Cl-CH_2Cl$ ($1$,$2$-dichloroethane),each carbon atom is bonded to two hydrogen atoms,one chlorine atom,and one other carbon atom.
Since each carbon atom forms $4$ sigma bonds and has no lone pairs,the steric number is $4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridization.

(A) The central atom $Xe$ has $8$ valence electrons.
It forms $4$ bonds with $F$ atoms and has $2$ lone pairs of electrons.
Total electron pairs = $4 \text{ (bond pairs)} + 2 \text{ (lone pairs)} = 6$.
Therefore,the hybridization is $sp^{3}d^{2}$ and the shape is square planar.

(B) In $HCHO$ (formaldehyde),the carbon atom is bonded to two hydrogen atoms by single bonds and one oxygen atom by a double bond.
The steric number of the carbon atom is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 3 + 0 = 3$.
$A$ steric number of $3$ corresponds to $sp^2$ hybridization.
Therefore,the correct option is $(B)$.

(D) is the correct answer.
In $NH_3$ (Ammonia),the central $N$ atom is $sp^3$ hybridized.
In $CH_4$ (Methane),the central $C$ atom is $sp^3$ hybridized.
In $H_2O$ (Water),the central $O$ atom is $sp^3$ hybridized.
In $CO_2$ (Carbon dioxide),the central $C$ atom is $sp$ hybridized because it forms two double bonds with oxygen atoms.

(A) Hybridization is a process of mixing atomic orbitals of comparable energy to form new hybrid orbitals. These hybrid orbitals are more stable and possess lower energy compared to the pure atomic orbitals from which they are formed. Therefore,the correct option is $A$.

(B) $CCl_4$ has $sp^3$ hybridization,which results in a tetrahedral geometry.
In a regular tetrahedral geometry,the bond angle is approximately $109.5^o$,which is commonly rounded to $109^o$.

(C) In the $NH_3$ molecule,the central nitrogen atom has $5$ valence electrons.
It forms $3$ sigma bonds with $3$ hydrogen atoms and has $1$ lone pair of electrons.
The total number of electron pairs (steric number) is $3 + 1 = 4$.
For a steric number of $4$,the hybridization is $sp^3$.

(B) Ethene $(CH_2=CH_2)$ consists of two carbon atoms that are $sp^2$ hybridized.
In an $sp^2$ hybridized system,the geometry around each carbon atom is trigonal planar.
The bond angle associated with this geometry is $120^o$.

(D) In methane $(CH_4)$,the central carbon atom is $sp^3$ hybridized.
Due to $sp^3$ hybridization,the molecule adopts a tetrahedral geometry.
The bond angle in a regular tetrahedral geometry is $109.5^o$.

(B) The three equivalent $sp^2$ hybrid orbitals arrange themselves in a trigonal planar geometry to minimize electron repulsion.
In this configuration,the carbon nucleus lies at the center of an equilateral triangle,and the three $sp^2$ orbitals point towards the corners of the triangle.
Therefore,the bond angle between any two $sp^2$ orbitals is $120^{\circ}$.

(C) To determine the hybridization,we use the formula: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs}$.
For $H_3C^{+}$,the carbon atom is bonded to $3$ hydrogen atoms and has $0$ lone pairs. $\text{Steric Number} = 3 + 0 = 3$,which corresponds to $sp^2$ hybridization.
For $NH_3$,$PH_3$,and $SbH_3$,the central atoms are bonded to $3$ atoms and have $1$ lone pair. $\text{Steric Number} = 3 + 1 = 4$,which corresponds to $sp^3$ hybridization (though in $PH_3$ and $SbH_3$,the bond angles are close to $90^{\circ}$ due to the lack of significant hybridization,often described as using pure $p$-orbitals).

(A) Diamond is the hardest material known on Earth.
Each carbon atom is surrounded by four other carbon atoms,which lie at the corners of a regular tetrahedron.
Valence bonds connect each carbon atom with four others.
Carbon atoms in diamond are $sp^3$ hybridized,resulting in a tetrahedral geometry.

(B) The hybridization of the central atom can be determined using the formula: $\text{Steric Number} = (\text{Number of bonded atoms}) + (\text{Number of lone pairs})$.
For $BeF_2$: $\text{Steric Number} = 2 + 0 = 2$ ($sp$ hybridization).
For $BCl_3$: $\text{Steric Number} = 3 + 0 = 3$ ($sp^2$ hybridization).
For $C_2H_2$: The carbon atoms have a steric number of $2$ ($sp$ hybridization).
For $NH_3$: $\text{Steric Number} = 3 + 1 = 4$ ($sp^3$ hybridization).
Thus,the molecule with $sp^2$ hybridization is $BCl_3$.

