Hybridisation Questions in English

(A) $sp^3$ hybridisation involves the mixing of one $s$ and three $p$ orbitals to form four equivalent $sp^3$ hybrid orbitals directed towards the corners of a tetrahedron.
Common geometries resulting from $sp^3$ hybridisation include tetrahedral (e.g.,$CH_4$),trigonal pyramidal (e.g.,$NH_3$),and bent/angular (e.g.,$H_2O$).
Triangular planar,See-Saw,$T$-shape,and linear geometries are associated with other hybridisation types (such as $sp^2$,$sp^3d$,or $sp^3d^2$) and are not produced by $sp^3$ hybridisation.
Therefore,all the given options are not produced by $sp^3$ hybridisation,but in the context of standard multiple-choice questions,this question is often flawed as multiple options are correct.

(B) $1$. Acetic acid $(CH_3COOH)$ contains a carbonyl group $(C=O)$,which involves an $sp^2$ hybridized carbon atom.
$2$. $C_3H_6$ can exist as propene $(CH_3-CH=CH_2)$,which contains $sp^2$ hybridized carbon atoms,or cyclopropane,where all carbons are $sp^3$ hybridized.
$3$. Acetonitrile $(CH_3CN)$ contains a cyano group $(-C \equiv N)$,where the carbon atom is $sp$ hybridized.
$4$. $C_2H_3Cl$ (vinyl chloride,$CH_2=CHCl$) contains $sp^2$ hybridized carbon atoms.
$5$. Cyclopropane $(C_3H_6)$ is a cyclic alkane where each carbon atom is bonded to two other carbons and two hydrogens,resulting in $sp^3$ hybridization for all carbon atoms.

(D) $A)$ Due to the small size and high charge density of $Be^{2+}$,its polarizing power is high,which leads to the sharing of electrons between $Be$ and $Cl$. This gives $BeCl_2$ significant covalent character.
$B)$ The $Be$ atom has only $4$ electrons in its valence shell,which does not satisfy the octet rule. Therefore,it is an electron-deficient molecule.
$C)$ $BeCl_2$ exists as a polymeric chain in the solid state and can form a dimer in the vapor phase.
$D)$ In solid $BeCl_2$,the $Be$ atom is $sp^3$ hybridized (due to coordinate bonding in the polymeric structure),and in the monomeric gaseous state,it is $sp$ hybridized. Thus,the statement that it is $sp^2$ is incorrect.

(D) In histamine,there are three nitrogen atoms labeled as $(I)$,$(II)$,and $(III)$:
$1$. Nitrogen $(I)$ is part of the imidazole ring and is bonded to one $H$ atom and two $C$ atoms. It has one lone pair. Its hybridization is $sp^2$,resulting in a trigonal planar geometry.
$2$. Nitrogen $(II)$ is part of the imidazole ring and is double-bonded to a $C$ atom. It has one lone pair. Its hybridization is $sp^2$,resulting in a trigonal planar geometry.
$3$. Nitrogen $(III)$ is an aliphatic primary amine nitrogen bonded to two $H$ atoms and one $C$ atom. It has one lone pair. Its hybridization is $sp^3$,resulting in a pyramidal (tetrahedral electron geometry) shape.
Since all statements $(A)$,$(B)$,and $(C)$ are correct,the correct option is $(D)$.

(C) $AlCl_3 + 6H_2O \to [Al(H_2O)_6]^{3+} + 3Cl^-$
In the complex ion $[Al(H_2O)_6]^{3+}$,the Aluminium ion is coordinated to $6$ water molecules.
The hybridization of $Al^{3+}$ in this octahedral complex is $sp^3d^2$.

(C) Hybridization is a concept used to explain the geometry of molecules by mixing atomic orbitals.
$(i)$ $BCl_3$: Boron $(Z=5)$ has ground state configuration $1s^2 2s^2 2p^1$. To form $3$ bonds,it promotes an electron to $2p$ to form $sp^2$ hybrid orbitals.
$(ii)$ $NH_3$: Nitrogen $(Z=7)$ has $1s^2 2s^2 2p^3$. It uses $sp^3$ hybridization in the ground state to form $3$ bonds and hold a lone pair.
$(iii)$ $PCl_3$: Phosphorus $(Z=15)$ has $3s^2 3p^3$. It uses $sp^3$ hybridization in the ground state to form $3$ bonds and hold a lone pair.
$(iv)$ $BeF_2$: Beryllium $(Z=4)$ has $1s^2 2s^2$. It promotes an electron to $2p$ to form $sp$ hybrid orbitals.
Therefore,$(ii)$ and $(iii)$ undergo hybridization in their ground state electronic configurations without needing promotion of electrons to higher energy subshells.

(A) The $S_8$ molecule has a crown-shaped puckered ring structure.
$I.$ Since all $S-S$ bonds are single covalent bonds,there are no $p_{\pi}-p_{\pi}$ multiple bonds present.
$II.$ Each sulfur atom has $6$ valence electrons. It forms $2$ single bonds with adjacent $S$ atoms,leaving $4$ electrons as $2$ lone pairs per $S$ atom. With $8$ sulfur atoms,the total number of lone pairs is $8 \times 2 = 16$. Thus,statement $II$ is incorrect.
$III.$ Each sulfur atom is bonded to $2$ other sulfur atoms and possesses $2$ lone pairs,resulting in a steric number of $4$. Therefore,each $S$ atom is $sp^3$ hybridized. This statement is correct.
Thus,statements $I$ and $III$ are true.

(B) In $SF_6$,the sulfur atom undergoes $sp^3d^2$ hybridization.
This hybridization involves one $s$,three $p$,and two $d$ orbitals.
The total number of orbitals involved is $1+3+2 = 6$.
The percentage of $d-$character is calculated as $\frac{\text{number of } d \text{ orbitals}}{\text{total number of orbitals}} \times 100 = \frac{2}{6} \times 100 = 33.33\% \approx 33\%$.
The geometry of $SF_6$ is octahedral,where all bond angles between adjacent $F-S-F$ bonds are $90^{\circ}$.

$(A)$ To determine the hybridisation of carbon atoms, we count the number of sigma bonds and lone pairs (steric number).
$1$. $sp^3$ hybridisation: $4$ sigma bonds.
$2$. $sp^2$ hybridisation: $3$ sigma bonds.
$3$. $sp$ hybridisation: $2$ sigma bonds.
In option $A$, $CH_2=CH-C\equiv N$:
- The first carbon $(CH_2=)$ has $3$ sigma bonds $\rightarrow sp^2$.
- The second carbon $(-CH-)$ has $3$ sigma bonds $\rightarrow sp^2$.
- The third carbon $(-C\equiv)$ has $2$ sigma bonds $\rightarrow sp$.
- The fourth carbon $(\equiv N)$ is nitrogen, but the carbon atom attached to nitrogen has $2$ sigma bonds $\rightarrow sp$.
Thus, the sequence is $sp^2, sp^2, sp, sp$.

(D) The relationship between the bond angle $(\theta)$ and the percentage of $s-$ character $(s\%)$ in hybrid orbitals is given by the formula: $\cos \theta = \frac{s}{s-1}$.
For a bond angle of $105^o$,$\cos(105^o) \approx -0.2588$.
Substituting this into the formula: $-0.2588 = \frac{s}{s-1}$.
Solving for $s$: $-0.2588(s-1) = s \implies -0.2588s + 0.2588 = s \implies 1.2588s = 0.2588 \implies s \approx 0.2056$.
This corresponds to approximately $20.56\%$ $s-$ character.
Among the given options,the range $23 - 25\%$ is the closest to the calculated value,as bond angles in molecules like $H_2O$ $(104.5^o)$ involve $sp^3$ hybridization with slightly less than $25\%$ $s-$ character due to Bent's rule.

(A) The size of a hybrid orbital depends on the percentage of $s$ and $p$ character.
As the $p-$character increases,the size of the hybrid orbital increases.
The $p-$character in $sp$ is $50\%$,in $sp^2$ is $66.6\%$,and in $sp^3$ is $75\%$.
Therefore,the increasing order of size is $sp < sp^2 < sp^3$.

(A) $1$. The central atom in $ICl_4^-$ is Iodine $(I)$.
$2$. The valence shell configuration of $I$ is $5s^2 5p^5$.
$3$. The number of electron pairs around $I$ is calculated as: $\frac{1}{2} \times (7 + 4 + 1) = 6$.
$4$. This corresponds to $sp^3d^2$ hybridization.
$5$. In $sp^3d^2$ hybridization,the two $d-$ orbitals involved are $d_{z^2}$ and $d_{x^2-y^2}$ to accommodate the square planar geometry with two lone pairs in the axial positions.

(C) In diborane $(B_2H_6)$,each boron atom is bonded to four hydrogen atoms.
Two hydrogen atoms are terminal,while two are bridging hydrogen atoms.
Each boron atom undergoes $sp^3$ hybridization to form four $sp^3$ hybrid orbitals.
Three of these orbitals contain one electron each,and one is empty,which helps in forming the $3c-2e$ (three-center two-electron) bonds.

(D) In $Si(CH_3)_4$ (tetramethylsilane),the central silicon atom is bonded to four methyl groups.
Silicon has $4$ valence electrons and forms $4$ sigma bonds with the methyl groups.
The steric number is $4 + 0 = 4$,which corresponds to $sp^3$ hybridization.
$A$ molecule with $4$ bonding pairs and no lone pairs on the central atom adopts a tetrahedral geometry.

(B) In graphite,each carbon atom is bonded to three other carbon atoms in a hexagonal planar structure,resulting in $sp^2$ hybridization.
In diamond,each carbon atom is bonded to four other carbon atoms in a tetrahedral geometry,resulting in $sp^3$ hybridization.
Therefore,the hybridization of carbon in graphite and diamond is $sp^2$ and $sp^3$ respectively.

(D) Carbon suboxide $(C_3O_2)$ has the chemical structure $O=C=C=C=O$.
In this molecule,each carbon atom is $sp$ hybridized.
The central carbon atom is bonded to two other carbon atoms,and the terminal carbon atoms are bonded to one carbon and one oxygen atom.
Due to $sp$ hybridization,the bond angle is $180^{\circ}$,which results in a linear structure.

(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 of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $NH_4^+$,the central atom is $N$ $(V = 5)$,$M = 4$ (four $H$ atoms),$C = 1$,and $A = 0$.
$H = \frac{1}{2} (5 + 4 - 1 + 0) = \frac{8}{2} = 4$.
$A$ value of $4$ corresponds to $sp^3$ hybridization.

(A) $1$. In $SO_2$,the central atom $S$ has one lone pair and two bond pairs (double bonds). The steric number is $1 + 2 = 3$,which corresponds to $sp^2$ hybridization.
$2$. In $CH_4$,the central atom $C$ has four bond pairs. The steric number is $4$,which corresponds to $sp^3$ hybridization.
$3$. In $NH_3$,the central atom $N$ has one lone pair and three bond pairs. The steric number is $1 + 3 = 4$,which corresponds to $sp^3$ hybridization.
$4$. In $SO_4^{2-}$,the central atom $S$ has four bond pairs. The steric number is $4$,which corresponds to $sp^3$ hybridization.
Therefore,$SO_2$ does not have $sp^3$ hybridization.

(B) The hybridization of an atom can be 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 $SO_2$,the central atom is sulfur $(S)$.
Valence electrons of $S = 6$.
Number of monovalent atoms ($O$ is divalent) = $0$.
$H = \frac{1}{2} (6 + 0 - 0 + 0) = 3$.
$A$ steric number of $3$ corresponds to $sp^2$ hybridization.
In $SO_2$,sulfur has one lone pair and two sigma bonds,resulting in $sp^2$ hybridization.

(A) The hybridization of an atom can be 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 charge on the cation,and $A$ is the charge on the anion.
For $SO_3$,the central atom is sulfur $(S)$.
Valence electrons of $S$ $(V)$ = $6$.
Number of monovalent atoms $(M)$ = $0$ (since oxygen is divalent).
$H = \frac{1}{2}(6 + 0) = 3$.
$A$ hybridization value of $3$ corresponds to $sp^2$ hybridization.
Thus,the sulfur atom in $SO_3$ is $sp^2$ hybridized.

(A) 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 $ClO_4^-$: $H = \frac{1}{2} [7 + 0 - 0 + 1] = 4$,which corresponds to $sp^3$ hybridization.
For $ClO_3^-$: $H = \frac{1}{2} [7 + 0 - 0 + 1] = 4$,which corresponds to $sp^3$ hybridization.
For $ClO_2^-$: $H = \frac{1}{2} [7 + 0 - 0 + 1] = 4$,which corresponds to $sp^3$ hybridization.
In all three ions,the chlorine atom is $sp^3$ hybridized due to the presence of lone pairs and bonding pairs.

(C) In $XeF_2$,the central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms and has $3$ lone pairs of electrons.
Total electron pairs = $2$ (bond pairs) + $3$ (lone pairs) = $5$.
For $5$ electron pairs,the hybridization is $sp^3d$.
In $XeF_4$,the central atom $Xe$ forms $4$ bonds with $F$ atoms and has $2$ lone pairs of electrons.
Total electron pairs = $4$ (bond pairs) + $2$ (lone pairs) = $6$.
For $6$ electron pairs,the hybridization is $sp^3d^2$.
Therefore,the correct statement is $XeF_2, sp^3d$.

(C) $S + 2Cl_2 \to SCl_4$
Hydrolysis of $SCl_4$ gives $H_2SO_3$ $(Y)$.
$SCl_4 + 3H_2O \to H_2SO_3 + 4HCl$
The anion of $Y$ is $SO_3^{2-}$.
In $SO_3^{2-}$,the central $S$ atom has $1$ lone pair and $3$ bond pairs,resulting in $sp^3$ hybridisation.
Due to the presence of one lone pair,the shape of $SO_3^{2-}$ is pyramidal.

(A) In $CH_3^+$,the carbon atom is bonded to $3$ hydrogen atoms and has a positive charge,resulting in $3$ sigma bonds and $0$ lone pairs. The steric number is $3$,which corresponds to $sp^2$ hybridisation.
In $CH_4$,the carbon atom is bonded to $4$ hydrogen atoms,resulting in $4$ sigma bonds and $0$ lone pairs. The steric number is $4$,which corresponds to $sp^3$ hybridisation.

(D) Benzene $(C_{6}H_{6})$ consists of $6$ carbon atoms,each of which is $sp^{2}$ hybridized.
Each $sp^{2}$ hybridized carbon atom forms $3$ $sp^{2}$ hybrid orbitals.
Therefore,the total number of $sp^{2}$ hybrid orbitals in a molecule of benzene is $6 \times 3 = 18$.

(N/A) Hybridization is defined as the process of intermixing of atomic orbitals of slightly different energies,thereby forming a new set of orbitals having equivalent energies and shapes.
$1$. Shape of $sp$ hybrid orbitals: $sp$ hybrid orbitals are formed by the intermixing of one $s$ and one $p$ orbital. They have a linear shape with a bond angle of $180^{\circ}$.
$2$. Shape of $sp^{2}$ hybrid orbitals: $sp^{2}$ hybrid orbitals are formed by the intermixing of one $s$ and two $p$ orbitals. They are oriented in a trigonal planar arrangement with bond angles of $120^{\circ}$.
$3$. Shape of $sp^{3}$ hybrid orbitals: $sp^{3}$ hybrid orbitals are formed by the intermixing of one $s$ and three $p$ orbitals. They are arranged in a tetrahedral geometry with bond angles of $109.5^{\circ}$.

