MeridianJunior College

2007JC2 Chemical BondingTime trial (Group A)

1Time allocation : 12 minutes

Draw dot-and-cross diagram to show the electronic structure of the following species. Draw and name the shape of the species. Indicate the polarity for molecules only.

Species / Dot-and-cross / Shape of Molecule / Polarity
(a) ClF5 /
/
Square pyramidal / Polar
(b)C2O42- /
/
Trigonal planar about
each C atom / NA
(c) ClF4- / /
square planar / NA
(d) NO3- / /

Trigonal planar / NA
(e) SbF52- /
/
square pyramidal / NA
(f) ClO2 / /
bent / Polar
(g) SO42- / /
tetrahedral / NA

[8]

HCJC Prelim : 10 mins

2N2O can be made in the laboratory by heating molten ammonium nitrate. The reaction is thought to occur by the following steps. ( NJC Prelim 2002modified)

NH4NO3 NH3 + HNO3

2HNO3 NO2+ + H2O + NO3-

NH3 + NO2+  [H3N - NO2]+  N2O + H3O+

H3O+ + NO3-  HNO3 + H2O

(a)Write a balanced equation for the overall equation.

Method : Add all the 4 equations together and cancel common intermediates

NH4NO3 N2O + 2H2O[1]

(b)Draw a displayed formulae showing the bonds and the bond angles in

(i)NO2+

(ii)N2O

(iii)[H3N - NO2]+

[7]

Time : 10 mins

3The following lists the boiling points of fluorine andsomefluoride compounds. By reference to their chemical structures and types of bonding, explain as fully as you can the differences in their boiling points.

Element / compound / Boiling point / oC
Hydrogen sulphide / -50
Hydrogen fluoride / 50
Methanoic acid with apparent Mr doubled / 100
Aluminium oxide / 1257
Alanine / 150

Al2O3 and Alanine

  • Aluminium oxide, Al2O3 has a giant ionic structure[1/2] while alanine (H2NCH(CH3)COOH) is an amino acid and exists as a zwitterions. [1/2]
  • Boiling these 2 compounds involves breaking the strong electrostatic forces of attraction between oppositely-charged ions[1]
  • A large amount of energy[½]is required, hence high melting points.

Methanoic acid and hydrogen fluoride

  • Methanoic acid (HCOOH), hydrogen sulphide and hydrogen fluoride all are simple molecular [1/2]structures.
  • For methanoic acid hydrogen fluoride, boiling involves breaking the weaker (when compared to ionic bond but stronger when compared to VDW’s ) intermolecular hydrogen bonds[½]
  • Methanoic acidexists as a dimer[½]and forms 2 hydrogen bonds per molecule[½]while hydrogen fluoride can only form 1 hydrogen bond per molecule[1/2]
  • Extent of intermolecular hydrogen bonds: methanoic acid > hydrogen fluoride[½]
  • Energy required to break intermolecular hydrogen bonds: methanoic acid > hydrogen fluoride[½]

H2S

  • For hydrogen sulphide, boiling involves the weak intermolecular Van der Waals’ forces,[1] hence require least amount of energy, [½]thus lowest boiling point.

[8]

RJC Prelim : 15 mins

4Explain the following observations in terms of structure and bonding

(a)There are two N2F2 with different boiling points

CONCEPT : N2F2 exists as 2 isomers; cis and trans isomers

[½]
Trans isomer
  • Dipole moment cancels out hence no net dipole moment
  • Trans isomer is non-polar[½]
  • Less energy is needed to overcome to weaker temporary dipole-dipole attraction[½]between molecules
/ [½]
Cis isomer
  • There is a net dipole moment
  • Cis isomer is polar[½]
  • More energy is needed t overcome to stronger permanentdipole-dipole attraction[½]between molecules

Note : This is an example of a situation where you cannot state weakVDW’s forces – but must explicitly compare strength of temporary and permanent dipole attractions

(b)Both beryllium difluoride and boron trifluoride can react with arsenic chloride but the mole ratios are different.

