Question

Number of \[\left( \pi \right)\] bonds in the following compound $\dfrac{{C{H_3}C \equiv CN}}{3}$
A.$1$
B.$2$
C.$3$
D.$none$

Answer 1

Hint: Different atoms undergo overlapping to form different types of bonds between them. Bonding formed between atoms is either sigma bond $\left( \sigma \right)$or pi bond $\left( \pi \right)$. Pi bond is formed when an atom undergoes side by side overlapping of two half-filled $\left( p \right)$ orbitals.

Complete answer:
Carbon atom has electronic configuration $\left( {1{s^2}2{s^2}2{p_x}^12{p_y}^1} \right)$ and has two unpaired electrons. Carbon transfers one electron from $\left( {2s} \right)$orbital to $\left( {2p} \right)$ orbital to make valency of 4.
Electronic configuration of nitrogen atom is $\left( {1{s^2}2{s^2}2{p_x}^12{p_y}^12{p_z}^1} \right)$ and has a valency of 3.
Now, $\left( {s - p} \right)$orbitals of nitrogen atom $\left( N \right)$and carbon atom undergoes overlapping to form a pi bond between both the atoms $\left( {C - N} \right)$.
It shows that the total number of pi bonds $\left( {C\underline{\underline \pi } N} \right)$between both the atoms are two.
Remaining orbitals undergo end-to-end overlapping to form sigma bonds in order to satisfy the valency of both carbon and nitrogen $\left( {C\underline {\sigma N} } \right)$.
Hence, cyanide group $\left( {C \equiv N} \right)$contains two pi bonds $\left( {C\underline{\underline \pi } N} \right)$and one sigma bond $\left( {C\underline \sigma N} \right)$.
Similarly, two atoms of carbon undergo lateral overlapping of orbitals to form two pi bonds and one sigma bond.
$\left( {C \equiv C} \right)$have two pi bonds $\left( {C\underline{\underline \pi } C} \right)$and one sigma bond $\left( {C\underline \sigma C} \right)$.
In the given molecule three atoms of hydrogen bonded with carbon atoms with sigma bond.
On calculating all the pi bond formed in the given molecule, it includes two pi bonds between $\left( {C\underline{\underline \pi } N} \right)$ and two pi bonds between $\left( {C\underline{\underline \pi } C} \right)$.
Number of \[\left( \pi \right)\]bonds in the following compound $\dfrac{{C{H_3}C \equiv CN}}{3}$
On solving it we get, $\pi = \dfrac{4}{3}$

Note:
Sigma bonds are quite stronger than pi bonds in a molecule. Sigma bonds are formed always in the same plane because of end-to-end orbital overlapping.
Pi bonds are not in a one plane because they are formed due to side-to-side overlapping which results in distribution of bonds half above the plane and remaining below the plane.