Hybridisation: Definition, Formula, Examples, Questions, C2, BF3, Water
Have you ever thought about how carbon manages to form such a vast variety of compounds, from the soft graphite in pencils to the hardest diamond? How do atoms arrange their orbitals so perfectly to form stable bonds in different shapes? The answer is by hybridisation. Hybridisation is a process in which atomic orbitals mix to create new, equivalent orbitals that determine the shape and stability of molecules.
- Hybridization
- Types Of Hybridisation
- How To Find Hybridisation
- sp Hybridization
- $\mathbf{s p}^2$Hybridization
- $\mathbf{s p}^3$Hybridization
- $s p^3 d$ Hybridisation
- $s p^3 d^2$Hybridization
- $\mathrm{d}^2 \mathrm{sp}^3$ Hybridisation
- $s p^3 d^3$ Hybridization
- Some Solved Examples
- Summary
In this article, we will study the hybridisation and the detail types of hybridisation structures and geomatry of compounds, how we can find hybridization. Some solved examples are also given in this article, and some extra practice question links are provided. To read more about hybridisation, scroll down.
Hybridization
Hybridization means mixing atomic orbitals to recombine into a new set of hybrid orbitals. Types of hybridization include $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3, \mathrm{sp}^3 \mathrm{~d}$, and $\mathrm{sp}^3 \mathrm{~d}^2$. These hybridizations correspond to different molecular geometries and bonding patterns. For instance, the sp3 hybridization gives a tetrahedral geometry, typical for molecules like methane, and CH4, while the hybridization of sp2 forms provides a trigonal planar geometry, seen for ethene, $\mathrm{C}_2 \mathrm{H}_4$. Hybrid orbitals are formed by mixing the s- and p- orbitals of an atom (and sometimes the d orbitals).
The Salient Features And Conditions For Hybridization
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Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.
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Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.
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A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.
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All orbitals in a set of hybrid orbitals are equivalent in shape and energy.
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The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.
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Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds.
Types Of Hybridisation
The hybridization can be of several types depending on the number of hybrid orbitals involved in the formation of molecules. The table given below describes all types of hybridization and their geometries.
How To Find Hybridisation
The hybridization depends upon sigma bonds and a lone pair of electrons.
Thus,
Hybridization = Number of sigma bonds + Number of lone pairs present on the central atom
For example, hybridization for NH3 is sp3 and its molecular geometry is tetrahedral.
NH3 has 3 sigma bonds and 1 lone pair, thus hybridization for NH3:
3 sigma bonds + 1 lone pair = 4
Thus hybridization for NH3 is sp3 and its geometry is tetrahedral.
sp Hybridization
This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry as shown in the figure. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180oC from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled.
$\mathbf{s p}^2$Hybridization
When 1 s-orbital and 2 p-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as sp2 hybrid orbitals. These hybrid orbitals arrange themselves at an angle of 120oC as shown in the figure.
$\mathbf{s p}^3$Hybridization
When 1 s-orbital and 3 p-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as sp3 hybrid orbitals. The bond angle between these hybrid orbitals is 109oC as shown in the figure.
$s p^3 d$ Hybridisation
When 1 s-orbital, 3 p-orbitals, and 1 d-orbital are involved in the molecule formation then the equivalent set of orbitals are known as sp3d hybrid orbitals. There are two kinds of bonds formed for sp3d hybridization, i.e., 2 axial bonds and 3 equatorial bonds. The angle between the axial bond and the equatorial plane is 90oC while the bond angle between the equatorial bonds is 120oC as shown in the figure given below:
$s p^3 d^2$Hybridization
When 1 s-orbital, 3 p-orbitals and 2 d-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as sp3d2 hybrid orbitals. There are two kinds of bonds formed for sp3d2 hybridisation, i.e., 2 axial bonds and 4 equatorial bonds. The angle between the axial bond and the equatorial plane is 90oC while the bond angle between the equatorial bonds is 90oC as shown in the figure given below:
$\mathrm{d}^2 \mathrm{sp}^3$ Hybridisation
When 2 d-orbital, 1 s-orbital and 3 p-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as d2sp3 hybrid orbitals. There are two kinds of bonds formed for sp3d2 hybridisation, i.e, 2 axial bonds and 4 equatorial bonds. The angle between the axial bond and the equatorial plane is 90oC while the bond angle between the equatorial bonds is 90oC as shown in the figure given below:
$s p^3 d^3$ Hybridization
When 1 s-orbital, 3 p-orbitals and 3 d-orbitals are involved in molecule formation then the equivalent set of orbitals are known as sp3d3 hybrid orbitals. The sp3d3 hybridization has a pentagonal bipyramidal geometry i.e., five bonds in a plane, one bond above the plane and one below it.