(N/A) In $AlCl_{3}$,the aluminium atom is in the $sp^{2}$ hybridised state,resulting in a trigonal planar geometry.
When $AlCl_{3}$ reacts with $Cl^{-}$ to form $AlCl_{4}^{-}$,the lone pair from the chloride ion is donated into the empty $3p_{z}$ orbital of the aluminium atom.
Consequently,the hybridisation of the $Al$ atom changes from $sp^{2}$ to $sp^{3}$,and the geometry changes from trigonal planar to tetrahedral.

(N/A) In $BF_{3}$,the Boron atom is $sp^{2}$ hybridized. It has an empty $2p_{z}$ orbital.
In $NH_{3}$,the Nitrogen atom is $sp^{3}$ hybridized,possessing one lone pair of electrons.
During the reaction,the lone pair from the Nitrogen atom is donated to the empty $2p_{z}$ orbital of the Boron atom to form a coordinate covalent bond.
As a result,the Boron atom changes its hybridization from $sp^{2}$ to $sp^{3}$ to accommodate the four bonding pairs.
The hybridization of the Nitrogen atom remains $sp^{3}$ as it still maintains four electron domains (three bonding pairs and one lone pair).

(N/A) Both $C_1$ and $C_2$ are $sp^3$ hybridized.
$(b)$ $C_1$ is $sp^3$ hybridized,while $C_2$ and $C_3$ are $sp^2$ hybridized.
$(c)$ Both $C_1$ and $C_2$ are $sp^3$ hybridized.
$(d)$ $C_1$ is $sp^3$ hybridized and $C_2$ is $sp^2$ hybridized.
$(e)$ $C_1$ is $sp^3$ hybridized and $C_2$ is $sp^2$ hybridized.

(N/A) The ground state and excited state outer electronic configurations of phosphorus $(Z=15)$ are:
Ground state: $3s^{2} 3p^{3}$
Excited state: $3s^{1} 3p^{3} 3d^{1}$
Phosphorus atom is $sp^{3}d$ hybridized in the excited state. These orbitals are filled by the electron pairs donated by five $Cl$ atoms to form $PCl_{5}$.
The five $sp^{3}d$ hybrid orbitals are directed towards the five corners of a trigonal bipyramid.
There are five $P-Cl$ sigma bonds in $PCl_{5}$. Three $P-Cl$ bonds lie in one plane and make an angle of $120^{\circ}$ with each other. These are called equatorial bonds.
The remaining two $P-Cl$ bonds lie above and below the equatorial plane and make an angle of $90^{\circ}$ with the plane. These are called axial bonds.
As the axial bond pairs suffer more repulsion from the three equatorial bond pairs,the axial bonds are slightly longer than the equatorial bonds.

(N/A) $(i)$ $BF_3$: Boron in $BF_3$ is $sp^2$ hybridised. It has a planar triangular geometry. The three $sp^2$ hybrid orbitals of boron overlap with the $p$-orbitals of three fluorine atoms.
$(ii)$ $BH_4^-$: Boron in $BH_4^-$ is $sp^3$ hybridised. It has a tetrahedral geometry,where the four $sp^3$ hybrid orbitals of boron overlap with the $1s$ orbitals of four hydrogen atoms.

(A) The state of hybridisation of carbon is determined by the number of sigma bonds and lone pairs around the carbon atom.
$(a)$ In $CO_3^{2-}$,the carbon atom is bonded to three oxygen atoms with one double bond and two single bonds (resonance hybrid). It has $3$ sigma bonds and $0$ lone pairs,resulting in $sp^2$ hybridisation.
$(b)$ In diamond,each carbon atom is bonded to four other carbon atoms via single covalent bonds. It has $4$ sigma bonds and $0$ lone pairs,resulting in $sp^3$ hybridisation.
$(c)$ In graphite,each carbon atom is bonded to three other carbon atoms in a hexagonal planar layer. It has $3$ sigma bonds and $0$ lone pairs,resulting in $sp^2$ hybridisation.

