Hybridisation Questions in English

(C) The structure of the compound $(CN)_4C_2$ is tetracyanoethylene,which can be represented as $(NC)_2C=C(CN)_2$.
In this molecule,the two central carbon atoms are bonded to two other carbon atoms via a double bond and two single bonds,making them $sp^2$ hybridized.
The four carbon atoms in the cyano $(-CN)$ groups are each bonded to one carbon atom and one nitrogen atom via a triple bond,making them $sp$ hybridized.
Therefore,the carbon atoms in the compound are both $sp$ and $sp^2$ hybridized.

(D) The structure of vinyl acetylene is $CH_2=CH-C \equiv CH$.
In this molecule:
- The two $sp^2$ hybridized carbon atoms each have $1$ unhybridised $p$-orbital.
- The two $sp$ hybridized carbon atoms each have $2$ unhybridised $p$-orbitals.
Total unhybridised $p$-orbitals $= (2 \times 1) + (2 \times 2) = 2 + 4 = 6$.

(C) $sp^2$ hybridization involves the mixing of one $s$ and two $p$ orbitals to form three equivalent hybrid orbitals.
These orbitals are oriented at an angle of $120^{\circ}$ to each other,resulting in a trigonal planar geometry.
Therefore,$sp^2$ hybridization is known as trigonal hybridization.

(A) In $CH_4$,the carbon atom undergoes $sp^3$ hybridization.
This results in a tetrahedral geometry for the molecule.
In a perfect tetrahedral geometry,the bond angle between any two $H-C-H$ bonds is $109^{\circ} 28^{\prime}$.

(C) In ethylene $(CH_2=CH_2)$,each carbon atom is $sp^2$ hybridized.
The geometry around each carbon atom is trigonal planar with bond angles of approximately $120^\circ$.

(A) The structure of acetylene is $HC \equiv CH$.
In this molecule,each carbon atom is bonded to one hydrogen atom by a single bond and to the other carbon atom by a triple bond.
Since each carbon atom is involved in one triple bond (one $\sigma$ and two $\pi$ bonds) and one single $\sigma$ bond,it has two $\sigma$ bonds and zero lone pairs.
Therefore,the steric number is $2$,which corresponds to $sp$-hybridisation.

(C) In a $CH_4$ molecule,the carbon atom is bonded to four hydrogen atoms through single covalent bonds.
According to the valence bond theory,the carbon atom undergoes $sp^3$ hybridisation in its excited state to form four equivalent $sp^3$ hybrid orbitals.
These four $sp^3$ hybrid orbitals overlap with the $1s$ orbitals of four hydrogen atoms to form four $C-H$ sigma bonds.
Therefore,carbon makes use of $sp^3$-hybridised orbitals.

(C) In graphite,each carbon atom is $sp^2$ hybridised.
Each carbon atom is covalently bonded to three other carbon atoms in a hexagonal planar arrangement.

(C) In diamond,each carbon atom is bonded to four other carbon atoms via $sp^3$ hybridised orbitals.
This results in a tetrahedral geometry around each carbon atom.

(B) The bond length depends on the hybridization of the carbon atoms involved.
As the $s$-character increases,the bond length decreases.
$C_2H_2$ ($sp$ hybridization,$1.20 \ \mathring{A}$) < $C_2H_4$ ($sp^2$ hybridization,$1.34 \ \mathring{A}$) < $C_6H_6$ (delocalized $sp^2$ hybridization,$1.39 \ \mathring{A}$) < $C_2H_6$ ($sp^3$ hybridization,$1.54 \ \mathring{A}$).
Thus,the correct order is $C_2H_2 < C_2H_4 < C_6H_6 < C_2H_6$.

(D) In $CH_4$,the carbon atom undergoes $sp^3$ hybridization.
It forms four equivalent $C-H$ sigma bonds directed towards the corners of a regular tetrahedron.
Therefore,the shape of the methane molecule is tetrahedral with a bond angle of $109.5^\circ$.