(B) In an $sp$ hybridised orbital,one $s$ orbital and one $p$ orbital combine to form two $sp$ hybrid orbitals.
Therefore,the $s-$character is calculated as $\frac{1}{1+1} = \frac{1}{2}$ or $50 \,\%$.
$s-$orbitals hold electrons closer to the nucleus,so a hybrid orbital with $50 \,\%\, s-$character holds electrons more tightly.

(B) The carbonyl carbon in $CH_3COCl$ is bonded to three atoms (one $CH_3$ group,one $O$ atom via a double bond,and one $Cl$ atom).
It has $3$ sigma bonds and $0$ lone pairs.
The steric number is $3 + 0 = 3$.
Therefore,the hybridization of the carbonyl carbon is $sp^2$.

(C) $s-$character in $sp = \frac{1}{2} \times 100 = 50\%$
$s-$character in $sp^2 = \frac{1}{3} \times 100 = 33.3\%$
$s-$character in $sp^3 = \frac{1}{4} \times 100 = 25\%$
Hence,maximum $s-$character is found in $sp-$hybridisation.

(B) The central phosphorus atom in $PCl_5$ has $5$ valence electrons.
It forms $5$ covalent bonds with $5$ chlorine atoms.
Therefore,the number of hybrid orbitals required is $5$.
This corresponds to $sp^3d$ hybridisation,which results in a trigonal bipyramidal geometry.

(B) $Hybridisation$ is defined as the process of mixing atomic orbitals of slightly different energies to redistribute their energies,resulting in the formation of a new set of orbitals of equivalent energies and shape,known as hybrid orbitals.

(C) The bond angle of $sp^3$ hybrid orbitals is $109.5^o$.
The bond angle of $sp^2$ hybrid orbitals is $120^o$.
The bond angle of $sp$ hybrid orbitals is $180^o$.
An $sp$ orbital has $50\%$ $s$-character,$sp^2$ has $33.33\%$ $s$-character,and $sp^3$ has $25\%$ $s$-character.
As the $s$-character increases,the bond angle increases.
Therefore,the hybrid orbital that forms a bond at an angle of $120^o$ is $sp^2$.

(A) As the $p-$character increases,the bond angle decreases.
For $sp$ hybridization,the $p-$character is $\frac{1}{2}$ $(50\%)$,and the bond angle is $180^{\circ}$.
For $sp^2$ hybridization,the $p-$character is $\frac{2}{3}$ $(\approx 66.7\%)$,and the bond angle is $120^{\circ}$.
For $sp^3$ hybridization,the $p-$character is $\frac{3}{4}$ $(75\%)$,and the bond angle is $109.5^{\circ}$.
Thus,as the $p-$character increases,the bond angle decreases.

(A) $sp^3$ hybridization involves the mixing of one $s$ and three $p$ orbitals to form four equivalent $sp^3$ hybrid orbitals.
These orbitals are directed towards the corners of a regular tetrahedron to minimize electron repulsion.
Therefore,the geometry of a molecule with $sp^3$ hybridization is tetrahedral.

(A) The central atom $S$ in $SF_6$ undergoes $sp^3d^2$ hybridization.
Since there are $6$ bonding pairs and $0$ lone pairs of electrons around the sulfur atom,the geometry is octahedral.

(B) To determine the hybridisation,we calculate the number of electron pairs around the central atom using the formula: $\text{Number of hybrid orbitals} = \text{Number of sigma bonds} + \text{Number of lone pairs}$.
In $BeCl_2$,$Be$ has $2$ sigma bonds and $0$ lone pairs,so it is $sp$ hybridised.
In $BCl_3$,$B$ has $3$ sigma bonds and $0$ lone pairs,so it is $sp^2$ hybridised.
In $CCl_4$,$C$ has $4$ sigma bonds and $0$ lone pairs,so it is $sp^3$ hybridised.
Therefore,the hybrid orbitals are $sp$,$sp^2$ and $sp^3$ respectively.

(D) The central atom in $CCl_4$ is carbon $(C)$.
Carbon has $4$ valence electrons.
Each chlorine $(Cl)$ atom forms one single bond with carbon,contributing $4$ electrons for bonding.
The total number of electron pairs around the central carbon atom is calculated as: $\frac{1}{2} \times (V + M - C + A)$,where $V=4$ (valence electrons of $C$),$M=4$ (monovalent atoms),$C=0$ (cation charge),and $A=0$ (anion charge).
Number of electron pairs = $\frac{1}{2} \times (4 + 4) = 4$.
Since there are $4$ bond pairs and $0$ lone pairs,the hybridization is $sp^3$,which corresponds to a tetrahedral geometry.