(N/A) The $B-F$ bond length in $BF_{3}$ is shorter than the $B-F$ bond length in $BF_{4}^{-}$.
$BF_{3}$ is an electron-deficient species. With a vacant $p$-orbital on boron, the fluorine and boron atoms undergo $p\pi-p\pi$ back-bonding to remove this deficiency. This imparts a partial double bond character to the $B-F$ bond.
This double-bond character causes the bond length to shorten in $BF_{3}$ $(130 \ pm)$.
However, when $BF_{3}$ coordinates with the fluoride ion, a change in hybridisation from $sp^{2}$ (in $BF_{3}$) to $sp^{3}$ (in $BF_{4}^{-}$) occurs.
Boron now forms $4 \sigma$ bonds and the double-bond character is lost. This accounts for a $B-F$ bond length of $143 \ pm$ in the $BF_{4}^{-}$ ion.

(N/A) $H_2C=O$: The carbon atom is bonded to two hydrogen atoms and one oxygen atom via a double bond. It has $3$ sigma bonds and $0$ lone pairs,resulting in $sp^2$ hybridisation. The shape is trigonal planar.
$(b)$ $CH_3F$: The carbon atom is bonded to three hydrogen atoms and one fluorine atom via single bonds. It has $4$ sigma bonds and $0$ lone pairs,resulting in $sp^3$ hybridisation. The shape is tetrahedral.
$(c)$ $HC \equiv N$: The carbon atom is bonded to one hydrogen atom via a single bond and one nitrogen atom via a triple bond. It has $2$ sigma bonds and $0$ lone pairs,resulting in $sp$ hybridisation. The shape is linear.

(B) In graphite,each carbon atom is covalently bonded to three other carbon atoms in a hexagonal planar arrangement.
Each carbon atom forms three sigma bonds using $sp^2$ hybrid orbitals.
The fourth valence electron of each carbon atom remains in an unhybridized $p$-orbital,which participates in $\pi$-bonding.
Therefore,the hybridization of $C$ atoms in graphite is $sp^2$.

(N/A) The covalent bond is formed by the overlapping of atomic orbitals.
The molecule of hydrogen is formed due to the overlap of $1s$ orbitals of two $H$ atoms.
In the case of polyatomic molecules like $CH_4$,$NH_3$,and $H_2O$,the geometry of the molecules is also important in addition to the bond formation.
In $CH_4$,the shape is tetrahedral and the $H-C-H$ bond angle is $109.5^{\circ}$. $NH_3$ has a pyramidal shape.
To explain these directional properties,the concept of hybridization of atomic orbitals is required.

(N/A) In order to explain the characteristic geometrical shapes of polyatomic molecules like $CH_{4}$,$NH_{3}$,and $H_{2}O$,Pauling introduced the concept of hybridization.
Definition: The process of intermixing of 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,is known as hybridization.
Example: In $CH_{4}$,the carbon atom undergoes $sp^{3}$ hybridization,where one $2s$ orbital and three $2p$ orbitals mix to form four equivalent $sp^{3}$ hybrid orbitals,resulting in a tetrahedral geometry.

(N/A) To explain the characteristic geometric shapes of polyatomic molecules like $CH_4$,$NH_3$,and $H_2O$,Pauling introduced the concept of hybridization of atomic orbitals.
Pauling's definition: According to him,atomic orbitals combine to form a set of equivalent orbitals,which are known as 'hybrid orbitals'.
These hybrid orbitals are used in bond formation instead of pure atomic orbitals. The process of forming a new set of orbitals with equal energy and identical shape is called 'hybridization'.
Example: One $2s$ and three $2p$ orbitals of carbon mix and hybridize to form four new equivalent $sp^3$ hybrid orbitals of identical shape.

(N/A) The main characteristics of hybridization are as follows:
$i$. The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridized.
$ii$. The hybridized orbitals are always equivalent in energy and shape.
$iii$. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
$iv$. These hybrid orbitals are directed in space in some preferred direction to have minimum repulsion between electron pairs and thus a stable arrangement.
Therefore,the type of hybridization indicates the geometry of the molecules.

