ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 14, Problem 14B.3BE
Interpretation Introduction

Interpretation:

The potential energy of the interaction between two parallel linear quadrupoles which are separated by a distance r has to be calculated.

Concept introduction:

The dipole forces are attractive forces between two permanent dipoles.  The interaction is between the positive end of one polar molecule and the negative end of another polar molecule.  The induced dipole interaction occurs when an ion or a dipole induces a temporary dipole in an atom or a molecule with no dipole.  The induced dipole forces are the attraction forces between molecules having temporary dipoles.

Expert Solution & Answer
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Answer to Problem 14B.3BE

The potential energy of the interaction between two parallel linear quadrupoles which are separated by a distance r is q24πε0r2(68r2l2+2r24l2).

Explanation of Solution

The two parallel linear quadrupoles which are separated by a distance r are shown below.

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 14, Problem 14B.3BE

Figure 1

The potential energy due to point dipole-point charge interaction is shown below.

    V=Q1Q2l4πε0r2

Where,

  • Q1 is the charge on each ion of dipole.
  • Q2 is the charge on point ion.
  • l is the length of the dipole.
  • ε0 is the permittivity.
  • r is the distance between point dipole and point charge.

In case of two parallel linear quadrupoles which are separated by a distance r, there are total nine interactions are there.

The positive charge is 2q.

The negative charge is q.

The potential energy of the interaction between two parallel linear quadrupoles which are separated by a distance r calculated as shown below.

    V=14πε0(q2r2q2(r2+l2)1/2+q2(r2+4l2)1/22q2(r2+l2)1/2+4q2r2q2(r2+l2)1/2+q2r+2q2(r2+l2)1/2q2(r2+4l2)1/2)=q24πε0(1r2(r2+l2)1/2+1(r2+4l2)1/22(r2+l2)1/2+4r2(r2+l2)1/2+1r+2(r2+l2)1/21(r2+4l2)1/2)

The above equation can be written in terms of lr.

    V=q24πε0(1r2r(1+lr22)1/2+1r(1+4lr22)1/22r(1+lr22)1/2+4r2r(1+lr22)1/2+1r2r(1+lr22)1/2+1r(1+4lr22)1/2)V=q24πε0r(12(1+lr22)1/2+1(1+4lr22)1/22(1+lr22)1/2+42(1+lr22)1/2+12(1+lr22)1/2+1(1+4lr22)1/2)

The term lr can be supposed as x.

  V=q24πε0r(12(1+x2)1/2+1(1+4x2)1/22(1+x2)1/2+42(1+x2)1/2+12(1+x2)1/2+1(1+4x2)1/2)=q24πε0r(62(1+x2)1/2+1(1+4x2)1/22(1+x2)1/22(1+x2)1/22(1+x2)1/2+1(1+4x2)1/2)

The above equation can be further simplified as shown below.

  V=q24πε0r(68(1+x2)1/2+2(1+4x2)1/2)

The above expression can be rationalized as shown below.

    V=q24πε0r(68(1x2)1/2(1+x2)1/2(1x2)1/2+2(14x2)1/2(1+4x2)1/2(14x2)1/2)=q24πε0r(68(1x2)1/2(1x4)1/2+2(14x2)1/2(116x4)1/2)=q24πε0r(6(1x4)(116x4)8(116x4)(1x2)+2(1x4)(14x2)(1x4)(116x4))

The value of r is very much greater than l.  Therefore, x1 and the denominator can be assumed as 1.

    V=q24πε0r(6(1(116x4)x4(116x4))8(1(1x2)16x4(1x2))+2(1(14x2)x4(14x2))1)=q24πε0r(6116x4x4+16x881x216x4+16x6+214x2x44x61)

The terns with a higher value of x can be neglected.

  V=q24πε0r(6181x2+214x21)

The above equation after expanding the terms is shown below.

    V=q24πε0r(6181x2+214x21)=q24πε0r(681x2+214x21)

Substitute the value of x=lr in the above expression.

    V=q24πε0r(681l2r2+214l2r2)=q24πε0r(68r2l2r2+2r24l2r2)=q24πε0r2(68r2l2+2r24l2)

Therefore, the potential energy of the interaction between two parallel linear quadrupoles which are separated by a distance r is q24πε0r2(68r2l2+2r24l2).