(D) In the $NH_4^+$ ion,the nitrogen atom is $sp^3$ hybridized.
Due to $sp^3$ hybridization,the geometry of the $NH_4^+$ ion is tetrahedral.
Therefore,the $4$ hydrogen atoms are situated at the corners of a tetrahedron.

(B) An $sp^2$ hybridized atomic orbital is formed by the mixing of $1$ $s$ orbital and $2$ $p$ orbitals.
An $sp^2$ hybridized atom typically forms $3$ $\sigma$ bonds.
Due to the trigonal planar geometry,the bond angle between these orbitals is $120^o$.

(C) The bond angle depends on the hybridization of the central atom.
For $sp^3$ hybridization,the bond angle is approximately $109.5^o$.
For $sp^2$ hybridization,the bond angle is $120^o$.
For $sp$ hybridization,the bond angle is $180^o$.
Therefore,the increasing order of bond angle is $sp^3 < sp^2 < sp$.

(B) In $BF_3$,the central atom $B$ has $3$ valence electrons.
It forms $3$ single bonds with $3$ $F$ atoms,resulting in $3$ bond pairs and $0$ lone pairs.
The steric number is $3 + 0 = 3$,which corresponds to $sp^2$ hybridization.
These $3$ $sp^2$-hybrid orbitals are arranged in a trigonal planar geometry with a bond angle of $120^{\circ}$.

(A) The central nitrogen atom in $NH_4^+$ undergoes $sp^3$ hybridization.
Since there are $4$ bonding pairs and $0$ lone pairs of electrons around the nitrogen atom,the geometry of the ammonium ion is tetrahedral.

(D) In $sp$ hybridization,one $s$ orbital and one $p$ orbital mix to form two equivalent $sp$ hybrid orbitals.
These orbitals are oriented at an angle of $180^{\circ}$ to minimize repulsion.
Therefore,the resulting molecular geometry is linear.

(B) The bond angle depends on the hybridisation state of the carbon atom.
For $sp^3$ hybridisation,the bond angle is $109.5^\circ$.
For $sp^2$ hybridisation,the bond angle is $120^\circ$.
For $sp$ hybridisation,the bond angle is $180^\circ$.
Therefore,as the hybridisation state changes from $sp^3$ $\rightarrow sp^2$ $\rightarrow sp$,the bond angle increases gradually.

(D) In $Si(CH_3)_4$,the central silicon atom is bonded to $4$ methyl groups.
Since there are $4$ bond pairs and $0$ lone pairs around the central $Si$ atom,the steric number is $4+0 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation and a tetrahedral geometry.

(C) In diborane $(B_2H_6)$,each boron atom is bonded to four other atoms (two terminal hydrogen atoms and two bridging hydrogen atoms).
According to the valence bond theory,each boron atom undergoes $sp^3$ hybridisation to form four $sp^3$ hybrid orbitals.
These orbitals are used to form bonds with the hydrogen atoms.

(A) $CH_2 = CH_2$ (ethene) has a planar geometry because the carbon atoms are $sp^2$-hybridized,resulting in a bond angle of approximately $120^{\circ}$.
$(B)$ $HC \equiv CH$ (ethyne) is linear due to $sp$-hybridization.
$(C)$ $BeCl_2$ is linear due to $sp$-hybridization.
$(D)$ $CO_2$ is linear due to $sp$-hybridization.
Therefore,the compound that does not possess linear geometry is $CH_2 = CH_2$.

(C) To determine the hybridization of the central atom,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 $NO_2^+$:
Central atom $N$ $(V=5)$,$M=0$,$C=1$,$A=0$.
$\text{Steric Number} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. Hybridization is $sp$.
$2$. For $SF_4$:
Central atom $S$ $(V=6)$,$M=4$,$C=0$,$A=0$.
$\text{Steric Number} = \frac{1}{2} [6 + 4 - 0 + 0] = 5$. Hybridization is $sp^3d$.
$3$. For $PF_6^-$:
Central atom $P$ $(V=5)$,$M=6$,$C=0$,$A=1$.
$\text{Steric Number} = \frac{1}{2} [5 + 6 - 0 + 1] = 6$. Hybridization is $sp^3d^2$.
Thus,the correct hybridization modes are $sp$,$sp^3d$,and $sp^3d^2$.

(A) The central atom $P$ has $5$ valence electrons.
In $PF_3$,it forms $3$ $P-F$ sigma bonds and has $1$ lone pair.
The steric number is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 3 + 1 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridization.