Chemistry 9701/21 — October/November 2013

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

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Topics Chemical Bonding · Hydrocarbons · Introduction to Organic Chemistry · Electrochemistry · Group 17 · Chemical Periodicity · +7 more

Q1Chemical BondingFree sample

Valence Shell Electron Pair Repulsion theory (VSEPR) is a model of electron-pair repulsion (including lone pairs) that can be used to deduce the shapes of, and bond angles in, simple molecules.

(a)

Complete the table below by using simple hydrogen-containing compounds. One example has been included.

number of bond pairs	number of lone pairs	shape of molecule	formula of a molecule with this shape
3	0	trigonal planar	$\text{BH}_3$
4	0
3	1
2	2

DifficultyEasy

Worked solution

Answer

number of bond pairs	number of lone pairs	shape of molecule	formula of a molecule with this shape
3	0	trigonal planar	$\text{BH}_3$
4	0	tetrahedral	$\text{CH}_4$
3	1	pyramidal	$\text{NH}_3$
2	2	non-linear	$\text{H}_2\text{O}$

Final answer

Row 2: tetrahedral, $\text{CH}_4$ ; Row 3: pyramidal, $\text{NH}_3$ ; Row 4: non-linear, $\text{H}_2\text{O}$

Detailed explanation

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory states that electron pairs (both bonding and lone pairs) in the valence shell of a central atom repel each other and will arrange themselves to be as far apart as possible. This arrangement minimises repulsion and determines the molecular geometry. Lone pairs occupy space and repel bonding pairs more strongly than bonding pairs repel each other, but for basic shape determination, we count the total number of electron domains (bond pairs + lone pairs) to find the basic geometry, then ignore the lone pairs when naming the molecular shape.

Understanding the Question

The question asks to complete a table by providing the molecular shape and a simple hydrogen-containing example for three combinations of bond pairs and lone pairs, using the VSEPR model. The first row (3 bond pairs, 0 lone pairs) is given as an example (trigonal planar, $\text{BH}_3$ ).

Approach

For each row, count the total number of electron pairs (bond pairs + lone pairs) to determine the basic electron geometry. Then, identify the molecular shape by considering only the positions of the atoms (bond pairs). Finally, select a simple hydride from the appropriate group in the periodic table that matches the bonding arrangement.

Step-by-Step Reasoning

Row 2 (4 bond pairs, 0 lone pairs): Total electron pairs = 4. The electron geometry is tetrahedral. With 0 lone pairs, the molecular shape is also tetrahedral. A simple Group IV hydride is methane, $\text{CH}_4$ (or $\text{SiH}_4$ , etc.).
Row 3 (3 bond pairs, 1 lone pair): Total electron pairs = 4. The electron geometry is tetrahedral. With 1 lone pair, the molecular shape is pyramidal (or trigonal pyramidal). A simple Group V hydride is ammonia, $\text{NH}_3$ (or $\text{PH}_3$ , etc.).
Row 4 (2 bond pairs, 2 lone pairs): Total electron pairs = 4. The electron geometry is tetrahedral. With 2 lone pairs, the molecular shape is non-linear (or bent or V-shaped). A simple Group VI hydride is water, $\text{H}_2\text{O}$ (or $\text{H}_2\text{S}$ , etc.).

Key Takeaways

VSEPR theory uses the total number of electron pairs to predict the basic geometry, and lone pairs modify the molecular shape. Common shapes for 4 electron pairs include tetrahedral (0 lone pairs), pyramidal (1 lone pair), and non-linear (2 lone pairs).

Common Mistakes

Confusing electron geometry (which includes lone pairs) with molecular shape (which only considers atoms).
Providing molecules that do not match the group (e.g., providing $\text{HCl}$ for 2 bond pairs and 2 lone pairs is incorrect as Cl has 3 lone pairs and 1 bond pair in $\text{HCl}$ , but $\text{H}_2\text{O}$ is correct).
Using incorrect terminology for shapes (e.g., saying 'bent' instead of 'non-linear' is usually accepted, but 'tetrahedral' for a molecule with lone pairs is wrong).

Things to Be Careful About

Ensure the formula provided actually has the stated number of bond and lone pairs on the central atom. For example, $\text{CH}_4$ has 4 bond pairs and 0 lone pairs on C, not on H.
Acceptable alternatives for shapes include 'trigonal pyramidal' for pyramidal, and 'bent' or 'V-shaped' for non-linear.

Techniques used

apply VSEPR theorydeduce molecular shape from electron pairsprovide examples of molecules with specific geometries

(b)

Tellurium, $\text{Te}$ , proton number 52, is used in photovoltaic cells.

When fluorine gas is passed over tellurium at $150\text{ }^\circ\text{C}$ , the colourless gas $\text{TeF}_6$ is formed.

(i)

Draw a ‘dot-and-cross’ diagram of the $\text{TeF}_6$ molecule, showing outer electrons only.

DifficultyMedium-Easy

Worked solution

Answer

Final answer

Dot-and-cross diagram of TeF6: Te central, 6 bonding pairs, 3 lone pairs on each F

Detailed explanation

Background Concept

A dot-and-cross diagram (Lewis structure) represents the valence electrons in a molecule. Dots and crosses are used to distinguish electrons from different atoms. A single covalent bond is represented by one dot and one cross shared between two atoms. Lone pairs are pairs of electrons that belong to a single atom and are not shared.

