Q. 234.0( 18 Votes )

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane

(ii) I2 and CCl4

(iii) NaClO4and water

(iv) methanol and acetone

(v) acetonitrile(CH3CN) and acetone(C3H6O)

Answer :

(i) Both the compounds are non-polar and they do not attract each other because they do not form any polar ions. Vanderwaals forces of attraction will be dominant in between them as vanderwaals forces of attraction are not a result of any chemical or electronic bond.

(ii) now here both the compounds are non-polar because in I2 both the atoms are same so they have same electronegativity and hence there will be no displacement of electron cloud, it will be in the centre. In case of CCl4 molecule, it has tetrahedral shape so two Cl atoms will cancel the attraction effect from two opposite Cl atoms, hence molecule as a whole is non polar. Therefore they will also have vanderwaals forces of attraction.

(iii) NaClO4 is ionic in nature as Na, Cl and O all have different electronegativity and their shape is also not symmetric. So molecular will be ionic in nature and we know that water is polar because O will attract the electron cloud towards it(more electronegative) hence there will be formation of dipole(two oppositely charged ions separated by a short distance). Therefore there will be ion-dipole interaction between them.

(iv) methanol and acetone are both polar molecules because of the presence of electron withdrawing O atom in methanol and ketone group in acetone. So they will have dipole-dipole interaction.

(v) acetonitrile is polar compound due to presence of electronegative N atom and acetone due to ketone group. So, dipole-dipole interaction.

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