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B
C2H5−
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C
BeCl2
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D
C2H2
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Solution
The correct option is AC2H5− In C2H6, both carbons have sp3 hybridisation.
In C2H5−, both carbons have sp2 hybridisation.
In sp2 hybridization, one s orbital and two p orbitals hybridize to form three sp2 orbitals, each consisting of 33% s character and 67% p character. This type of hybridization is required whenever an atom is surrounded by three groups of electrons.
In BeCl2(Cl−Be−Cl), beryllium has sp hybridisation.