Hybridisation Questions in English

(C) The hybridization of a central atom is determined by the number of electron pairs (bonding pairs + lone pairs) around it.
For $sp^3d^2$ hybridization,the number of hybrid orbitals is $1 + 3 + 2 = 6$.
According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,a molecule with $6$ electron pairs arranged around a central atom adopts an octahedral geometry to minimize inter-electronic repulsions.
Therefore,the geometry of the molecule with $sp^3d^2$ hybridized central atom is octahedral.

(A) The correct option is $A$. The bond angle increases as the $s$-character of the hybrid orbital increases.

$Hybridization$	$s\%$	$Bond \ Angle$
$sp$	$50\%$	$180^{\circ}$
$sp^2$	$33.3\%$	$120^{\circ}$
$sp^3$	$25\%$	$109.5^{\circ}$

(A) To determine the hybridization,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.
$A$. For $PCl_3$: $\text{Steric Number} = \frac{1}{2} [5 + 3] = 4$. This corresponds to $sp^3$ hybridization.
$B$. For $SO_3$: $\text{Steric Number} = \frac{1}{2} [6 + 0] = 3$. This corresponds to $sp^2$ hybridization.
$C$. For $BF_3$: $\text{Steric Number} = \frac{1}{2} [3 + 3] = 3$. This corresponds to $sp^2$ hybridization.
$D$. For $NO_3^-$: $\text{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. This corresponds to $sp^2$ hybridization.
Therefore,$PCl_3$ is the correct answer.

(C) The $BCl_3$ molecule undergoes $sp^2$ hybridization at the central boron atom.
It has a trigonal planar geometry with a bond angle of $120^{\circ}$.
$NH_3$,$H_3O^{+}$,and $PCl_3$ all exhibit $sp^3$ hybridization with a trigonal pyramidal geometry due to the presence of one lone pair of electrons.

(C) For $BeCl_2$: The central atom $Be$ has $2$ valence electrons and forms $2$ bonds with $Cl$. It undergoes $sp$-hybridization and has a linear shape.
For $BCl_3$: The central atom $B$ has $3$ valence electrons and forms $3$ bonds with $Cl$. It undergoes $sp^2$-hybridization and has a trigonal planar (triangular planar) shape.
Therefore,the correct set is $BCl_3$,$sp^2$,triangular planar.

(A) In $CH_4$,the carbon atom is $sp^3$ hybridized. The percentage of $s$-character is $\frac{1}{4} \times 100 = 25\%$.
In $C_2H_4$ (ethene),the carbon atom is $sp^2$ hybridized. The percentage of $s$-character is $\frac{1}{3} \times 100 \approx 33.3\%$.
In $C_2H_2$ (ethyne),the carbon atom is $sp$ hybridized. The percentage of $s$-character is $\frac{1}{2} \times 100 = 50\%$.
Therefore,the correct sequence is $25, 33, 50$.

(C) $SO_2$ has $sp^2$ hybridization and a bent ($V$-shaped) structure with a bond angle of approximately $119^o$ due to one lone pair of electrons on the $S$ atom.
$CO_2$ and $N_2O$ are linear molecules with $sp$ hybridization.
$CO$ is also a linear molecule with $sp$ hybridization.

(A) $1$. $BF_3$ has a trigonal planar geometry with $sp^2$ hybridization,but it is non-polar due to the cancellation of dipole moments.
$2$. $SiF_4$ has a tetrahedral geometry with $sp^3$ hybridization and is non-polar.
$3$. $HClO_2$ has a central $Cl$ atom with $sp^3$ hybridization.
$4$. $H_2CO_3$ (carbonic acid) has a central carbon atom with $sp^2$ hybridization. Due to the asymmetric arrangement of atoms and the difference in electronegativity between $C$,$O$,and $H$,the molecule is polar.

(B) The hybridization of carbon in $C_2H_2$ $(HC \equiv CH)$ is $sp$.
Due to $sp$ hybridization,the bond angle is $180^{\circ}$,which results in a linear molecular geometry.
$CCl_4$ has a tetrahedral structure,$SO_2$ has a bent structure,and $C_2H_4$ has a planar trigonal structure around each carbon atom.

(D) The $sp^3d^2$ hybridization involves the mixing of one $s$,three $p$,and two $d$ orbitals.
This combination results in an octahedral geometry with bond angles of $90^{\circ}$.

(A) To determine the structure,we calculate the hybridization and number of lone pairs for each species:
$1$. $BF_4^-$: Boron has $3$ valence electrons. It forms $4$ bonds with $F$ atoms and gains $1$ electron from the negative charge. Total electron pairs = $(3+4+1)/2 = 4$. Hybridization is $sp^3$,which corresponds to a regular tetrahedral geometry.
$2$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $1$ lone pair. Total electron pairs = $5$ ($sp^3d$ hybridization). Due to the lone pair,it has a see-saw shape.
$3$. $XeF_4$: Xenon has $8$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $2$ lone pairs. Total electron pairs = $6$ ($sp^3d^2$ hybridization). Due to the two lone pairs,it has a square planar geometry.
$4$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has a $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridization is $dsp^2$,resulting in a square planar geometry.
Thus,only $BF_4^-$ has a regular tetrahedral structure.

(A) In boric acid $(H_3BO_3)$,the boron atom is bonded to three oxygen atoms via single bonds. Boron has $3$ valence electrons,all of which are involved in bonding,resulting in a trigonal planar geometry with $sp^2$ hybridization.
Each oxygen atom is bonded to one boron atom and one hydrogen atom,and it also possesses two lone pairs of electrons. Thus,each oxygen atom has $4$ electron domains (two bond pairs and two lone pairs),resulting in $sp^3$ hybridization.

(B) In the $BF_3$ molecule,the central Boron atom has $3$ valence electrons.
It forms $3$ single bonds with $3$ Fluorine atoms.
Using the formula for steric number: $\text{Steric Number} = \frac{1}{2} \times (V + M - C + A)$,where $V = 3$ (valence electrons of $B$),$M = 3$ (monovalent atoms),$C = 0$,and $A = 0$.
$\text{Steric Number} = \frac{1}{2} \times (3 + 3) = 3$.
$A$ steric number of $3$ corresponds to $sp^2$ hybridisation.

(D) To determine the hybridization of the central atom,we use 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.
$1$. For $BF_3$: $H = \frac{1}{2}(3 + 3) = 3$ ($sp^2$ hybridization).
$2$. For $NCl_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$ hybridization).
$3$. For $H_2S$: $H = \frac{1}{2}(6 + 2) = 4$ ($sp^3$ hybridization).
$4$. For $SF_4$: $H = \frac{1}{2}(6 + 4) = 5$ ($sp^3d$ hybridization).
$5$. For $BeCl_2$: $H = \frac{1}{2}(2 + 2) = 2$ ($sp$ hybridization).
Comparing the results,both $NCl_3$ and $H_2S$ have $sp^3$ hybridization. Therefore,the correct option is $(D)$.

(C) The central carbon atom in $CO_2$ is bonded to two oxygen atoms by double bonds.
According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,the carbon atom has two bond pairs and zero lone pairs,resulting in a linear geometry.
In terms of hybridisation,the carbon atom undergoes $sp$ hybridisation,where one $2s$ orbital and one $2p$ orbital mix to form two equivalent $sp$ hybrid orbitals.
These two $sp$ hybrid orbitals are oriented at an angle of $180^{\circ}$ to minimize repulsion,which explains the linear shape and $180^{\circ}$ bond angle of the $CO_2$ molecule.

(D) In $NH_3$,the central nitrogen atom undergoes $sp^3$ hybridization.
It has three bond pairs and one lone pair,resulting in a pyramidal geometry.

(B) The hybridisation 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.
$1$. For $NO_2$: $H = \frac{1}{2} [5 + 0 - 0 + 0] = 2.5$. Since $NO_2$ has an odd electron,it is considered $sp$ hybridised.
$2$. For $SF_4$: $H = \frac{1}{2} [6 + 4 - 0 + 0] = 5$,which corresponds to $sp^3d$ hybridisation.
$3$. For $PF_6^-$: $H = \frac{1}{2} [5 + 6 - 0 + 1] = 6$,which corresponds to $sp^3d^2$ hybridisation.
Thus,the correct set is $sp$,$sp^3d$,$sp^3d^2$.

(B) The atomic number of Boron $(B)$ is $5$. Its ground state electronic configuration is $1s^2 2s^2 2p^1$.
In the excited state,one electron from the $2s$ orbital is promoted to the $2p$ orbital,resulting in the configuration $1s^2 2s^1 2p_x^1 2p_y^1$.
These three orbitals $(2s, 2p_x, 2p_y)$ undergo $sp^2$ hybridisation to form three equivalent $sp^2$ hybrid orbitals.
These three $sp^2$ hybrid orbitals overlap with the $3p$ orbitals of three $Cl$ atoms to form three $B-Cl$ bonds.
Therefore,the hybridisation of $B$ in $BCl_3$ is $sp^2$.

(D) 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 $SO_3$,the central atom is $S$ (valence electrons $V = 6$). Oxygen is a divalent atom,so $M = 0$.
$H = \frac{1}{2}(6 + 0 - 0 + 0) = 3$.
$A$ value of $H = 3$ corresponds to $sp^2$ hybridization.
Thus,the $S$ atom in $SO_3$ is $sp^2$ hybridized,resulting in a trigonal planar structure.

(A) The central atom $I$ (Iodine) has the valence shell electronic configuration $5s^2 5p^5$.
In the excited state,$I$ promotes electrons to the $5d$ orbitals to form $7$ unpaired electrons to bond with $7$ fluorine atoms.
This results in the involvement of one $s$,three $p$,and three $d$ orbitals,leading to $sp^3d^3$ hybridization.
Therefore,the hybridization of $IF_7$ is $sp^3d^3$,and its geometry is pentagonal bipyramidal.

(B) In the ammonia molecule $(NH_3)$,the $N$ atom is $sp^3$-hybridized.
Due to the presence of one lone pair of electrons on the nitrogen atom,there is greater $Lp-Bp$ (lone pair-bond pair) repulsion compared to $Bp-Bp$ repulsion.
This causes the bond angles to decrease from the ideal tetrahedral angle of $109.5^\circ$ to $107^\circ$,resulting in a distorted tetrahedral or pyramidal geometry.

(A) The atomic number of $Be$ is $4$. The ground state electronic configuration is $1s^2, 2s^2$.
In the excited state,one electron from the $2s$ orbital is promoted to the $2p$ orbital,resulting in the configuration $1s^2, 2s^1, 2p^1$.
These two orbitals ($2s$ and $2p$) hybridize to form two equivalent $sp$ hybrid orbitals.
This type of hybridization is known as diagonal or linear hybridization,which results in a linear geometry for the $BeCl_2$ molecule.

(A) The trigonal bipyramidal geometry is associated with $sp^3d$ hybridization.
In this geometry,the central atom is surrounded by $5$ electron pairs,resulting in a trigonal bipyramidal arrangement,as seen in $PCl_5$.

(B) Carbon has only two unpaired electrons in its ground state configuration $(1s^2 2s^2 2p_x^1 2p_y^1)$.
Hybridization is the concept that explains how carbon achieves a valency of $4$ by mixing its $2s$ and $2p$ orbitals to form four equivalent $sp^3$ hybrid orbitals.

(C) Hybridization is the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies,resulting in the formation of a new set of orbitals of equivalent energies and shape. Therefore,it is due to the overlapping of orbitals of same energy content.

(D) For a molecule $MX_3$ to have a zero dipole moment,it must possess a trigonal planar geometry.
In a trigonal planar geometry,the central atom $M$ undergoes $sp^2$ hybridization.
These $3$ $sp^2$ hybrid orbitals form $3$ $\sigma$ bonds with $X$ atoms,resulting in a symmetric structure where the individual bond dipoles cancel each other out.

