Chemistry 9701/43 — May/June 2015

Background Concept

Molecular formulas for simple covalent compounds can be predicted using the valencies (group numbers) of the constituent elements. Oxygen is in Group 16 and typically forms two covalent bonds. Fluorine is in Group 17 and forms one covalent bond.

Understanding the Question

You are told the molecule contains exactly three atoms and is made of fluorine and oxygen. You need to predict its chemical formula.

Approach

Since oxygen forms two bonds and fluorine forms one, the oxygen atom must be the central atom bonded to two fluorine atoms to satisfy all valencies. This gives a 1:2 ratio of oxygen to fluorine.

Step-by-Step Reasoning

• Oxygen needs 2 electrons to complete its octet, so it forms 2 bonds.
• Each fluorine needs 1 electron, so each forms 1 bond.
• Connecting two fluorine atoms to one oxygen atom uses all available bonding capacity: F–O–F.
• The molecular formula is therefore $\text{OF}_2$ (or written as $\text{F}_2\text{O}$).

Key Takeaways

When predicting formulas for simple binary molecules, the less electronegative element (or the one capable of forming more bonds) is usually the central atom.

Common Mistakes

• Guessing $\text{FO}$ or $\text{OF}$, which would leave unpaired electrons and not satisfy octets.
• Writing $\text{F}_3\text{O}$, which would exceed fluorine's valency of 1.

Things to Be Careful About

Both $\text{OF}_2$ and $\text{F}_2\text{O}$ are acceptable. The mark scheme specifically allows either.

Background Concept

A dot-and-cross diagram shows the valence electrons in a molecule, using dots for one atom's electrons and crosses for the other's. Shared pairs form covalent bonds, while non-bonding pairs are lone pairs. Every atom in $\text{OF}_2$ achieves a full octet.

Understanding the Question

Draw the dot-and-cross diagram for $\text{OF}_2$, showing all bonding and lone pairs.

Approach

Oxygen has 6 valence electrons; each fluorine has 7. Oxygen shares one electron with each fluorine to form two single covalent bonds. The remaining electrons are placed as lone pairs to complete the octets.

Step-by-Step Reasoning

• Central atom: Oxygen. Place 6 valence electrons around it (e.g., 2 dots and 4 crosses, or just use one symbol for O's electrons and another for F's).
• Bonding: Draw two fluorine atoms on either side. Draw one pair of electrons (one dot, one cross) between O and each F to represent the shared bonding pairs.
• Lone pairs on Oxygen: After sharing 2 electrons, oxygen has 4 non-bonding electrons left. Place these as two lone pairs (e.g., two dots above and two dots below).
• Lone pairs on Fluorine: Each fluorine has 7 valence electrons. One is shared in the bond, leaving 6 non-bonding electrons. Place these as three lone pairs around each F atom (e.g., three pairs of dots).
• Check: Oxygen has 2 bonding pairs + 2 lone pairs = 8 electrons. Each fluorine has 1 bonding pair + 3 lone pairs = 8 electrons. All octets are satisfied.

Key Takeaways

In dot-and-cross diagrams, clearly distinguish between the electrons contributed by each atom (using dots and crosses) and ensure all atoms have a complete outer shell.

Common Mistakes

• Forgetting lone pairs on the central oxygen atom.
• Placing too many or too few lone pairs on the fluorine atoms (must be exactly 3 per F).
• Drawing ionic bonds instead of covalent shared pairs.

Things to Be Careful About

The diagram must show shared pairs (one dot and one cross in each bond) and all lone pairs. The shape is bent, but the diagram itself is a 2D representation of electron distribution.

Background Concept

The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the 3D shape of molecules based on the repulsion between electron pairs in the valence shell of the central atom. Both bonding pairs and lone pairs repel each other, but lone pairs repel more strongly, compressing bond angles.

Understanding the Question

Predict the shape of the $\text{OF}_2$ molecule.

Approach

Determine the number of bonding pairs and lone pairs around the central oxygen atom, then use VSEPR to find the shape.