Understanding the Question

Part (b)(i) asks for a dot-and-cross diagram of $\text{TeF}_6$ , showing only outer (valence) electrons. Tellurium (Te) is in Group VI, so it has 6 outer electrons. Fluorine (F) is in Group VII, so each has 7 outer electrons.

Approach

Determine the total number of outer electrons: Te (6) + 6 × F (7) = 48 electrons.
Place Te in the center and arrange the 6 F atoms around it.
Form 6 single covalent bonds between Te and each F. This uses 12 electrons (6 from Te, 6 from F).
Complete the octets of the F atoms by adding 3 lone pairs (6 electrons) to each F. This uses 36 electrons.
Total electrons used = 12 (bonds) + 36 (lone pairs on F) = 48. Te has no lone pairs.

Step-by-Step Reasoning

Te has 6 valence electrons. Each F has 7.
Te forms 6 single bonds with 6 F atoms. In the diagram, show one cross (from Te) and one dot (from F) in each bond.
Each F atom now has 2 electrons in the bond and needs 6 more to complete its octet. Add 3 lone pairs (6 dots) around each F atom.
Te has used all 6 of its valence electrons in the 6 bonds, so it has no lone pairs.
The resulting diagram has Te in the center, surrounded by 6 F atoms, each with 3 lone pairs and 1 bonding pair to Te.

Key Takeaways

In dot-and-cross diagrams, ensure all atoms (except H) have a full outer shell (octet). The central atom's valence electrons are typically shown as crosses, and the surrounding atoms' electrons as dots.

Common Mistakes

Forgetting to add lone pairs to the fluorine atoms.
Showing lone pairs on the central Te atom (it has 6 bonds, using all 6 valence electrons).
Incorrectly counting electrons or not showing the correct number of bonding pairs.

Things to Be Careful About

The diagram must show outer electrons only. Inner shell electrons are not included.
Use different symbols (dots and crosses) to distinguish electrons from Te and F, though in some mark schemes, just showing pairs is acceptable if the context is clear. Always follow the specific mark scheme guidance for dot-and-cross diagrams.

Techniques used

draw a dot-and-cross diagram for a central atom with six bonding pairs

(ii)

What will be the shape of the $\text{TeF}_6$ molecule?

DifficultyEasy

Worked solution

Answer

octahedral

Final answer

octahedral

Detailed explanation

Background Concept

VSEPR theory predicts molecular geometry based on the number of electron pairs around the central atom. For 6 electron pairs (all bonding), the geometry that minimises repulsion is octahedral.

Understanding the Question

Part (b)(ii) asks for the shape of the $\text{TeF}_6$ molecule. From the dot-and-cross diagram in (b)(i), Te has 6 bond pairs and 0 lone pairs.

Approach

Count the total number of electron domains around the central Te atom. There are 6 bond pairs and 0 lone pairs, giving a total of 6 electron domains. According to VSEPR theory, 6 electron domains arrange themselves in an octahedral geometry.

Step-by-Step Reasoning

Central atom: Te
Bond pairs: 6 (one to each F)
Lone pairs: 0
Total electron domains: 6
Electron geometry: octahedral
Molecular shape: octahedral (since there are no lone pairs to distort the shape)

Key Takeaways

Molecules with 6 bonding pairs and 0 lone pairs have an octahedral shape.

Common Mistakes

Confusing octahedral with other shapes like tetrahedral or trigonal bipyramidal.
Forgetting that 'square-based bipyramid' is an acceptable alternative name for octahedral.

Things to Be Careful About

Ensure the answer is 'octahedral' and not 'octahedron' (the shape is octahedral, the geometry is an octahedron, but 'octahedral' is the standard term for the molecular shape).

Techniques used

apply VSEPR theory to a molecule with 6 bonding pairs and 0 lone pairs

(iii)

What is the $\text{F}-\text{Te}-\text{F}$ bond angle in $\text{TeF}_6$ ?

DifficultyEasy

Worked solution

Answer

90°

Final answer

90°

Detailed explanation

Background Concept

In an octahedral molecule, the 6 bonding pairs are arranged symmetrically around the central atom. The angle between adjacent bonds (e.g., axial-equatorial or equatorial-equatorial) is 90°. The angle between opposite bonds (axial-axial) is 180°.

Understanding the Question

Part (b)(iii) asks for the F-Te-F bond angle in $\text{TeF}_6$ . Since the molecule is octahedral, the angle between adjacent fluorine atoms is 90°.

Approach

Recall the standard bond angles for an octahedral geometry. The primary bond angle between adjacent ligands is 90°.

Step-by-Step Reasoning

Molecular shape: octahedral
In an octahedral geometry, there are 6 positions: 4 equatorial and 2 axial.
The angle between any two adjacent bonds (e.g., equatorial-equatorial or axial-equatorial) is 90°.
The angle between opposite bonds (axial-axial) is 180°, but the standard bond angle asked for is typically the adjacent angle, which is 90°.

Key Takeaways

Octahedral molecules have bond angles of 90° (and 180° for opposite bonds). The standard answer for 'the bond angle' is 90°.

Common Mistakes

Stating 109.5° (which is for tetrahedral) or 120° (which is for trigonal planar).
Forgetting to include the degree symbol (°).

Things to Be Careful About

Ensure the answer is '90°' and not just '90'. The degree symbol is required.
If asked for all bond angles, include both 90° and 180°, but typically '90°' is the expected answer for 'the bond angle' in an octahedral molecule.

Techniques used

determine bond angles in an octahedral molecule

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