Recommended topic video on(Hybridisation)
Some Solved Examples
Example 1: The type of hybridization and number of lone pair (s) of electrons of Xe in XeOF4, respectively, are :
1) sp3d2 and 1
2) sp3d and 2
3) sp3d2and 2
4) sp3d and 1
Solution
As we have learned in Hybridisation:
Hybridisation is sp3d2sp3 d2
It contains $5 \sigma$ 5σ bond and $1 \pi$ 1π bond.
It has 1 lone pair of electrons.
Hence, the correct answer is Option (1)
Example 2: The correct statement about ICl5 and ICl4-is:
1)Both are isostructural
2)ICl5 is trigonal bipyramidal and ICl4-is tetrahedral
3)ICl5 is square bipyramidal and ICl4- is tetrahedral
4)ICl5 is square pyramidal and ICl4-is square planar.
Solution
ICl5 is square bipyramidal and Cl4−is square planar.
ICl5:−
Square Pyramidal
Lone Pair =1
Bond Pair=5
hybridisation= sp3d2
ICl4- :-
Square planar
Lone Pairs = 2
Bond Pairs = 4
hybridisation = sp3d2
Hence, the correct answer is Option (4)
Example 3: The orbitals undergoing Hybridization involve
1) Orbitals of the same atom with almost similar energies
2)Orbitals of different atoms but with equal energies
3)Orbitals of different atoms with different energies
4)Orbitals of the same atoms with exactly equal energies
Solution
The orbitals of the same atom having similar energies undergo hybridization to form hybrid orbitals which have the same energy.
Hence, the answer is option (1).
Example 4: In which pair of species, both species have a similar geometry?
1)CO2, SO2
2)CO23- and SO32−
3) SO42- and ClO4-
4)PH3 and BH3
Solution
The geometry of CO2 is linear.(O=C=O)(O=C=O)
The geometry of SO2 is V-shape
The geometry of CO32- is trigonal planar.
The geometry of SO32− is a pyramidal shape
The geometry of SO42− is tetrahedral
The geometry of ClO4− is tetrahedral
The geometry of NH3 is a pyramidal shape
The geometry of BH3 is trigonal planar
Hence, the answer is option (3).
Practice More Question With The Link Given Below
Summary
Hybridization describes mixing atomic orbitals to form hybrid orbitals, which allows for the determination of the shape and bonding properties of molecules. It was introduced by Linus Pauling, and it includes types such as sp, sp2, sp3, sp3d, and sp3d2—each being associated with specific molecular geometries. For example, sp3 hybridization would give a tetrahedral shape and sp2 results in a trigonal planar structure. In forming hybrid orbitals, the 's' and 'p' orbitals of an atom combine, and sometimes the 'd' orbitals, thereby forming orbitals that become equal in energy and shape.
Frequently Asked Questions (FAQs)
In phosphine, the phosphorus atom is sp3 hybridized. This results in four sp3 hybrid orbitals, three of which form bonds with hydrogen atoms, while the fourth contains a lone pair of electrons, similar to ammonia.
In the nitrate ion, the nitrogen atom is sp2 hybridized. This results in a trigonal planar geometry with three equivalent N-O bonds. The remaining p orbital on nitrogen participates in resonance, distributing the negative charge among all three oxygen atoms.
In methane, the carbon atom undergoes sp3 hybridization, forming four equivalent sp3 hybrid orbitals. These orbitals are arranged tetrahedrally, each forming a sigma bond with a hydrogen atom, resulting in the molecule's tetrahedral geometry.
In benzene, each carbon atom is sp2 hybridized. This results in three sp2 hybrid orbitals per carbon that form sigma bonds (two with adjacent carbons and one with hydrogen). The remaining p orbital on each carbon overlaps to form a delocalized pi system, explaining the planar, hexagonal structure.
Both oxygen in H2O and sulfur in H2S are sp3 hybridized. However, the bond angle in H2O (104.5°) is closer to the ideal tetrahedral angle (109.5°) than in H2S (92°). This is because oxygen's smaller size and higher electronegativity lead to greater repulsion between electron pairs, widening the angle.
Hybridization is the process of mixing atomic orbitals to form new hybrid orbitals with different shapes, energies, and electron distribution. This concept explains molecular geometry and bond formation in many compounds.
Hybridization is necessary because it helps explain molecular geometries and bond angles that cannot be accounted for by simple atomic orbital theory. It provides a framework for understanding how atoms can form multiple equivalent bonds in molecules like methane (CH4) or boron trifluoride (BF3).
Promotion refers to the theoretical excitation of an electron from a lower energy orbital to a higher energy orbital before hybridization occurs. For example, in carbon, an electron is promoted from the 2s to the 2p orbital before sp3 hybridization.