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Chapter 14 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 14 - Prob. 14A.3DQCh. 14 - Prob. 14A.1AECh. 14 - Prob. 14A.1BECh. 14 - Prob. 14A.2AECh. 14 - Prob. 14A.2BECh. 14 - Prob. 14A.3AECh. 14 - Prob. 14A.3BECh. 14 - Prob. 14A.4AECh. 14 - Prob. 14A.4BECh. 14 - Prob. 14A.5AECh. 14 - Prob. 14A.5BECh. 14 - Prob. 14A.6AECh. 14 - Prob. 14A.6BECh. 14 - Prob. 14A.7AECh. 14 - Prob. 14A.7BECh. 14 - Prob. 14A.8AECh. 14 - Prob. 14A.8BECh. 14 - Prob. 14A.9AECh. 14 - Prob. 14A.9BECh. 14 - Prob. 14A.1PCh. 14 - Prob. 14A.2PCh. 14 - Prob. 14A.3PCh. 14 - Prob. 14A.4PCh. 14 - Prob. 14A.5PCh. 14 - Prob. 14A.6PCh. 14 - Prob. 14A.7PCh. 14 - Prob. 14A.8PCh. 14 - Prob. 14A.10PCh. 14 - Prob. 14A.12PCh. 14 - Prob. 14A.13PCh. 14 - Prob. 14B.1DQCh. 14 - Prob. 14B.2DQCh. 14 - Prob. 14B.3DQCh. 14 - Prob. 14B.4DQCh. 14 - Prob. 14B.5DQCh. 14 - Prob. 14B.1AECh. 14 - Prob. 14B.1BECh. 14 - Prob. 14B.2AECh. 14 - Prob. 14B.2BECh. 14 - Prob. 14B.3AECh. 14 - Prob. 14B.3BECh. 14 - Prob. 14B.4AECh. 14 - Prob. 14B.4BECh. 14 - Prob. 14B.5AECh. 14 - Prob. 14B.5BECh. 14 - Prob. 14B.6AECh. 14 - Prob. 14B.6BECh. 14 - Prob. 14B.1PCh. 14 - Prob. 14B.2PCh. 14 - Prob. 14B.3PCh. 14 - Prob. 14B.4PCh. 14 - Prob. 14B.5PCh. 14 - Prob. 14B.6PCh. 14 - Prob. 14B.7PCh. 14 - Prob. 14B.8PCh. 14 - Prob. 14B.10PCh. 14 - Prob. 14C.1DQCh. 14 - Prob. 14C.2DQCh. 14 - Prob. 14C.1AECh. 14 - Prob. 14C.1BECh. 14 - Prob. 14C.2AECh. 14 - Prob. 14C.2BECh. 14 - Prob. 14C.3AECh. 14 - Prob. 14C.3BECh. 14 - Prob. 14C.4AECh. 14 - Prob. 14C.4BECh. 14 - Prob. 14C.1PCh. 14 - Prob. 14C.2PCh. 14 - Prob. 14D.1DQCh. 14 - Prob. 14D.2DQCh. 14 - Prob. 14D.3DQCh. 14 - Prob. 14D.4DQCh. 14 - Prob. 14D.5DQCh. 14 - Prob. 14D.1AECh. 14 - Prob. 14D.1BECh. 14 - Prob. 14D.2AECh. 14 - Prob. 14D.2BECh. 14 - Prob. 14D.3AECh. 14 - Prob. 14D.3BECh. 14 - Prob. 14D.4AECh. 14 - Prob. 14D.4BECh. 14 - Prob. 14D.5AECh. 14 - Prob. 14D.5BECh. 14 - Prob. 14D.6AECh. 14 - Prob. 14D.6BECh. 14 - Prob. 14D.8AECh. 14 - Prob. 14D.8BECh. 14 - Prob. 14D.9AECh. 14 - Prob. 14D.9BECh. 14 - Prob. 14D.2PCh. 14 - Prob. 14D.3PCh. 14 - Prob. 14D.4PCh. 14 - Prob. 14D.6PCh. 14 - Prob. 14D.7PCh. 14 - Prob. 14D.8PCh. 14 - Prob. 14D.9PCh. 14 - Prob. 14D.10PCh. 14 - Prob. 14E.1DQCh. 14 - Prob. 14E.2DQCh. 14 - Prob. 14E.3DQCh. 14 - Prob. 14E.4DQCh. 14 - Prob. 14E.5DQCh. 14 - Prob. 14E.1AECh. 14 - Prob. 14E.1BECh. 14 - Prob. 14E.1PCh. 14 - Prob. 14E.3PCh. 14 - Prob. 14.1IACh. 14 - Prob. 14.2IACh. 14 - Prob. 14.6IACh. 14 - Prob. 14.8IA
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