DP IB Chemistry: HL

Topic Questions

Syllabus Edition

First teaching 2014

Last exams 2024

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14.1 More Structures & Shapes

1a2 marks

Two types of covalent bond are sigma and pi bonds.

i)
Describe how a sigma (straight sigma) bond is formed

[1]

ii)
Describe how a pi (straight pi) bond is formed

[1]

1b2 marks

Describe the difference in the location of the electron dense regions in sigma (straight sigma) and pi (straight pi) bonds.

1c2 marks

Deduce the number of sigma (straight sigma) and pi (straight pi) bonds in methane, CH4.

1d2 marks

Deduce the number of sigma (straight sigma) and pi (straight pi) bonds in oxygen, O2.

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2a2 marks

Sulfur can form bonds with six fluorine atoms to form sulfur hexafluoride, SF6.

i)
How many electrons are in the outer shell of the sulfur in SF6?

[1]

ii)
State the minimum and maximum numbers of electrons possible in the outer shell of sulfur.

[1]

2b1 mark

Sulfur has no lone pairs when bonded to fluorines in SF6. Predict the molecular geometry of sulfur hexafluoride, SF6.

2c1 mark

State the F-S-F bond angles in SF6.

2d3 marks

Phosphorus pentafluoride, PF5, is also a molecule with an expanded octet around the central atom.

i)
Draw a Lewis (electron dot) structure for PF5

[1]

ii)
Predict the molecular geometry of PF5

[1]

iii)
State the F-P-F bond angle(s)

[1]

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3a2 marks

Although noble gases do not normally react, a few compounds are possible. One is xenon tetrafluoride.
Draw the Lewis structure (electron dot) for XeF4.

3b2 marks

Predict the molecular geometry and electron domain geometry for the XeF4 molecule.

3c2 marks

Predict and explain the F-Xe-F bond angle in XeF4

3d
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2 marks

The formal charge on an atom can be calculated by the following: 

FC = (Number of valence electrons) - ½(Number of bonding electrons) - (Number of non-bonding electrons)

Calculate the formal charge on the xenon and the fluorines in xenon tetrafluoride, XeF4.

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4a2 marks

Draw a Lewis (electron dot) structure for carbon dioxide, CO2.

4b2 marks

Predict the molecular geometry and the O-C-O bond angle in carbon dioxide, CO2.

4c2 marks

An alternative way to draw the carbon dioxide molecule is:

co2-resonance-lewis-structure-ib-hl-sq-14-1-4c

Identify the formal charge on each of the oxygen atoms.

4d2 marks

State which of the Lewis structures, that from part a) or part c), is preferable and explain your choice.

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1a4 marks

Deduce the number of possible resonance structures for the carbonate ion, CO32-, and draw two of them.

Include the formal charges for each oxygen.

1b2 marks

An alternative structure for the carbonate ion is proposed:

ib-hl-sq-14-1-h-proposed-incorrect-carbonate-structure-q1b

Explain why this structure is not accepted as another resonance structure for the carbonate ion.

1c2 marks

Deduce the number of sigma (σ) and pi (π) bonds present in any of the resonance structures of the carbonate ions shown in part a).

1d
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1 mark
Deduce the bond order of the C-O bond in the carbonate ion, CO32-.

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2a6 marks

Silicon can form silicon tetrachloride SiCl4 and also silicon hexachloride, SiCl62-.

i)
Draw the Lewis structure for SiCl4 and SiCl62-.
[2]
ii)
Use VSEPR theory to deduce the Cl-Si-Cl bond angles in both the SiCl4 and SiCl62- molecules.
[2]
iii)
Predict the molecular geometry of each molecule.
[2]
2b3 marks

Carbon can form CCl4 but cannot form CCl62-. Explain why.

2c2 marks

Deduce which, if any, of SiCl4 and SiCl62-, are polar molecules and explain your choice.

2d6 marks

Formal charge can be used to decide on the most stable, and therefore most likely, form a molecule can take. Resonance structures occur when more than one Lewis diagram describes a structure equally well.

i)
Deduce the formal charge on the silicon and each chlorine within SiCl4 and SiCl62-

[2]

ii)
Predict which will be the most stable molecule and explain your answer.

[2]

iii)
Predict if any resonance structures are possible for SiCl62- and explain your answer.

[2]

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3a4 marks

Natural rubber, polyisoprene, forms a flexible polymer in the following reaction:

 ib-hl-sq-14-1-h-structures-of-isopropene-and-polyisopropene-q3a

i)
Deduce the number of sigma (σ) and pi (π) bonds in the monomer.

[2]

ii)
Deduce the number of sigma (σ) and pi (π) bonds in the repeating unit.

[2]

3b2 marks

Deduce the number of carbons with a tetrahedral geometry in both the monomer, isoprene, and the repeating unit of the polymer, polyisoprene.

3c4 marks

Polymer formation involves a radical intermediate to lengthen the polymer chain.

