Question

How many sigma and pi bonds are generally part of a triple bond?

Answer 1

Hint:With the head to head overlapping of orbitals a single bond is formed which is represented by a single $( - )$ . On the other hand, when two orbitals combine in such a way that they sidewise overlap it will form a pi bond represented as $( = )$ . Compounds are termed as saturated and unsaturated according to the type of bond they have.


Complete step by step answer:
In the bond formation, orbitals of two atoms must overlap to such an extent that there is maximum overlap between them. This will make a bond and they will combine, there are generally two types of overlapping which we see. If we take an example of a p orbital having two lobes, it will combine in two ways or there are two types of overlapping that we see. If an orbital combines like head to head that bond formed is called a sigma bond. It is represented as $C - C$
But when orbitals combine in sidewise overlapping fashion, it means without bending or turning for head to head overlapping, they are called to known as pi- bond and represented as $C = C$
Now the question arises that unsaturated compounds which are having double or triple bonds have how many sigma and pi bonds?
In a double bonded compound for example any olefin ${H_2}C = C{H_2}$ there is one sigma bond and one pi bond. In acetylene ${C_2}{H_2}$ which is also an unsaturated compound having a triple bond. In a triple bond there is a sigma bond formed by head to head overlapping and two pi bonds which are perpendicular to each other.


Note:A triple bond is always formed by two pi bonds, don’t write it as one pi bond. These two pi bonds are such that one pi bond is formed by ${p_y} - {p_y}$ overlapping and another pi bond is formed by overlapping of ${p_x} - {p_x}$ . These two orbitals are perpendicular to each other and also perpendicular to the sigma bond formed by head to head overlapping between ${p_z} - {p_z}$ .