Questions on electronegativity, polarity and intermolecular forces

Question 1

Explain the trend in the boiling points of the Group 6 hydrides given in the table.

Accepted Answer

Indicative content

from H₂S to H₂Te the number of electrons in the molecule increases, from 18 to 54, as the Group 6 atom gets larger
so the induced dipole–dipole forces between the molecules become stronger, because a larger electron cloud is more easily distorted
and more energy is needed to overcome them, so the boiling point rises steadily from H₂S to H₂Te
water is anomalous: it has the fewest electrons, only ten, but by far the highest boiling point of the four
oxygen is small and highly electronegative, so water molecules form hydrogen bonds between a δ+ hydrogen and a lone pair on oxygen
hydrogen bonds are much stronger than the other intermolecular forces, so far more energy is needed to boil water

Levels

Level 3 (5–6 marks) the rise from H₂S to H₂Te is explained through more electrons and stronger induced dipole–dipole forces, water is identified as the anomaly, and hydrogen bonding is explained with its cause in the small, highly electronegative oxygen and its strength relative to the other forces; both the steady rise and the break at water are accounted for
Level 2 (3–4 marks) the rise from H₂S to H₂Te is explained and water is said to hydrogen bond; what causes the hydrogen bond, or why it outweighs the other forces, is not said, or water is never named as the break in the trend
Level 1 (1–2 marks) the boiling point is said to rise down the group or water to be unusual, often with the O–H bonds in water said to be harder to break than S–H

Question 2

Explain why electronegativity increases across Period 3 and decreases down Group 7.

Accepted Answer

B1 across the period the nuclear charge increases B1 the bonding electrons stay in the same shell with similar shielding, so they are attracted more strongly B1 down the group the atoms have more shells, so the bonding pair is further from the nucleus B1 and is more shielded by inner electrons, so it is attracted less strongly despite the greater nuclear charge

Question 3

Explain how induced dipole–dipole forces arise between two bromine molecules.

Accepted Answer

B1 the electrons in a Br₂ molecule are constantly moving, so at an instant they can be unevenly distributed B1 this gives the molecule an instantaneous dipole B1 the instantaneous dipole induces a dipole in a neighbouring molecule by attracting or repelling its electrons B1 the δ+ end of one dipole attracts the δ− end of the other, so the molecules attract

Question 4

Explain why water, H₂O, has a much higher boiling point than hydrogen sulfide, H₂S.

Accepted Answer

B1 oxygen is much more electronegative than sulfur, so the O–H bond is very polar and the hydrogen is strongly δ+ B1 water molecules form hydrogen bonds between a δ+ hydrogen and a lone pair on the oxygen of another molecule B1 H₂S molecules cannot form hydrogen bonds, only weaker permanent dipole–dipole and induced dipole–dipole forces B1 more energy is needed to overcome the hydrogen bonds between water molecules

Question 5

Describe the features of a diagram showing a hydrogen bond between two water molecules.

Accepted Answer

B1 partial charges shown: δ− on each oxygen and δ+ on each hydrogen B1 at least one lone pair shown on the oxygen that accepts the hydrogen bond B1 the hydrogen bond drawn as a dashed line from a δ+ hydrogen of one molecule to the lone pair on the oxygen of the other B1 the O–H···O atoms in a straight line, at 180°

Question 6

Compare the boiling points of butan-1-ol (118 °C) and pentane (36 °C), which have the same number of electrons.

Accepted Answer

B1 both have similar induced dipole–dipole forces, since they have the same number of electrons B1 butan-1-ol has an O–H group, so its molecules form hydrogen bonds to each other B1 pentane is non-polar and has only induced dipole–dipole forces B1 the hydrogen bonds need extra energy to overcome, so butan-1-ol boils about 80 °C higher

Question 7

Justify the order of the boiling points of the hydrogen halides given in the table.

Accepted Answer

B1 from HCl to HI the boiling point rises because each molecule has more electrons, so the induced dipole–dipole forces between molecules are stronger B1 the permanent dipole weakens from HCl to HI as the halogen becomes less electronegative, but the gain in induced dipole–dipole forces outweighs the loss in permanent dipole–dipole forces B1 HF boils highest although it has the fewest electrons because hydrogen bonds form between its molecules: fluorine is small and very electronegative and carries a lone pair B1 a hydrogen bond is much stronger than either of the other intermolecular forces, so more energy is needed to separate HF molecules; reject 'H–F bonds break'

Question 8

Explain why tetrachloromethane, CCl₄, is not a polar molecule even though its C–Cl bonds are polar.

Accepted Answer

B1 chlorine is more electronegative than carbon, so each C–Cl bond is polar, Cδ+–Clδ− B1 the molecule is tetrahedral and symmetrical B1 the four bond dipoles point in directions that cancel, so there is no overall dipole

Question 9

Explain why the boiling points of the halogens increase from fluorine to iodine.