There are three main types of hybridization commonly encountered in organic chemistry: sp3, sp2, and sp. These correspond to the mixing of one s orbital with three, two, or one p orbital(s), respectively.
Atomic orbitals are the electron distributions in an isolated atom, while hybrid orbitals are mathematical combinations of atomic orbitals that better describe bonding in molecules. Hybrid orbitals have different shapes and energies compared to the original atomic orbitals.
The strength of carbon-carbon bonds is influenced by hybridization. Generally, the bond strength increases with increasing s-character: C(sp)-C(sp) > C(sp2)-C(sp2) > C(sp3)-C(sp3). This is because orbitals with more s-character hold electrons closer to the nucleus, resulting in stronger overlap.
In sulfur dioxide, the sulfur atom is sp2 hybridized. This results in a bent molecular geometry with an O-S-O bond angle of approximately 119°. The remaining p orbital on sulfur contains a lone pair of electrons.
Hybridization affects bond length: C(sp3)-C(sp3) bonds are longest, followed by C(sp2)-C(sp2), with C(sp)-C(sp) being shortest. This is due to increasing s-character from sp3 to sp, which pulls electrons closer to the nucleus, resulting in shorter, stronger bonds.
In acetylene, each carbon atom undergoes sp hybridization. This results in two sp hybrid orbitals (used for sigma bonding) and two unhybridized p orbitals (used for pi bonding) per carbon, explaining the linear geometry and triple bond.
In carbon dioxide, the carbon atom is sp hybridized. This results in two sp hybrid orbitals that form sigma bonds with the oxygen atoms, while the remaining two p orbitals participate in pi bonding, explaining the linear structure of the molecule.
Terminal alkynes are weakly acidic due to the sp hybridization of the carbon atom bonded to the hydrogen. The high s-character (50%) of the sp hybrid orbital makes the C-H bond more polarized, allowing for easier proton removal compared to sp2 or sp3 hybridized carbons.
The reactivity difference is related to hybridization: alkanes (sp3) are least reactive due to strong, localized sigma bonds. Alkenes (sp2) are more reactive due to the presence of a pi bond. Alkynes (sp) are most reactive because of the high s-character of the hybrid orbitals and the reactivity of the triple bond.
Hybridization affects carbocation stability. A tertiary carbocation (with an sp2 hybridized carbon) is more stable than a secondary or primary carbocation because the empty p orbital can better overlap with adjacent C-H bonds, allowing for hyperconjugation and greater electron delocalization.
In ethene, each carbon atom undergoes sp2 hybridization. This results in three sp2 hybrid orbitals and one unhybridized p orbital per carbon atom, allowing for the formation of a double bond between the carbons.
In boron trifluoride (BF3), the boron atom undergoes sp2 hybridization. This results in a trigonal planar geometry with 120° bond angles between the three B-F bonds.
In graphene, each carbon atom undergoes sp2 hybridization. This results in three sp2 hybrid orbitals that form sigma bonds with neighboring carbons in a planar hexagonal structure, while the remaining p orbital forms delocalized pi bonds above and below the plane.
In a carbonyl group, the oxygen atom is sp2 hybridized. This allows for the formation of a sigma bond with carbon using one sp2 hybrid orbital, while another sp2 orbital holds a lone pair. The remaining p orbital participates in pi bonding with carbon.
Hybridization affects reactivity by influencing factors such as bond angles, molecular shape, and electron distribution. For example, the sp-hybridized carbons in alkynes are more reactive in certain addition reactions compared to sp2 or sp3 hybridized carbons.
In water, the oxygen atom undergoes sp3 hybridization. Although there are only two bonding pairs and two lone pairs, the sp3 hybridization explains the observed bond angle of approximately 104.5°, which is close to the tetrahedral angle of 109.5°.
In ammonia, the nitrogen atom undergoes sp3 hybridization. This results in four sp3 hybrid orbitals, three of which form bonds with hydrogen atoms, while the fourth contains a lone pair of electrons.
Hybridization determines the arrangement of electron pairs around the central atom, which in turn dictates the molecular geometry. For example, sp3 hybridization leads to a tetrahedral arrangement, while sp2 hybridization results in a trigonal planar shape.
Hybridization can affect bond strength by changing the amount of s-character in the hybrid orbital. Generally, bonds formed by orbitals with more s-character (e.g., sp) are stronger than those formed by orbitals with less s-character (e.g., sp3).
Hybridization can influence an atom's electronegativity. As the s-character of hybrid orbitals increases (from sp3 to sp2 to sp), the electronegativity of the atom also increases due to the electrons being held more tightly.