(C) Bond length is inversely proportional to the $s$-character of the hybrid orbital of the carbon atom.
$1.$ In $C_2H_2$ (Ethyne),the hybridization of carbon is $sp$,which has $50 \% \ s$-character.
$2.$ In $C_2H_4$ (Ethene),the hybridization of carbon is $sp^2$,which has $33.33 \% \ s$-character.
$3.$ In $C_2H_6$ (Ethane),the hybridization of carbon is $sp^3$,which has $25 \% \ s$-character.
Since $C_2H_6$ has the lowest $s$-character $(25 \%)$,the $C-H$ bond length is the greatest in $C_2H_6$.

(B) In ethylene $(CH_2=CH_2)$,each carbon atom is bonded to two hydrogen atoms and one other carbon atom via a double bond.
Each carbon atom is surrounded by three electron domains (two single bonds and one double bond),which corresponds to $sp^2$ hybridisation.
The geometry around each carbon atom is trigonal planar.

(B) In ethane $(CH_3-CH_3)$,each carbon atom is bonded to four other atoms (three hydrogen atoms and one carbon atom) by single bonds.
Therefore,each carbon atom undergoes $sp^3$ hybridization.
According to $VSEPR$ theory,$sp^3$ hybridized carbon atoms exhibit a tetrahedral geometry around the central atom.

(C) In the ethylene molecule $(C_2H_4)$,each carbon atom is bonded to two hydrogen atoms and one other carbon atom via a double bond.
Each carbon atom undergoes $sp^2$ hybridization.
Due to $sp^2$ hybridization,the bond angles are approximately $120^{\circ}$,and all atoms lie in the same plane.
Therefore,the shape of the ethylene molecule is planar.

(A) The shapes of methane $(CH_4)$,ethene $(C_2H_4)$ and ethyne $(C_2H_2)$ are tetrahedral,planar and linear respectively.
In methane,the $C$ atom is $sp^3$ hybridized,resulting in a tetrahedral geometry.
In ethene,the $C$ atoms are $sp^2$ hybridized,resulting in a planar geometry.
In ethyne,the $C$ atoms are $sp$ hybridized,resulting in a linear geometry.

(D) The bond length depends on the hybridization of the carbon atoms involved in the bond.
$sp^{3}-sp^{3}$ single bond length is $1.54 \, \mathring{A}$.
$sp^{2}-sp^{2}$ double bond length is $1.34 \, \mathring{A}$.
$sp-sp$ triple bond length is $1.20 \, \mathring{A}$.
In propyne $(CH_{3}-C \equiv CH)$,there is a $C \equiv C$ bond,which has the shortest bond length among the given options.

(C) The strength of a $C - C$ bond depends on the percentage of $s$-character in the hybrid orbitals.
Greater $s$-character leads to shorter and stronger bonds.
For $sp$ hybridization,$s$-character is $50\%$.
For $sp^2$ hybridization,$s$-character is $33.3\%$.
For $sp^3$ hybridization,$s$-character is $25\%$.
Therefore,the order of bond strength is $sp - sp > sp^2 - sp^2 > sp^3 - sp^3$.

(C) The correct answer is $C$.
In ethylene $(CH_2=CH_2)$,the carbon atoms are $sp^2$ hybridized,resulting in a bond angle of $120^{\circ}$.
In benzene $(C_6H_6)$,each carbon atom is also $sp^2$ hybridized,resulting in a bond angle of $120^{\circ}$.
Therefore,both ethylene and benzene have the same bond angle of $120^{\circ}$.

(A) In methane $(CH_4)$,the carbon atom forms four single bonds,resulting in $sp^3$ hybridization.
In ethene $(C_2H_4)$,each carbon atom is bonded to two hydrogen atoms and one other carbon atom via a double bond,resulting in $sp^2$ hybridization.
In ethyne $(C_2H_2)$,each carbon atom is bonded to one hydrogen atom and one other carbon atom via a triple bond,resulting in $sp$ hybridization.
Therefore,the correct sequence is $sp^3$,$sp^2$,and $sp$.

(C) In acetylene $(HC \equiv CH)$,both carbon atoms are $sp$-hybridized.
Due to $sp$-hybridization,the bond angle is $180^{\circ}$,which results in a linear structure.

(C) The correct answer is $C$.
In $C_6H_6$ (Benzene),all $6$ carbon atoms are $sp^2$-hybridized,which results in a trigonal planar geometry for each carbon atom.
Consequently,the entire molecule of Benzene is planar.

(B) $1$. For $NH_3$: The central atom $N$ has $3$ bond pairs and $1$ lone pair,so the hybridisation is $sp^3$.
$2$. For $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $d^8$ metal ion. With strong field ligands like $Cl^-$ (in square planar geometry),it undergoes $dsp^2$ hybridisation.
$3$. For $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs,resulting in $sp^3d$ hybridisation.
$4$. For $BCl_3$: The central atom $B$ has $3$ bond pairs and $0$ lone pairs,resulting in $sp^2$ hybridisation.
Therefore,the order is $sp^3, dsp^2, sp^3d, sp^2$.

(D) In the $CO_2$ molecule,the carbon atom undergoes $sp$ hybridization,resulting in a linear geometry.
In the $H_2CO_3$ molecule (carbonic acid),the carbon atom undergoes $sp^2$ hybridization.
Therefore,the statement that they have the same hybridization is incorrect.

(A) In $C_{60}$ (Buckminsterfullerene),each carbon atom is $sp^2$-hybridized.
It consists of $20$ hexagonal and $12$ pentagonal rings,which results in a truncated icosahedral structure,often referred to as a soccer ball shape.

(B) In graphite,each carbon atom is $sp^2$ hybridized,forming a layered structure with delocalized electrons.
In diamond,each carbon atom is $sp^3$ hybridized,forming a rigid three-dimensional tetrahedral network.

(A) In an $sp^3$ hybrid orbital,one $s$-orbital and three $p$-orbitals are mixed to form four equivalent hybrid orbitals.
Therefore,the $s$-character is $\frac{1}{1+3} = \frac{1}{4}$,which is equal to $25\%$.

(D) In $CCl_4$,the carbon atom undergoes $sp^3$ hybridization.
Therefore,it adopts a tetrahedral geometry where the four valencies are directed towards the four corners of a regular tetrahedron.

(C) Hybridization is the concept of mixing atomic orbitals of the same atom to form new hybrid orbitals of equivalent energy. It is not the same as the formation of molecular orbitals,which involves the overlap of atomic orbitals from different atoms.
In $dsp^2$ hybridization,the geometry is square planar. While adjacent hybrid orbitals are at $90^o$ to each other,the orbitals opposite to each other are at $180^o$. Therefore,the statement that all hybrid orbitals are at $90^o$ to each other is incorrect.

(C) In $CO_2$,the bond angle is $180^\circ$ due to $sp$ hybridization.
The structure is linear: $O=C=O$.

(D) For a central atom with $sp^3d$ hybridization,the electron pairs are arranged in a trigonal bipyramidal geometry to minimize repulsion.
Therefore,the correct geometry is $Trigonal \text{ } bipyramidal$.

(D) In $dsp^2$ hybridization,the molecular geometry is square planar.

(A) In $C_2H_2$ (acetylene),the structure is $HC \equiv CH$.
Each carbon atom is $sp$ hybridized,resulting in a linear geometry with a bond angle of $180^{\circ}$.

(B) In $BCl_3$,the boron atom undergoes $sp^2$ hybridization.
This results in a trigonal planar geometry for the molecule.

(B) The central atom in $SO_2$ is $S$. The number of hybrid orbitals is calculated as $1/2(V + M - C + A)$,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 $SO_2$: $V = 6$ (for $S$),$M = 0$ (since $O$ is divalent).
Number of hybrid orbitals = $1/2(6 + 0) = 3$.
$3$ hybrid orbitals correspond to $sp^2$ hybridization.

(C) The octahedral shape is associated with $sp^3d^2$ hybridization.
In $TeF_6$,the central atom $Te$ has $6$ valence electrons and forms $6$ bonds with $F$ atoms,resulting in $6$ bond pairs and $0$ lone pairs.
This corresponds to $sp^3d^2$ hybridization,which gives an octahedral geometry.

(C) In diamond,each carbon atom is bonded to four other carbon atoms in a tetrahedral geometry,resulting in $sp^3$ hybridization.
In graphite,each carbon atom is bonded to three other carbon atoms in a planar hexagonal arrangement,resulting in $sp^2$ hybridization.
In acetylene $(C_2H_2)$,each carbon atom is bonded to one hydrogen atom and one other carbon atom via a triple bond,resulting in $sp$ hybridization.
Therefore,the hybridization states are $sp^3, sp^2, sp$ respectively.

