Match the following :

A) In ${\text{PH}}_{\text{3}}$ molecule only London dispersion forces are present

E.M of phosphorous = 2.1

E.N of hydrogen = 2.1

Difference in E.N of “P” and $\text{H}=\text{2.1}-\text{2.1}=\text{0}$

E.N. difference is zero.  In ${\text{PH}}_{\text{3}}$ molecule no dipoles formed.  In the ${\text{PH}}_{\text{3}}$ molecule London dispersion forces are present.

B) E.N. of carbon = 2.5

E.N of hydrogen = 2.1

Difference in carbon and hydrogen $=\text{2.5}-2.1$

                                                                $=0.4$ 

A non-polar molecule may be polarized by the presence of polarmolecule near their by making it an induced dipole.

i) A dipole induced dipole interaction is due to the approach of a molecule with no permanent dipole.

ii) This approach causes electrons of non-polar molecule to polarized toward away from dipole of approaching molecule.

Dipole-dipole interactions:

In a dipole-dipole interaction the attraction between two polar molecules.  When they approach each others the negative end one molecule attract the positive end of other molecule.

Ex: ${\text{NH}}_{\text{3}}\text{ molecule}$

E.N of nitrogen = 3.0

E.N of hydrogen = 2.1

Difference is 0.9.