Reaction of BeF2 with AsCl3

[½]
(Shape: tetrahedral about B and As atom)
  • BF3 needs to react with only 1 AsCl3 molecule in order for B to achieve stable octet configuration[½]
  • Mole ratio: BF3  AsCl3[1/2]
/ Reaction of BF3 with AsCl3

[½]
(Shape: tetrahedral about Be and As atom)
  • BeF2 needs to react with only 2 AsCl3 molecule in order for Be to achieve stable octet configuration[½]
  • Mole ratio: BeF3  2AsCl3[1/2]

(c)I3- exists but not F3-

I3-

  • I is in period 5 hence it can expand beyond octet configuration[1/2]due toavailability of energetically accessible d orbitals to accommodate extra electrons[1]

F3-

  • F is in period 2, hence it has no energetically accessibly d orbitals[½]are available to accommodate more than 8 electrons
  • Hence F3- does not exist.

(d)Liquid IF5 can conduct electricity

  • IF5 undergoes self-ionisation[½] to form ions:

IF5 + IF5IF4 + + IF6 -[½]

  • Can conduct electricity due to presence of free mobile ions, IF4 + and IF6 - .[1]

(e) Chlorofluorocarbons can be used as aerosol propellants. Draw the displayed formulae of the three isomers of C2H2FCl. Predict which isomer has the smallest overall dipole moment . (JJC Prelim 2002)

[½][½][½]

TransCis

ABC

Structure A has overall smallest dipole[½]

[12]

Time allocation: 8 mins

5Phosphine, PH3, is a poisonous gas which is used to kill insects and vermin in grain stores. It can be made by treating aluminium phosphide, AlP, with dilute sulphuric acid.

(a)Write a balanced equation for the action of dilute sulphuric acid on aluminium phosphide.

2AlP + 3H2SO4  2PH3 + Al2(SO4)3

[1]

(b)Aluminium phosphide has a similar structure to diamond. Draw a section of the structure of aluminium phosphide, indicating in the compound the approximate bond angles.

Structure of AlP

[1/2 for shape

½ for correct bond angle]

(c)Suggest, with reasoning, whether aluminium phosphide conducts electricity.

AlP does not conduct electricity[½]due to absence of delocalised electrons and free mobile ions[½].

(d) Phosphine is less soluble in water than ammonia. Explain the .difference in their solubilities in water.

Hydrogen bondingis formed between NH3 and H2O molecules. [1]

Hence ammonia is soluble in water.

The hydrogen bonding between water molecules is stronger than the Van der Waals forces of attraction between PH3 molecules[½] hence cannot displace the hydrogen bonding [½] and is thus less soluble in water.

Hence, phosphine is less soluble in water than ammonia.

[5]

Additional Question

HCJC Prelim 2002 modified : 8 mins

6Oxygen-oxygen bond lengths in some molecules are given below :

Oxygen / 0.121 nm
Ozone / 0.128 nm
Hydrogen peroxide / 0.149 nm

(a)Draw the shape and show the approximate bond angle in a hydrogen peroxide molecule, oxygen molecule and ozone molecule.

[½][½][½]

Bent [½] Bent about each O atom[½]Trigonal planar[½]

(b)Predict and explain the relative stability of these three molecules.

  • Oxygen-oxygen bond length in O2 < O3 < H2O2[½]
  • Oxygen-oxygen bond strength in O2 > O3 > H2O2[½]
  • Energy required to break oxygen-oxygen bond in O2 > O3 > H2O2[½]
  • Relative stability of O2 > O3 > H2O2[½]

(c)Hydrogen peroxide and chlorine dioxide are often used as bleaches. Explain why hydrogen peroxide is a liquid but chlorine dioxide is a gas at room temperature.

H2O2 and ClO2havesimple molecular structures.[½].

At room temperature,

there is insufficient energy[½] to overcome the strong hydrogen bondingbetween H2O2 molecules[½]and hence H2O2 is a liquid.

there is sufficient energy to overcome the weaker Van der Waal’s forces betweenClO2 molecules[½] and hence ClO2 is a gas.

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