
Number of and bonds in molecule is/are:
(A)
(B)
(C)
(D)
Answer
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Hint: Recall the molecular orbital theory (MOT) and write the electronic configuration of molecule according to MOT. You will find that the molecule has two sets of paired orbitals in the degenerate pi-bonding orbitals and bond order comes out to be 2. Thus, 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 and written as . It is a component of carbon vapour and is unstable at ambient temperature. Its IUPAC name is ethenediylidene or dicarbon.
Bonding in molecule: Configuration of molecule according to molecular orbital theory (MOT) is:
The bond order of molecule is:
Bond order=
Therefore, the bond order of molecule is two. This means there should exist a double bond between the two carbons in a 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. and . These 4 electrons are in the pi orbitals and thus the two bonds in the 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 molecules.
Thus, the number of and bonds in molecule will be zero and two respectively.
Therefore, the correct option is C.
Note: Usually most people think that molecule, having 8 valence electrons, does not exist. But it does exist at very high temperatures and in the gaseous state. At low temperatures, aggregates to form many allotropic forms of carbon like buckyballs, nanotubes, graphene sheets, graphite, soot and so on. 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
Bonding in
The bond order of
Bond order=
Therefore, the bond order of
Thus, the number of
Therefore, the correct option is C.
Note: Usually most people think that
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