The frequency of stretching vibrations is correlated to the strength and stiffness of the bond between two atoms. This can be thought of as a ball-and-spring model. Using this knowledge, rank the following bonds in each part of the question by increasing frequency.

(a) Alkyne

(b) Alkane

(c) alkene

a > c > b

Explanation:

As higher is the strength and stiffness of the bond between two atoms, more stable it is, and more difficult is to these bonds vibrate. So, the stretching vibration decreases when the strength and stiffness increases.

As more bonds are done between the atoms, more strength, and stiffness they have. So, the order of increase is:

simple bond > double bond > triple bond

And the increased frequency of vibration is:

triple bond > double bond > simple bond

An alkane is a hydrocarbon that has only simple bonds between carbons, an alkene is a hydrocarbon with one double bond between carbon, and an alkyne is a hydrocarbon with one triple bond. So, the increase in vibration of them is:

alkyne (a) > alkene (c) > alkane (b)

Answer: (A) < (C) < (B)

Ranking in order of increasing frequency.