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Question
What is the total number of sigma and pi bonds in the following molecules?
C2H4
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Solution 1
A single bond is a result of the axial overlap of bonding orbitals. Hence, it contributes a sigma bond. Multiple bonds (double or triple bonds) are always formed as a result of the sidewise overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-bonds and one sigma bond.
The structure of C2H4 can be represented as:
Hence, there are five sigma bonds and one pi-bond in C2H4.
Solution 2
\[\begin{array}{cc}
\ce{H}\phantom{......}\ce{H}\\
\backslash\phantom{......}/\\
\ce{C = C}\\
/\phantom{......}\backslash\\
\ce{H}\phantom{.......}\ce{H}
\end{array}\]
Sigma bond = 5
π bonds = 1
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\[\begin{array}{cc}
\phantom{.....}\ce{O}\\
\phantom{.....}||\\
\ce{\overset{∗}{C}H2 = CH - \overset{∗}{C} - O - H}
\end{array}\]
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\[\begin{array}{cc}
\phantom{..........}\ce{O}\\
\phantom{..........}||\\
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\end{array}\]
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\[\ce{\overset{∗}{C}H3 - CH = CH - CH3}\]
What is the type of hybridisation of carbon atoms marked with star.
\[\ce{CH3 - \overset{∗}{C} ≡ CH}\]
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(I) (SO3)
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The hybridisation of carbanion is:
