
Number of $\sigma $ and $\pi $ bonds in ${C_2}$ molecule is/are:
(A) $1\sigma {\text{ and 1}}\pi $
(B) $1\sigma {\text{ and 2}}\pi $
(C) $2\pi {\text{ only}}$
(D) $1\sigma {\text{ and 3}}\pi $
Answer
220.8k+ views
Hint: Recall the molecular orbital theory (MOT) and write the electronic configuration of ${C_2}$ molecule according to MOT. You will find that the ${C_2}$ molecule has two sets of paired orbitals in the degenerate pi-bonding orbitals and bond order comes out to be 2. Thus, ${C_2}$ molecule will form two bonds and only these 4 electrons in the degenerate pi-bonding orbitals will be involved in bonding.
Complete step by step solution:
Diatomic carbon is a green-greyish inorganic compound. It has a chemical formula ${C_2}$ and written as $C = C$. It is a component of carbon vapour and is unstable at ambient temperature. Its IUPAC name is ethenediylidene or dicarbon.
Bonding in ${C_2}$ molecule: Configuration of ${C_2}$ molecule according to molecular orbital theory (MOT) is: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}$
The bond order of ${C_2}$ molecule is:
Bond order= $\dfrac{{{\text{no}}{\text{. of bonding electrons - no}}{\text{. of antibonding electrons }}}}{2} = \dfrac{{8 - 4}}{2} = 2$
Therefore, the bond order of ${C_2}$ molecule is two. This means there should exist a double bond between the two carbons in a ${C_2}$ molecule. But some studies show that a quadruple bond exists in dicarbon. MO theory also shows that the last two paired sets of electrons enter in the degenerate (having same energy) pi-bonding set of orbitals i.e. $\pi 2{p_x}$ and $\pi 2{p_y}$. These 4 electrons are in the pi orbitals and thus the two bonds in the ${C_2}$ molecule will be pi bonds only and no sigma bond. Usually, whenever there is a double bond, one is a sigma bond before a pi-bond. But this is not the case in ${C_2}$ molecules.
Thus, the number of $\sigma $ and $\pi $ bonds in ${C_2}$ molecule will be zero and two respectively.
Therefore, the correct option is C.
Note: Usually most people think that ${C_2}$ molecule, having 8 valence electrons, does not exist. But it does exist at very high temperatures and in the gaseous state. At low temperatures, ${C_2}$ aggregates to form many allotropic forms of carbon like buckyballs, nanotubes, graphene sheets, graphite, soot and so on. ${C_2}$ or carbon is diamagnetic in nature because all the electrons are paired.
Complete step by step solution:
Diatomic carbon is a green-greyish inorganic compound. It has a chemical formula ${C_2}$ and written as $C = C$. It is a component of carbon vapour and is unstable at ambient temperature. Its IUPAC name is ethenediylidene or dicarbon.
Bonding in ${C_2}$ molecule: Configuration of ${C_2}$ molecule according to molecular orbital theory (MOT) is: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}$
The bond order of ${C_2}$ molecule is:
Bond order= $\dfrac{{{\text{no}}{\text{. of bonding electrons - no}}{\text{. of antibonding electrons }}}}{2} = \dfrac{{8 - 4}}{2} = 2$
Therefore, the bond order of ${C_2}$ molecule is two. This means there should exist a double bond between the two carbons in a ${C_2}$ molecule. But some studies show that a quadruple bond exists in dicarbon. MO theory also shows that the last two paired sets of electrons enter in the degenerate (having same energy) pi-bonding set of orbitals i.e. $\pi 2{p_x}$ and $\pi 2{p_y}$. These 4 electrons are in the pi orbitals and thus the two bonds in the ${C_2}$ molecule will be pi bonds only and no sigma bond. Usually, whenever there is a double bond, one is a sigma bond before a pi-bond. But this is not the case in ${C_2}$ molecules.
Thus, the number of $\sigma $ and $\pi $ bonds in ${C_2}$ molecule will be zero and two respectively.
Therefore, the correct option is C.
Note: Usually most people think that ${C_2}$ molecule, having 8 valence electrons, does not exist. But it does exist at very high temperatures and in the gaseous state. At low temperatures, ${C_2}$ aggregates to form many allotropic forms of carbon like buckyballs, nanotubes, graphene sheets, graphite, soot and so on. ${C_2}$ or carbon is diamagnetic in nature because all the electrons are paired.
Recently Updated Pages
The hybridization and shape of NH2 ion are a sp2 and class 11 chemistry JEE_Main

What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main

Aromatization of nhexane gives A Benzene B Toluene class 11 chemistry JEE_Main

Show how you will synthesise i 1Phenylethanol from class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

Which of the following Compounds does not exhibit tautomerism class 11 chemistry JEE_Main

Trending doubts
Understanding Atomic Structure for Beginners

Understanding Entropy Changes in Different Processes

Common Ion Effect: Concept, Applications, and Problem-Solving

What Are Elastic Collisions in One Dimension?

Free Radical Substitution and Its Stepwise Mechanism

Understanding Geostationary and Geosynchronous Satellites

Other Pages
JEE Advanced 2026 Revision Notes for Practical Organic Chemistry

NCERT Solutions For Class 11 Chemistry Chapter 7 Equilibrium in Hindi - 2025-26

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Devuthani Ekadashi 2025: Correct Date, Shubh Muhurat, Parana Time & Puja Vidhi

Difference Between Exothermic and Endothermic Reactions: Key Differences, Examples & Diagrams