Step-by-Step Reasoning

• The central oxygen atom has 2 bonding pairs (the two O–F single bonds) and 2 lone pairs.
• This gives a total of 4 electron domains, which arrange themselves in a tetrahedral electron geometry to minimize repulsion.
• Because two of these domains are lone pairs, the molecular shape (considering only the atoms) is bent (or non-linear).
• The bond angle is slightly less than the ideal tetrahedral angle of $109.5^\circ$ (approximately $103^\circ$) due to the greater repulsion from the lone pairs.

Key Takeaways

Molecules with 4 electron domains (2 bonding, 2 lone pairs) have a bent or non-linear shape, similar to water ($\text{H}_2\text{O}$).

Common Mistakes

• Saying 'tetrahedral' (this is the electron geometry, not the molecular shape).
• Saying 'linear' (this would only be true if there were no lone pairs on the central atom, like in $\text{CO}_2$).

Things to Be Careful About

The question asks for the 'shape', which refers to the arrangement of atoms only, not the electron pairs. 'Bent' or 'non-linear' are the accepted terms.

Background Concept

Standard electrode potentials ($E^\ominus$) measure the tendency of a species to gain electrons (be reduced). A more positive $E^\ominus$ value indicates a greater tendency to be reduced, meaning the oxidised form is a stronger oxidising agent. The Data Booklet provides $E^\ominus$ values for half-equations of the form $\text{X}_2 + 2\text{e}^- \rightleftharpoons 2\text{X}^-$.

Understanding the Question

Use $E^\ominus$ values to predict which is the stronger oxidising agent: fluorine or chlorine.

Approach

Look up the standard reduction potentials for $\text{F}_2$ and $\text{Cl}_2$ in the Data Booklet. Compare the values; the more positive value corresponds to the stronger oxidising agent.

Step-by-Step Reasoning

• From the Data Booklet: $E^\ominus(\text{F}_2/\text{F}^-) = +2.87 \text{ V}$ and $E^\ominus(\text{Cl}_2/\text{Cl}^-) = +1.36 \text{ V}$.
• Since $+2.87 \text{ V} > +1.36 \text{ V}$, fluorine has a greater tendency to gain electrons than chlorine.
• Therefore, fluorine is the stronger oxidising agent (it is more easily reduced).

Key Takeaways

The more positive the $E^\ominus$ value for a halogen/halide half-cell, the stronger the halogen as an oxidising agent.

Common Mistakes

• Confusing the direction of the trend (thinking a more negative value means stronger oxidising agent).
• Forgetting to state the actual $E^\ominus$ values; the mark scheme requires them to be cited.

Things to Be Careful About

Always quote the exact $E^\ominus$ values from the Data Booklet. The comparison must explicitly link the more positive value to fluorine being the stronger oxidising agent.

Background Concept

When a stronger oxidising agent (higher $E^\ominus$) encounters a halide ion of a weaker halogen (lower $E^\ominus$), a displacement reaction occurs. The stronger oxidising agent gains electrons (is reduced) and oxidises the halide ion to the halogen. This is a redox (reduction-oxidation) reaction.

Understanding the Question

Predict the type of reaction between chlorine fluoride ($\text{ClF}$) and potassium bromide ($\text{KBr}$) solution.

Approach

$\text{ClF}$ contains fluorine in a -1 oxidation state and chlorine in a +1 oxidation state. Fluorine is a much stronger oxidising agent than bromine, and chlorine in +1 is also a strong oxidising agent. Bromide ions ($\text{Br}^-$) will be oxidised to bromine ($\text{Br}_2$), while the chlorine/fluorine in $\text{ClF}$ will be reduced. This involves both oxidation and reduction, so it is a redox reaction.

Step-by-Step Reasoning

• Potassium bromide provides $\text{Br}^-$ ions.
• $\text{ClF}$ will act as an oxidising agent, oxidising $\text{Br}^-$ to $\text{Br}_2$.
• In $\text{ClF}$, both Cl and F will be reduced to their stable -1 halide ions ($\text{Cl}^-$ and $\text{F}^-$).
• Since there is a transfer of electrons (oxidation of bromide and reduction of $\text{ClF}$), the reaction type is redox.

Key Takeaways

Interhalogen compounds can act as oxidising agents and will undergo redox reactions with halide ions of less electronegative halogens.

Common Mistakes

• Saying 'displacement' instead of 'redox'. While it is a displacement, the fundamental reaction type is redox. The mark scheme specifically asks for 'type of reaction' and credits 'redox'.
• Saying 'substitution' or 'addition' (these are organic reaction types).

Things to Be Careful About

The question asks for the 'type of reaction'. 'Redox' is the most fundamental and correct classification here.

Background Concept

Interhalogen compounds like $\text{ClF}$ can react with halide salts. In $\text{ClF}$, fluorine is more electronegative, so it takes the -1 oxidation state, leaving chlorine in the +1 oxidation state. When $\text{ClF}$ reacts with $\text{KBr}$, the bromide ions ($\text{Br}^-$) are oxidised to bromine ($\text{Br}_2$), and both the chlorine (+1) and fluorine (-1) in $\text{ClF}$ end up as chloride ($\text{Cl}^-$) and fluoride ($\text{F}^-$) ions, respectively, pairing with the potassium ions.

Understanding the Question

Construct a balanced equation for the reaction between $\text{ClF}$ and $\text{KBr}$ solution.

Approach

1. Identify the reactants: $\text{ClF}$ and $\text{KBr}$.
2. Identify the products: $\text{Br}^-$ is oxidised to $\text{Br}_2$. $\text{K}^+$ remains a spectator ion. $\text{Cl}$ and $\text{F}$ from $\text{ClF}$ become $\text{Cl}^-$ and $\text{F}^-$. Thus, the products are $\text{KCl}$, $\text{KF}$, and $\text{Br}_2$.
3. Balance the equation.

Step-by-Step Reasoning

• Unbalanced equation: $\text{ClF} + \text{KBr} \rightarrow \text{KCl} + \text{KF} + \text{Br}_2$
• Balance bromine: There are 2 Br atoms in $\text{Br}_2$, so we need $2\text{KBr}$ on the left.
  $\text{ClF} + 2\text{KBr} \rightarrow \text{KCl} + \text{KF} + \text{Br}_2$
• Check potassium: 2 K on the left, 1 K in $\text{KCl}$ and 1 K in $\text{KF}$ = 2 K on the right. Balanced.
• Check chlorine and fluorine: 1 Cl and 1 F on the left, 1 Cl in $\text{KCl}$ and 1 F in $\text{KF}$ on the right. Balanced.
• Final balanced equation: $\text{ClF} + 2\text{KBr} \rightarrow \text{KCl} + \text{KF} + \text{Br}_2$

Key Takeaways

When writing equations for interhalogen reactions with halide salts, remember that the interhalogen breaks down into its constituent halide ions, and the less electronegative halogen in the interhalogen is reduced while the external halide is oxidised.

Common Mistakes

• Forgetting to balance the potassium or bromine atoms.
• Writing $\text{Cl}_2$ or $\text{F}_2$ as products instead of the metal halides $\text{KCl}$ and $\text{KF}$. (Fluorine is too reactive to exist as $\text{F}_2$ in aqueous solution with bromide; it will simply form $\text{F}^-$).
• Writing $\text{HCl}$ or $\text{HF}$ instead of the potassium salts (since the reaction is with $\text{KBr}$ solution, the cation is $\text{K}^+$).

Things to Be Careful About

Ensure the equation is fully balanced with correct formulae for all species. State symbols are not required unless specified, but if included, $\text{KBr}$, $\text{KCl}$, and $\text{KF}$ are (aq), $\text{ClF}$ is (g) or (aq), and $\text{Br}_2$ is (aq) or (l). The mark scheme does not require state symbols for this part.