(A) The central atom in $NF_3$ is nitrogen $(N)$.
Nitrogen has $5$ valence electrons.
In $NF_3$,nitrogen forms $3$ sigma bonds with $3$ fluorine atoms and has $1$ lone pair.
The total number of electron pairs around the nitrogen atom is $3 + 1 = 4$.
Since the steric number is $4$,the hybridisation of the nitrogen atom is $sp^3$.

(C) In $sp^3d$ hybridisation,the central atom undergoes mixing of one $s$,three $p$,and one $d$-orbital to form five equivalent hybrid orbitals.
Specifically,the $d$-orbital involved is the $d_{z^2}$ orbital,which aligns along the $z$-axis to facilitate the trigonal bipyramidal geometry.

(C) The compound $X$ is carbon tetrachloride $(CCl_4)$.
$CCl_4$ undergoes $sp^3$ hybridization and possesses a perfectly symmetrical tetrahedral geometry because all four substituents attached to the central carbon atom are identical ($Cl$ atoms).
In molecules like chloromethane $(CH_3Cl)$,chloroform $(CHCl_3)$,or iodoform $(CHI_3)$,the presence of different atoms attached to the central carbon leads to unequal bond pairs and lone pair-bond pair repulsions,causing the bond angles to deviate from the ideal tetrahedral angle of $109^o 28'$.

(C) The correct answer is $(C)$.
$C_2H_2$ (acetylene) has a linear structure because both carbon atoms are $sp$-hybridized.
In $sp$-hybridization,the bond angle is $180^\circ$,resulting in a linear geometry.

(A) As the $s-$character of hybridized orbitals decreases,the bond angle also decreases.
In $sp^3$ hybridization: $s-$character is $25\%$,bond angle is $109.5^\circ$.
In $sp^2$ hybridization: $s-$character is $33.3\%$,bond angle is $120^\circ$.
In $sp$ hybridization: $s-$character is $50\%$,bond angle is $180^\circ$.
Therefore,as the $s-$character decreases $(50\%$ $\rightarrow 33.3\%$ $\rightarrow 25\%)$,the bond angle decreases $(180^\circ$ $\rightarrow 120^\circ$ $\rightarrow 109.5^\circ)$.

(A) The relationship between the bond angle $(\theta)$ and the percentage of $s-$character $(s)$ in a hybrid orbital is given by Coulson's formula: $\cos \theta = \frac{s}{s-1}$.
Given $\theta = 105^{o}$,we have $\cos(105^{o}) = -0.2588$.
Substituting this into the formula: $-0.2588 = \frac{s}{s-1}$.
$-0.2588(s-1) = s \implies -0.2588s + 0.2588 = s$.
$1.2588s = 0.2588 \implies s = \frac{0.2588}{1.2588} \approx 0.2056$.
Thus,the percentage of $s-$character is approximately $20.56\%$.
However,considering the standard approximation for $sp^{3}$ hybridization where $s-$character is $25\%$ at $109.5^{o}$,as the bond angle decreases due to lone pair repulsion,the $s-$character decreases. For $105^{o}$,the value falls in the range of $20-21\%$. Therefore,option $A$ is the correct choice.

(C) In $CH_3^+$,the central carbon atom is bonded to three hydrogen atoms and has no lone pair of electrons.
The hybridization of the carbon atom is $sp^2$,which corresponds to a trigonal planar geometry.
Therefore,the shape of the $CH_3^+$ species is trigonal planar.

(D) The number of $90^{\circ}$ bond pair-bond pair angles for different geometries are as follows:
$1$. $dsp^2$ hybridization (Square planar): There are $4$ angles of $90^{\circ}$ between adjacent bond pairs.
$2$. $sp^3d$ hybridization (Trigonal bipyramidal): There are $6$ angles of $90^{\circ}$ (three between equatorial and axial bonds,and three more between the other equatorial and axial bonds).
$3$. $sp^3d^2$ hybridization (Octahedral): There are $12$ angles of $90^{\circ}$ between adjacent bond pairs.
Comparing these,$sp^3d^2$ hybridization has the maximum number of $90^{\circ}$ angles. Therefore,the correct option is $D$.

