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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.
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