In PCl5, the phosphorus atom undergoes sp3d hybridization. This results in five hybrid orbitals arranged in a trigonal bipyramidal geometry. Three equatorial positions are occupied by sp3d hybrid orbitals, while the two axial positions are occupied by hybrid orbitals with more d-character.
Hybridization can influence magnetic properties by affecting the number of unpaired electrons. For example, the sp2 hybridization in O2 leaves two unpaired electrons in separate p orbitals, making it paramagnetic. In contrast, the sp hybridization in CO results in all paired electrons, making it diamagnetic.
In sulfur hexafluoride, the sulfur atom undergoes sp3d2 hybridization. This involves the mixing of one s orbital, three p orbitals, and two d orbitals, resulting in six equivalent hybrid orbitals that form an octahedral arrangement.
sp3d hybridization involves the mixing of one s orbital, three p orbitals, and one d orbital, resulting in five hybrid orbitals and a trigonal bipyramidal geometry. sp3d2 hybridization mixes one s orbital, three p orbitals, and two d orbitals, creating six hybrid orbitals and an octahedral geometry.
In xenon tetrafluoride, the xenon atom undergoes sp3d2 hybridization. This involves the mixing of one s orbital, three p orbitals, and two d orbitals, resulting in six hybrid orbitals. Four of these form bonds with fluorine atoms, while two contain lone pairs, explaining the square planar geometry.
In iodine pentafluoride, the iodine atom undergoes sp3d2 hybridization. This results in six hybrid orbitals, five of which form bonds with fluorine atoms. The remaining orbital contains a lone pair of electrons, explaining the molecule's square pyramidal geometry.
In the gaseous state, phosphorus in PCl5 is sp3d hybridized, resulting in a trigonal bipyramidal geometry. However, in the solid state, it exists as an ionic compound [PCl4]+[PCl6]-, where phosphorus in [PCl4]+ is sp3 hybridized (tetrahedral) and in [PCl6]- is sp3d2 hybridized (octahedral).
In sulfuric acid, the sulfur atom is sp3 hybridized. This results in a tetrahedral arrangement around sulfur, with four sp3 hybrid orbitals forming sigma bonds (two with OH groups, two with oxygen atoms). The additional bonds to oxygen involve d-orbital participation, explaining the molecule's overall structure.
Hybridization can influence molecular polarity by affecting the distribution of electron density and molecular geometry. For example, the sp3 hybridization of carbon in methane (CH4) results in a tetrahedral structure with no net dipole moment, making the molecule non-polar.
In ethanol, the carbon atoms are sp3 hybridized, resulting in tetrahedral geometry around each carbon with bond angles close to 109.5°. The oxygen atom is also sp3 hybridized, explaining the bent shape of the O-H bond and its angle of approximately 108°.
In diborane, each boron atom is sp3 hybridized. However, there aren't enough electrons for all eight sp3 orbitals to form normal two-center two-electron bonds. This leads to the formation of three-center two-electron bonds, explaining the molecule's unique bridge structure.
In pyridine, the nitrogen atom is sp2 hybridized. This results in three sp2 hybrid orbitals (two forming bonds with adjacent carbons, one holding a lone pair) and one unhybridized p orbital that participates in the aromatic pi system of the ring.
In fullerene (C60), each carbon atom is sp2 hybridized. This results in a structure composed of hexagons and pentagons, where each carbon forms three sigma bonds using sp2 hybrid orbitals. The remaining p orbital contributes to a delocalized pi system across the molecule's surface.
In formaldehyde, the carbon atom is sp2 hybridized. This results in three sp2 hybrid orbitals arranged in a trigonal planar geometry. Two of these orbitals form bonds with hydrogen atoms, one forms a sigma bond with oxygen, and the remaining p orbital forms a pi bond with oxygen, resulting in a planar molecule.
In nitric acid, the nitrogen atom is sp2 hybridized. This results in a planar arrangement around nitrogen, with three sp2 hybrid orbitals forming sigma bonds (two with oxygen atoms, one with an OH group). The remaining p orbital participates in pi bonding with one of the oxygen atoms.
Ethane (C2H6) has sp3 hybridized carbons, allowing for free rotation around the C-C single bond. Ethene (C2H4) has sp2 hybridized carbons with a rigid double bond. This rigidity in ethene allows for stronger intermolecular forces (pi stacking), resulting in a higher boiling point compared to the more flexible ethane.
Cyclopropane's stability, despite its strained 60° bond angles, is partly explained by the concept of bent bonds. The carbon atoms use sp3-like hybrid orbitals that are slightly bent outwards, allowing for better orbital overlap. This "banana bond" structure helps distribute electron density and stabilize the molecule.
In cyclobutadiene, the carbon atoms are sp2 hybridized. This results in a square planar structure with alternating single and double bonds. However, the molecule is highly unstable due to antiaromaticity, which ar
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