The radical in the formation of polyisoprene is shown below, where X represents the existing chain:

 

X-CH2CCH3CHCH2

i)
Identify the atom that is the radical in the structure shown.
[1]
ii)
Deduce the formal charge on the radical atom.
[1]
iii)
Use the information above, and your knowledge of structure and bonding, to predict if the structure is stable or not.
[2]

3d2 marks

Isoprene is not produced directly by the rubber tree, but is the product of a series of biochemical reactions from the isopentenyl pyrophosphate molecules present in the tree.

 

The structure of isopentenyl pyrophosphate is shown below:

ib-hl-sq-14-1-h-structure-of-isopentenyl-pyrophosphate-q3d

Deduce the number of sigma (σ) and pi(π) bonds present in one molecule of isopentenyl pyrophosphate.

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4a4 marks

One interhalogen compound is IF5.

 
i)
Draw the Lewis structure for IF5.

[1]

ii)
Use VSEPR theory to deduce the bond angles in IF5.

[1]

iii)
Predict whether IF5 will be a polar molecule and explain your choice.

[2]

4b4 marks

Iodine can also form the triiodide ion, I3-.

 
i)
Draw the Lewis structure for I3-.
[1]
ii)
Use VSEPR theory to deduce the bond angles in I3-.
[1]
iii)
Explain the position of the lone pairs on the central iodine.
[2]
4c2 marks

Deduce the formal charge on each of the iodine atoms in the triiodide molecule, I3-.

4d3 marks

An alternative Lewis structure for the triiodide ion, I3-, is suggested:

ib-hl-sq-14-1-h-alternative-lewis-structure-for-the-triiodide-ion-q4d

Deduce the formal charges and use them to suggest if the structure is stable and likely to occur.

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1a2 marks
a)

Phosphorus tribromide and sulfur tetrafluoride are two colourless compounds which both react with water to form toxic products. 

Deduce the Lewis(electron dot) structure of both molecules.

1b2 marks
b)

Predict the shapes of the two molecules of phosphorus tribromide and sulfur tetrafluoride

1c2 marks
c)

Explain why both phosphorus tribromide and sulfur tetrafluoride are polar.

1d2 marks
d)

Compare the formation of a sigma (σ) and a pi (π) bond between two carbon atoms in a molecule.

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2a2 marks
a)

But-2-ene-1,4-dioic acid exists as both cis and trans isomers. The cis isomer is shown below

q2a_14-1_ib_hl_medium_sq

  

Describe the type of covalent bond between carbon and hydrogen in the molecule shown above and how it is formed.

 

2b2 marks
b)

Identify how many sigma bonds and how many pi (π) bonds are present in cis but-2-ene-1,4-dioc acid.

2c3 marks
c)

Draw the Lewis structures, predict the shape and deduce the bond angles for xenon tetrafluoride.

2d3 marks
d)
Compare the polarity of xenon tetrafluoride with chlorine trifluoride.

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3a2 marks
a)
Carbon dioxide can be represented by at least two resonance structures, I and II.
q3a_14-1_ib_hl_medium_sq

   

Calculate the formal charge on each oxygen atom in the two structures.


Structure

I

II

O atom labelled (1)

 

 

O atom labelled (2)

 

 

3b2 marks
b)

Deduce, giving a reason, the more likely resonance structure from part a)

3c3 marks
c)

Nitrous oxide can be represented by different Lewis (electron dot) structures. 

Deduce the formal charge (FC) of the nitrogen and oxygen atoms in three of these Lewis (electron dot) structures, A, B and C, represented below. 

LHS: atom on the left-hand side; RHS: atom on the right-hand side

 

Lewis (electron dot) structure

FC of O on LHS

FC of central N

FC of N on RHS

A

q3c1_14-1_ib_hl_medium_sq      

B

q3c2_ib_hl_medium_sq      

C

q3c_14-1_ib_hl_medium_sq      
3d3 marks
d)

Based on the formal charges assigned in part c), deduce which Lewis (electron dot) structure of N2O (A, B, or C) is the preferred.

Explain another factor that also must be taken into account in determining the preferred structure.

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4a2 marks
a)
Use the concept of formal charge to explain why BF3 is an exception to the octet rule.
4b2 marks
b)

Compounds containing two different halogen atoms bonded together are called interhalogen compounds. They are interesting because they contain halogen atoms in unusual oxidation states. One such compound is BrF3

Deduce the electron domain geometry and molecular geometry of BrF3.

4c2 marks
c)
Give the approximate bond angle(s) and a valid Lewis (electron dot) structure for BrF3
4d2 marks
d)
Explain why bromine trifluoride, BrF3 has its lone pairs of electrons located in equatorial positions. 

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5a2 marks
a)
Draw two different Lewis (electron dot) structures for SO42–, one of which obeys the octet rule for all its atoms, the other which has an octet for S expanded to 12 electrons. 
5b2 marks
b)

Explain which of the two SO42– structures is preferred using formal charges.

5c2 marks
c)

Consider the molecule shown below.

q5c_14-1_ib_hl_medium_sq

  Identify the number of sigma and pi bonds in this molecule.

5d3 marks
d)

One of the intermediates in the reaction between nitrogen monoxide and hydrogen is dinitrogen monoxide, N2O. This can be represented by the resonance structures belowq5d_14-1_ib_hl_medium_sq

 Analyse the bonding in dinitrogen monoxide in terms of sigma and pi bonds.

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