Show by calculation that the dipole moment of methane is zero.

Explanation

Methane molecule is tetrahedral
`therefore mu_(CH_(3))=3mu_(C-H)cos(180^(@)-109^(@)28')`
`=3mu_(C-H)cos(70^(@)32')`
`=(3mu_(C-H))/(3)=mu_(C-H)`
`mu_(CH_(3))=mu_(C-H)=mu_(1)(say) and theta=180^(@)`
`therefore` The resultant dipole moment `(mu)`
`=sqrt(mu_(C-H)+mu^(2)(CH_(3))+2mu_(C-H)*mu_(CH_(3))cos180^(@))`
`=sqrt(mu_(1)^(2)+mu_(1)^(2)+2mu_(1)mu_(1)(-1))=sqrt(2mu_(1)^(2)-2mu_(1)^(2))=0`