(A) In $AsF_5$,the hybridization of $As$ is $sp^3d$.
Therefore,the orbitals involved are one $s$,three $p$ $(p_x, p_y, p_z)$,and one $d$ $(d_{z^2})$ orbital.

(C) The given molecule is piperidine.
In piperidine,the nitrogen atom is bonded to two carbon atoms and one hydrogen atom,and it also possesses one lone pair of electrons.
Number of electron domains around $N = 2 \text{ (bonds to } C) + 1 \text{ (bond to } H) + 1 \text{ (lone pair)} = 4$.
Since the steric number is $4$,the hybridization of the nitrogen atom is $sp^3$.

(C) In $SO_4^{2-}$ and $BF_4^-$,the central atoms ($S$ and $B$) undergo $sp^3$ hybridization.
Both species have a tetrahedral geometry due to the absence of lone pairs on the central atom in the tetrahedral arrangement.

(A) In $sp^2$ hybridization,the three hybrid orbitals are oriented at an angle of $120^o$ to each other to minimize electron repulsion.
Therefore,the bond angle is $120^o$.

(D) In molecules with $sp^2$ hybridization,the $C - C$ bond length is $1.4 \ \mathring{A}$.

(C) The total number of valence electrons in $XeF_2$ is $8 + (2 \times 7) = 22$.
To find the hybridization,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,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $XeF_2$: $V = 8$,$M = 2$,$C = 0$,$A = 0$.
$X = \frac{1}{2} [8 + 2] = 5$.
$A$ steric number of $5$ corresponds to $sp^3d$ hybridization.

(A) In $SF_6$,the central sulfur atom is bonded to $6$ fluorine atoms with no lone pairs,resulting in $sp^3d^2$ hybridization.
This hybridization corresponds to an octahedral geometry.

(C) The molecule $CH_2 = C = CH_2$ (allene) is non-planar.
In this molecule,the central carbon atom is $sp$ hybridized,while the two terminal carbon atoms are $sp^2$ hybridized.
The two $CH_2$ groups lie in mutually perpendicular planes to minimize electronic repulsion,making the overall structure non-planar.

(A) To determine the hybridization of sulfur in $SO_2$,we calculate the steric number $(SN)$:
$SN = \frac{1}{2} [V + M - C + A]$
Where $V$ is the number of valence electrons of the central atom $(S = 6)$,$M$ is the number of monovalent atoms attached $(0)$,$C$ is the cationic charge $(0)$,and $A$ is the anionic charge $(0)$.
$SN = \frac{1}{2} [6 + 0 - 0 + 0] = 3$.
Since the steric number is $3$,the hybridization is $sp^2$.

(C) $PF_5$ undergoes $sp^3d$ hybridization and has a trigonal bipyramidal geometry.
In this structure,two fluorine atoms are at axial positions,while the other three are at equatorial positions.
The axial $P-F$ bonds are longer than the equatorial $P-F$ bonds due to greater repulsion experienced by the axial bonds from the equatorial bond pairs.

(A) In $IF_5$,the total number of valence electrons is $7 + (5 \times 7) = 42$.
Using the formula for steric number $(X)$:
$X = \frac{42}{8} = 5$ (quotient) and $2$ (remainder).
Then,$\frac{2}{2} = 1$ (quotient) and $0$ (remainder).
$X = 5 + 1 = 6$.
Since the steric number is $6$,the hybridization is $sp^3d^2$.

(C) The total number of valence electrons in $BeF_2$ is $2 + (2 \times 7) = 16$.
Using the formula for hybridization,$X = \frac{16}{8} = 2$.
$A$ steric number of $2$ corresponds to $sp$ hybridization.
Therefore,the beryllium atom in $BeF_2$ undergoes $sp$ hybridization.

(B) When one $s$ orbital and one $p$ orbital undergo hybridization,they form two equivalent $sp$ hybrid orbitals.
These two $sp$ hybrid orbitals are oriented in space at an angle of $180^o$ to each other,resulting in a linear geometry.