
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
213.3k+ 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
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions ForClass 11 Chemistry Chapter Chapter 5 Thermodynamics

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Collision: Meaning, Types & Examples in Physics

Atomic Structure: Definition, Models, and Examples