(D) 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,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1$. For $SO_4^{2-}$: $SN = \frac{1}{2} (6 + 0 - 0 + 2) = 4$,which corresponds to $sp^3$ hybridization.
$2$. For $CH_4$: $SN = \frac{1}{2} (4 + 4 - 0 + 0) = 4$,which corresponds to $sp^3$ hybridization.
$3$. For $H_2O$: $SN = \frac{1}{2} (6 + 2 - 0 + 0) = 4$,which corresponds to $sp^3$ hybridization.
$4$. For $CO_2$: $SN = \frac{1}{2} (4 + 0 - 0 + 0) = 2$,which corresponds to $sp$ hybridization.
Therefore,$CO_2$ does not contain $sp^3$ hybridization.

(B) $sp$-hybridization involves the mixing of one $s$ and one $p$ orbital to form two equivalent $sp$ hybrid orbitals.
These two hybrid orbitals are oriented at an angle of $180^\circ$ to each other,resulting in a linear geometry.

(A) The hybridization of the central atom is 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 $ClO_2^-$,$V = 7$ (for $Cl$),$M = 0$ (oxygen is divalent),$C = 0$,and $A = 1$.
$H = \frac{1}{2}(7 + 0 - 0 + 1) = \frac{8}{2} = 4$.
$A$ value of $4$ corresponds to $sp^3$ hybridization.
Therefore,the chlorine atom in $ClO_2^-$ is $sp^3$ hybridized,resulting in an angular (bent) shape due to the presence of two lone pairs.

(B) The central atom in $ClF_3$ is chlorine $(Cl)$.
Chlorine has $7$ valence electrons.
It forms $3$ single bonds with $3$ fluorine atoms,using $3$ valence electrons.
This leaves $4$ electrons,which form $2$ lone pairs.
Total number of electron pairs = $\text{Bond pairs} + \text{Lone pairs} = 3 + 2 = 5$.
An electron pair count of $5$ corresponds to $sp^3d$ hybridisation.

(C) The central iodine atom in $IF_4^+$ has $7$ valence electrons.
It forms $4$ bonds with fluorine atoms,leaving $1$ lone pair of electrons.
Total electron pairs = $4 \text{ (bond pairs)} + 1 \text{ (lone pair)} = 5$.
Thus,the hybridization is $sp^3d$.
According to $VSEPR$ theory,$5$ electron pairs with $1$ lone pair result in a see-saw or irregular tetrahedral geometry.

(A) is the correct answer because $BeF_3^-$ does not exist as a stable species under standard conditions.
In $OH_3^+$ (hydronium ion),the $O$ atom is $sp^3$ hybridized with one lone pair.
In $NH_2^-$,the $N$ atom is $sp^3$ hybridized with two lone pairs.
In $NF_3$,the $N$ atom is $sp^3$ hybridized with one lone pair.

(A) To determine the hybridization of the chlorine atom in $ClO_2^-$,we use the formula: $H = \frac{1}{2} [V + M - C + A]$
Where $V$ is the number of valence electrons of the central atom $(Cl = 7)$,
$M$ is the number of monovalent atoms attached ($0$ in this case,as oxygen is divalent),
$C$ is the cationic charge $(0)$,
$A$ is the anionic charge $(1)$.
$H = \frac{1}{2} [7 + 0 - 0 + 1] = \frac{8}{2} = 4$.
$A$ value of $4$ corresponds to $sp^3$ hybridization.
$ClO_2^-$ has two lone pairs on the chlorine atom,which results in a $V$-shaped or bent structure.