(B) The requirements for hybridization are as follows:
$1$. The orbitals present in the valence shell of the atom undergo hybridization.
$2$. The orbitals undergoing hybridization should have almost equal energy.
$3$. Promotion of an electron is not an essential condition prior to hybridization.
$4$. It is not necessary that only half-filled orbitals participate in hybridization; in some cases,even fully filled orbitals of the valence shell take part in hybridization.

(N/A)

Atomic orbitals	Hybrid orbitals
Atomic orbitals have different energies.	Hybrid orbitals are of the same type and have the same energy.
Atomic orbitals have different shapes.	Hybrid orbitals have the same shape.
Atomic orbitals can form both $\sigma$ and $\pi$ bonds.	Hybrid orbitals form only $\sigma$ bonds.
Atomic orbitals undergo hybridization before bond formation.	Hybrid orbitals directly participate in bond formation.
Atomic orbitals may contain lone pairs or unpaired electrons.	Hybrid orbitals can contain lone pairs or participate in bonding.
Rearrangement of electrons in atomic orbitals is possible in the excited state.	After hybridization,the electron configuration remains stable.

(N/A) $sp$ orbital: The two $sp$ hybrid orbitals are oriented in opposite directions along a straight line with an angle of $180^{\circ}$ between them.
$sp^2$ orbital: These three $sp^2$ hybrid orbitals are oriented in a trigonal planar arrangement,symmetrically at an angle of $120^{\circ}$ between them.
$sp^3$ hybridization: The four $sp^3$ hybrid orbitals are directed towards the four corners of a regular tetrahedron. The angle between any two $sp^3$ hybrid orbitals is $109^{\circ} 28^{\prime}$.

(N/A) In the gaseous state,$AlCl_3$ exists as a dimer $Al_2Cl_6$,where each $Al$ atom is $sp^3$ hybridized.
In the reaction $AlCl_3 + Cl^{-} \to AlCl_4^-$,the $Al$ atom in the product $AlCl_4^-$ also exhibits $sp^3$ hybridization.
Therefore,there is no change in the hybridization of the $Al$ atom during this reaction.

(N/A) In $BF_3$,the hybridization of $B$ is $sp^2$,but in the adduct $F_3B \cdot NH_3$,the hybridization of $B$ changes to $sp^3$.
In $NH_3$,the hybridization of $N$ is $sp^3$,and it remains $sp^3$ in the adduct $F_3B \cdot NH_3$.
Therefore,the hybridization of the $B$ atom changes from $sp^2$ to $sp^3$,while the hybridization of the $N$ atom remains $sp^3$.

Definition: The process of mixing one $s$ orbital and one $p$ orbital of the same atom to form two equivalent hybrid orbitals is called $sp$-hybridization.
Characteristics:
- In each $sp$-hybrid orbital,$50\%$ $s$-character and $50\%$ $p$-character is observed.
- Both equivalent $sp$-hybrid orbitals are arranged along the $Z$-axis at an angle of $180^{\circ}$ in opposite directions to each other.
- The two $sp$-hybrid orbitals point in opposite directions along the $Z$-axis with large positive lobes and small negative lobes. This provides more effective overlapping,resulting in the formation of stronger bonds.
- $A$ molecule in which the central atom is $sp$-hybridized and is linked directly to two other atoms possesses a linear geometry.

(N/A) In $BeCl_{2}$,the electronic configuration of $Be$ is $[He] 2s^{2}$.
It forms an excited state $Be^{*}: [He] 2s^{1} 2p^{1}$.
The divalency of $Be$ in the excited state occurs with one electron in the $2s$ orbital and one in the $2p$ orbital.
One $2s$ and one $2p$ orbital of the excited $Be$ atom undergo hybridization to form two $sp$ hybrid orbitals.
These two $sp$ hybrid orbitals are oriented in opposite directions,forming an angle of $180^{\circ}$.
The $Cl$ atom has the configuration $[Ne] 3s^{2} 3p_{x}^{2} 3p_{y}^{2} 3p_{z}^{1}$.
Each $sp$ hybrid orbital of $Be$ overlaps with the $3p_{z}$ orbital of a chlorine $(Cl)$ atom along the $z$-axis to form two $Be-Cl$ $\sigma$ bonds.
This axial overlapping results in a linear shape for the $BeCl_{2}$ molecule.

