14. Amongst the molecules given below, the total number of molecules that have at least one sp2 hybridized atom is _____.
Total Number of Molecules Having at Least One sp2 Hybridized Atom
Given molecules: C6H6, NO2, BF3, H2O2, SO2, C2H2, and L-tryptophan
Correct Answer: 5
To determine the total number of molecules having at least one sp2 hybridized atom, each species must be examined individually. The important wording in the question is “at least one”. This means that every atom in the molecule does not need to be sp2 hybridized. A molecule is counted even if only one of its atoms has sp2 hybridization.
Among the seven given species, C6H6, NO2, BF3, SO2, and L-tryptophan contain at least one sp2 hybridized atom. H2O2 and C2H2 do not satisfy this condition. Therefore, the required total is 5.
How to Identify an sp2 Hybridized Atom
An atom is generally considered sp2 hybridized when one s orbital and two p orbitals combine to form three equivalent sp2 hybrid orbitals. One unhybridized p orbital remains available for π-bond formation or electron delocalization. The ideal geometry associated with sp2 hybridization is trigonal planar, with bond angles close to 120°.
For many main-group atoms, hybridization can be estimated by considering the number of effective electron domains or the steric number around the atom. A steric number of three generally corresponds to sp2 hybridization. However, resonance, delocalization, and molecular structure must also be considered, particularly for species such as NO2, SO2, and complex organic molecules.
Detailed Hybridization Analysis of Each Molecule
1. C6H6 – Benzene
Benzene, C6H6, definitely contains sp2 hybridized atoms. Each of the six carbon atoms in the benzene ring forms three sigma bonds and retains one unhybridized p orbital. The six unhybridized p orbitals overlap continuously around the ring to form the delocalized aromatic π-electron system.
Each carbon atom has approximately trigonal planar geometry, and the entire benzene molecule is planar. Therefore, all six carbon atoms in C6H6 are sp2 hybridized.
Conclusion: C6H6 is counted.
2. NO2 – Nitrogen Dioxide
Nitrogen dioxide, NO2, is an odd-electron molecule. The central nitrogen atom is surrounded by two bonding regions and an unpaired electron, giving an approximately trigonal planar electron-domain arrangement. The molecule is bent rather than linear, and the bonding is described using resonance between the two N–O bonds.
The central nitrogen atom is treated as sp2 hybridized. An unhybridized p orbital participates in the delocalized π-bonding system involving nitrogen and oxygen. Therefore, NO2 contains at least one sp2 hybridized atom.
Conclusion: NO2 is counted.
3. BF3 – Boron Trifluoride
In BF3, the central boron atom forms three sigma bonds with three fluorine atoms. Boron has three regions of electron density and no lone pair in the conventional Lewis structure. Therefore, its steric number is three.
The boron atom uses three sp2 hybrid orbitals to form three B–F sigma bonds. As a result, BF3 has a trigonal planar geometry with bond angles of approximately 120°.
Conclusion: BF3 is counted.
4. H2O2 – Hydrogen Peroxide
Hydrogen peroxide has the structure H–O–O–H. Each oxygen atom forms two sigma bonds: one O–H bond and one O–O bond. In addition, each oxygen atom possesses two lone pairs of electrons.
Thus, each oxygen atom has four electron domains around it. Four electron domains correspond to sp3 hybridization. There is no carbon-carbon or other conventional π bond requiring an sp2 hybridized atom in H2O2.
Conclusion: H2O2 is not counted.
5. SO2 – Sulfur Dioxide
In sulfur dioxide, SO2, the central sulfur atom has two bonding regions associated with the two oxygen atoms and one lone pair. Thus, there are three principal electron domains around sulfur.
A steric number of three corresponds to sp2 hybridization. The electron-domain geometry around sulfur is approximately trigonal planar, while the molecular geometry is bent because one of the three positions is occupied by a lone pair. The bonding also involves resonance and electron delocalization between the two S–O bonds.
Conclusion: SO2 is counted.
6. C2H2 – Ethyne or Acetylene
Ethyne has the structure H–C≡C–H. Each carbon atom forms one sigma bond with hydrogen and one sigma bond with the other carbon atom. The remaining two unhybridized p orbitals on each carbon participate in the formation of two π bonds, producing the carbon-carbon triple bond.
Since each carbon atom has only two regions of electron density, both carbon atoms are sp hybridized rather than sp2 hybridized. The molecule is linear, with a bond angle of 180°.
The presence of a multiple bond alone does not automatically mean that an atom is sp2 hybridized. A carbon-carbon double bond generally contains sp2 hybridized carbon atoms, whereas the carbon atoms of a carbon-carbon triple bond are sp hybridized.
Conclusion: C2H2 is not counted.
7. L-Tryptophan
L-Tryptophan is an aromatic amino acid containing an indole ring system. The fused aromatic ring contains several carbon atoms that are sp2 hybridized. These atoms possess unhybridized p orbitals that participate in the extended aromatic π-electron system.
In addition to the aromatic sp2 atoms, the carboxyl group of L-tryptophan also contains sp2 hybridized atoms because of the planar and resonance-stabilized nature of the –COOH group. Therefore, L-tryptophan clearly contains more than one sp2 hybridized atom and must be included in the count.
Conclusion: L-Tryptophan is counted.
Final Counting of Molecules Containing sp2 Hybridized Atoms
The molecules that contain at least one sp2 hybridized atom are:
C6H6 – Yes
NO2 – Yes
BF3 – Yes
H2O2 – No
SO2 – Yes
C2H2 – No
L-Tryptophan – Yes
Therefore, the total number of molecules satisfying the condition is:
1 + 1 + 1 + 0 + 1 + 0 + 1 = 5
Why C2H2 Is Not Included Despite Having a Multiple Bond
One of the most important distinctions in this problem is between sp and sp2 hybridization. C2H2 contains a carbon-carbon triple bond, but the carbon atoms involved in the triple bond are sp hybridized. Each carbon uses two sp hybrid orbitals for sigma bonding and retains two unhybridized p orbitals for the two π bonds.
Therefore, C2H2 must not be counted merely because it contains π bonds. The question specifically requires the presence of an sp2 hybridized atom.
Why L-Tryptophan Is Included
L-Tryptophan is a larger molecule containing atoms with different hybridizations. Its α-carbon is sp3 hybridized, but the question does not require every atom to be sp2 hybridized. The aromatic indole ring contains several sp2 hybridized atoms, which is sufficient to satisfy the condition.
This illustrates the importance of reading the phrase “at least one sp2 hybridized atom” carefully. Even if most atoms in a molecule have other hybridizations, the presence of a single sp2 hybridized atom is enough for the molecule to be counted.
Final Answer
Among C6H6, NO2, BF3, H2O2, SO2, C2H2, and L-tryptophan, the species containing at least one sp2 hybridized atom are C6H6, NO2, BF3, SO2, and L-tryptophan.
H2O2 contains sp3 hybridized oxygen atoms, while the carbon atoms in C2H2 are sp hybridized. Hence, these two species are excluded.
Correct Answer: 5