(B) The central atom $Xe$ has $8$ valence electrons.
In $XeF_2$,$Xe$ forms $2$ single bonds with $F$ atoms,using $2$ electrons.
The remaining $6$ electrons form $3$ lone pairs.
The steric number is calculated as: $\text{Steric Number} = \text{Number of bond pairs} + \text{Number of lone pairs} = 2 + 3 = 5$.
$A$ steric number of $5$ corresponds to $sp^3d$ hybridisation.

(A) In $1, 1, 2, 2-$tetrachloroethene $(Cl_2C=CCl_2)$,each carbon atom is $sp^2$ hybridized,which corresponds to a trigonal planar geometry with a bond angle of approximately $120^{\circ}$.
In tetrachloromethane $(CCl_4)$,the central carbon atom is $sp^3$ hybridized,which corresponds to a tetrahedral geometry with a bond angle of $109.5^{\circ}$.
Therefore,the correct bond angles are $120^{\circ}$ and $109.5^{\circ}$ respectively.

(A) The structure is $CH_3^{III} - CH_2^{II} - C \equiv CH^I$.
Electronegativity is directly proportional to the $s$-character of the hybrid orbital.
$I$: The terminal carbon is $sp$ hybridized,which has $50\% \ s$-character.
$II$: The carbon in $CH_2$ is $sp^3$ hybridized,which has $25\% \ s$-character.
$III$: The carbon in $CH_3$ is $sp^3$ hybridized,which has $25\% \ s$-character.
Since $sp$ hybridization has the highest $s$-character,the carbon atom labeled $I$ is the most electronegative.

(B) In a $CH_4$ molecule,the carbon atom is $sp^3$ hybridized.
This hybridization results in four equivalent $C-H$ bonds directed towards the corners of a regular tetrahedron.
Therefore,the hydrogen atoms around the central carbon atom are arranged in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.

(C) .

Type	$s-\text{character}$
$sp^{3}$	$25\%$
$sp^{2}$	$33.33\%$
$sp$	$50\%$

The $s-$character is calculated as $\frac{1}{n+1} \times 100$,where $n$ is the number of $p-$orbitals involved in hybridization. For $sp$ hybridization,$n=1$,so $s-\text{character} = \frac{1}{1+1} \times 100 = 50\%$. This is the highest among the given options.

(A) In $C_2H_2$ (acetylene),each carbon atom is bonded to one hydrogen atom by a single bond and to the other carbon atom by a triple bond.
Each carbon atom is surrounded by two electron domains (one $C-H$ sigma bond and one $C-C$ sigma bond).
Therefore,the steric number is $2$,which corresponds to $sp$ hybridisation.
The structure is $\mathop {CH}\limits^{sp} \equiv \mathop {CH}\limits^{sp}$.

(A) In $CH_3 - C \equiv CH$,the carbon atom of the methyl group $(CH_3)$ is $sp^3$ hybridized,and the adjacent carbon atom involved in the triple bond is $sp$ hybridized. Therefore,the $C-C$ single bond between the methyl group and the alkyne carbon is formed by the overlap of an $sp^3$ orbital from the methyl carbon and an $sp$ orbital from the alkyne carbon. The structure is represented as: $\mathop{CH_3}\limits_{sp^3} - \mathop{C}\limits_{sp} \equiv CH$.

(A) In $CH_2Cl_2$,the central carbon atom is bonded to two hydrogen atoms and two chlorine atoms.
It undergoes $sp^3$ hybridization,which results in a tetrahedral geometry around the carbon atom.
Therefore,the correct option is $(A)$.

(D) In $CH_3^+$ (methyl carbocation),the central carbon atom is bonded to three hydrogen atoms and has a positive charge,resulting in three bonding pairs and zero lone pairs. The steric number is $3$,which corresponds to $sp^2$ hybridization.
In $C_2H_5^+$ (ethyl carbocation),the positively charged carbon atom is bonded to one methyl group and two hydrogen atoms,also resulting in three bonding pairs and zero lone pairs. The steric number is $3$,which corresponds to $sp^2$ hybridization.
Therefore,both $CH_3^+$ and $C_2H_5^+$ exhibit $sp^2$ hybridization.