0
407views
Conducting polymers.
1 Answer
0
2views

Polymeric materials has been synthesized which possess electrical conductivities on par with metallic conductors. Such polymers are called conducting polymers.

Conductivities as high as $1.5\times 10^7 ohm^{-1}m^{-1}$ have been attained in these polymeric materials.

On a volume basis, this value is equal to one-fourth of the conductivity of copper, or is twice it’s conductivity on the basis of weight.

INTRINSIC CONDUCTING POLYMER:

  • It is a polymer whose backbones or associated groups consist of delocalized electron-pair or residual charge.
  • Such polymers essentially contain conjugated$\pi$-electrons backbones, which is responsible for electrical charge.
  • Overlapping of orbitals over the entire backbone results in the formation of valence bands as well as conduction bands, which extends over the entire polymer molecule.
  • Presence of conjugated$\pi$-electron in a polymer increases its conductivity to a larger extent. For example: Poly-p-phenylene, polyquinoline, polyaniline, polypyrrole, etc.

DOPED CONDUCTING POLYMER:

  • It is obtained by exposing a polymer to a charged transfer agent in either gas phase or in solution.
  • Intrinsically conducting polymers possess low conductivity, but these possess low ionisation potential and high electron affinities, so these can be easily oxidised or reduced.

Consequently, the conductivity of intrinsically conducting polymer can be increased by creating either positive or negative charges on the polymer backbone by oxidation or reduction.

This technique is called doping. It is of the two types

  1. p-doping
  2. n-doping

Examples:

(i)

$\text{(C}_2\text{H}_2)_n + 2\text{FeCl}_3 \rightarrow \text{(C}_2\text{H}_2)_n^+\text{FeCl}_4^- + \text{FeCl}_2$

Where,

$\text{(C}_2\text{H}_2)_n$-Polyacetylene

$ \text{(C}_2\text{H}_2)_n^+\text{FeCl}_4^- $-Lewis acid

$2\text{(C}_2\text{H}_2)_n + 3\text{I}_2 \rightarrow 2[\text{(C}_2\text{H}_2)_n ^+\text{I}_3^-]$

(ii)

$\text{CH}=\text{CH}_\text{CH}=\text{CH}_ + \text{B} \xrightarrow{Reduction}_\text{CH}=\text{CH}-{\text{CH}}=\text{CH}$

Please log in to add an answer.