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The C ≡ N triple bond in acetonitrile has a dipole moment of about 3.6Dand a bond length of about 1.16Å. Calculate the amount of charge separation in this bond. How important is the charge separated resonance form in the structure of acetonitrile?

Short Answer

Expert verified

Amount of charge separation is δ=0.6465e .

There is 64.65% of the charge separated resonance structure in resonance equilibrium.

Step by step solution

01

Bond dipole moment

The polarity of an individual bond is measured as its dipole moment (μ) which can be defined as μ=δ × d, where the amount of charge at either end of the dipole is represented by δ and the distance between the charges is represented by d . When common units are used, it can be written as:

μ (in debyes) =4.8 × δ(electroncharge) × d(in angstroms) .

02

Resonance structures

A single structural formula sometimes cannot explain all the properties of a compound that is given. In such cases, the compound may be represented by two or more structural formulae which differ from each other only in the arrangement of electrons. None of this structural formula alone can explain all the observed properties of the compound. The compound is then said to show resonance. The various structures are known as resonating structures.

03

Calculation of charge separation and relative importance of the resonating form

As per given data,

Dipole moment (μ) = 3.6D

Distance between the charges(d )= 1.16 â„«

Now,

μ (in debyes) =4.8 × δ(electroncharge) × d(in angstroms)

⟹ 3.6D = 4.8 × δ × 1.16 Å

⟹ δ = 3.6 ∕ 4.8 × 1.16

⟹ δ = 3.6 ∕ 5.568

⟹ δ = 0.6465 e

The amount δ of charge separation is about 0.6465 e (electronic charge).

The amount of charge separation is found to be 0.6465 e which implies that there is 64.65% of the charge separated resonance structure in resonance equilibrium.

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