Mg2C3 reacts with water forming propyne. C4−3 has:
A
Two sigma and two pi bonds
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B
Three sigma and one pi bonds
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C
Two sigma and one pi bonds
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D
Two sigma and three pi bonds
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Solution
The correct option is A Two sigma and two pi bonds Mg2C3+4H2O→2Mg(OH)2+C3H4 When propyne looses its four hydrogen, C4−3 is obtained. The structure is: −C≡C−C3− ∴ It has two σ bonds and two π bonds.