Accepted Answer

B1 the molecules have more electrons down the group B1 so the induced dipole–dipole forces between the molecules are stronger B1 more energy is needed to overcome these forces, so the boiling point is higher

Question 10

Explain why pentane boils at 36 °C but its isomer 2,2-dimethylpropane boils at 10 °C.

Accepted Answer

B1 the two isomers have the same number of electrons B1 pentane is a straight chain, and its molecules touch each other over a larger area than the nearly spherical 2,2-dimethylpropane molecules B1 so the induced dipole–dipole forces between pentane molecules are stronger and need more energy to overcome

Question 11

Name the strongest type of intermolecular force between molecules of Br₂, of HCl and of CH₃OH.

Accepted Answer

B1 Br₂ has only induced dipole–dipole forces between its molecules B1 HCl has permanent dipole–dipole forces as its strongest force B1 CH₃OH forms hydrogen bonds through its O–H group

Question 12

Suggest whether trichloromethane, CHCl₃, is a polar molecule, and give a reason.

Accepted Answer

B1 trichloromethane is a polar molecule B1 the C–Cl bonds are polar and the C–H bond is almost non-polar B1 so the tetrahedral molecule is not symmetrical in charge, and the bond dipoles do not cancel

Question 13

Suggest why hydrazine, N₂H₄, boils at 114 °C while ethane, C₂H₆, which has the same number of electrons, boils at −89 °C.

Accepted Answer

B1 the induced dipole–dipole forces are similar, since both molecules have 18 electrons B1 hydrazine has H atoms bonded to N and lone pairs on N, so its molecules form hydrogen bonds; ethane cannot B1 the hydrogen bonds need much more energy to overcome, so hydrazine boils far higher

Question 14

Suggest why propanone, CH₃COCH₃, mixes with water although it cannot form hydrogen bonds between its own molecules.

Accepted Answer

B1 propanone has no hydrogen bonded to oxygen, so it cannot hydrogen bond to itself B1 its oxygen has lone pairs and is δ− B1 δ+ hydrogen atoms of water molecules form hydrogen bonds to those lone pairs, so propanone and water attract each other

Question 15

Evaluate the claim that water has a high boiling point because its O–H bonds are strong.

Accepted Answer

B1 the claim is wrong: no O–H bonds are broken when water boils, and the vapour is still H₂O molecules B1 boiling overcomes the forces between molecules, which in water are hydrogen bonds B1 the high boiling point is due to hydrogen bonding between water molecules, about two per molecule, not to the strength of the covalent bonds

Question 16

Explain why water boils higher than hydrogen fluoride, although an H–F bond is more polar than an O–H bond.

Accepted Answer

B1 each water molecule has two δ+ hydrogen atoms and two lone pairs B1 so it can form on average two hydrogen bonds; HF has only one hydrogen per molecule and averages one B1 more hydrogen bonds per molecule in water need more energy to overcome

Question 17

Explain why water has a much higher surface tension than hexane.

Accepted Answer

B1 a molecule in the body of a liquid is attracted equally in every direction, but a molecule at the surface has neighbours only beside and below it, so it is pulled inwards B1 in water the inward pull comes from hydrogen bonds; in hexane only from induced dipole–dipole forces B1 hydrogen bonds are stronger, so more energy is needed to enlarge the surface of water, and its surface tension is higher

Question 18

Suggest why 1-chlorobutane does not dissolve in water, although its molecules are polar.

Accepted Answer

B1 1-chlorobutane has no hydrogen bonded to N, O or F, so it cannot form hydrogen bonds with water B1 it can form only permanent dipole–dipole and induced dipole–dipole forces with water molecules B1 these are weaker than the hydrogen bonds between water molecules that would have to be broken, so it does not dissolve

Question 19

Identify which of NH₃, CH₄, HCl and HF form hydrogen bonds between their molecules.

Accepted Answer

B1 NH₃ and HF form hydrogen bonds, as each has H bonded to N or F and a lone pair on that atom B1 CH₄ and HCl do not, because carbon and chlorine are not electronegative enough

Question 20

Suggest why grease, a mixture of long-chain hydrocarbons, can be washed off a bicycle chain with hexane but not with water.

Accepted Answer

B1 grease and hexane are both non-polar, so the induced dipole–dipole forces they form with each other are like those broken in each, and the grease dissolves B1 grease can form only induced dipole–dipole forces with water, which cannot make up for the hydrogen bonds between water molecules that would have to break

Question 21

State what is meant by the term electronegativity.

Accepted Answer

B1 the power of an atom to attract the pair of electrons in a covalent bond towards itself

Hydrogen halide	Boiling point / °C
HF	20
HCl	−85
HBr	−67
HI	−35