(N/A) In $BeCl_{2}$,the ground state electronic configuration of $Be$ is $[He] 2s^{2}$.
To form two bonds,$Be$ undergoes excitation to $Be^{*}$ with configuration $[He] 2s^{1} 2p^{1}$.
One $2s$ and one $2p$ orbital of the excited $Be$ atom undergo $sp$ hybridization to form two equivalent $sp$ hybrid orbitals.
These two $sp$ hybrid orbitals are oriented in opposite directions to minimize repulsion,resulting in a bond angle of $180^{\circ}$.
Each $sp$ hybrid orbital of $Be$ overlaps axially with the half-filled $3p_{z}$ orbital of a chlorine $(Cl)$ atom,forming two $Be-Cl$ $\sigma$ bonds.
Due to this linear arrangement of the two $sp$ hybrid orbitals,the $BeCl_{2}$ molecule adopts a linear geometry.

(N/A) Definition: The process of mixing one $ns$ orbital and two $np$ orbitals of an atom with comparable energies to form three equivalent new orbitals is called $sp^{2}$ hybridization. The resulting orbitals are known as $sp^{2}$ hybrid orbitals.
Characteristics:
$1$. Each $sp^{2}$ hybrid orbital has $33.3\%$ $s$-character and $66.6\%$ $p$-character.
$2$. The three $sp^{2}$ hybrid orbitals are oriented in space at an angle of $120^{\circ}$ to minimize repulsion,resulting in a trigonal planar geometry.
$3$. Each $sp^{2}$ hybrid orbital consists of a large positive lobe and a small negative lobe.
$4$. These hybrid orbitals overlap with $s$,$p$,or other hybrid orbitals of adjacent atoms to form $\sigma$ bonds.
$5$. The three $\sigma$ bonds formed by these orbitals lie in the same plane at an angle of $120^{\circ}$.

(N/A) In $sp^2$ hybridization,one $s$ orbital and two $p$ orbitals of the same shell mix to form three equivalent $sp^2$ hybrid orbitals.
Example: $BCl_3$ molecule.
The electronic configuration of Boron $(Z=5)$ in the ground state is $[He] 2s^2 2p^1$. In the excited state,one $2s$ electron is promoted to the $2p$ orbital,resulting in $[He] 2s^1 2p_x^1 2p_y^1$.
These three orbitals $(2s, 2p_x, 2p_y)$ undergo $sp^2$ hybridization to form three $sp^2$ hybrid orbitals. These orbitals are oriented in a trigonal planar geometry at an angle of $120^{\circ}$.
Each of these three $sp^2$ hybrid orbitals overlaps with the half-filled $3p$ orbital of a Chlorine atom to form three $B-Cl$ $\sigma$ bonds. Thus,$BCl_3$ has a trigonal planar geometry with $Cl-B-Cl$ bond angles of $120^{\circ}$.

(N/A) In the ground state,the electron configuration of Boron $(Z=5)$ is $B: [He] 2s^{2} 2p^{1}$.
In the excited state,one electron from the $2s$ orbital is promoted to the $2p$ orbital,resulting in the configuration $B^{*}: [He] 2s^{1} 2p_{x}^{1} 2p_{y}^{1}$.
Boron now has three half-filled orbitals $(2s, 2p_{x}, 2p_{y})$,which undergo $sp^{2}$ hybridization to form three equivalent $sp^{2}$ hybrid orbitals.
These three $sp^{2}$ hybrid orbitals are arranged in a trigonal planar geometry with a bond angle of $120^{\circ}$.
Each of these three $sp^{2}$ hybrid orbitals overlaps with a half-filled $3p$ orbital of a Chlorine atom to form three $B-Cl$ $\sigma$ bonds.
Thus,$BCl_{3}$ has a trigonal planar shape with $Cl-B-Cl$ bond angles of $120^{\circ}$.