OCR A-Level Chemistry: Acids, Redox, Electrons & Bonding

Water and hydrogen sulfide are both simple molecular hydrides of Group 16 elements, yet their boiling points are very different.

Hydride	H₂O	H₂S
Boiling point / °C	100	-60

Both molecules are bent (non-linear), but the H–O–H bond angle in water (104.5°) is slightly larger than the H–S–H bond angle in hydrogen sulfide (92°).

Describe and explain why water has a much higher boiling point than hydrogen sulfide, and why the two molecules have the bent shape described. Your answer should refer to the intermolecular forces in each substance and to electron-pair repulsion in determining the shapes.

(6 marks)

A sample of an iron(II) salt is dissolved in dilute sulfuric acid and titrated against potassium manganate(VII). In acidic solution the manganate(VII) ion is reduced to the manganese(II) ion while iron(II) is oxidised to iron(III).

Quantity	Value
Volume of iron(II) solution	25.0 cm³
Concentration of KMnO₄	0.0200 mol dm⁻³
Mean titre of KMnO₄	22.40 cm³

(a) Using oxidation numbers, deduce the half-equation for the reduction of the manganate(VII) ion, $\text{MnO}_4^{-}$MnO4−​, to $\text{Mn}^{2+}$Mn2+ in acidic solution, and hence write the overall ionic equation for the reaction with iron(II) ions. (3 marks)

(b) Calculate the concentration, in mol dm⁻³, of iron(II) ions in the original solution. (3 marks)

The boiling points of the four hydrogen halides are shown below.

Hydrogen halide	HF	HCl	HBr	HI
Number of electrons per molecule	10	18	36	54
Boiling point / °C	20	-85	-67	-35

Explain the trend in boiling point shown by these four hydrogen halides. Your answer should account for why hydrogen fluoride does not fit the pattern set by the other three, and why the boiling point rises from HCl to HI.

(5 marks)

Boron trifluoride, $\text{BF}_3$BF3​, reacts with ammonia, $\text{NH}_3$NH3​, to form a single product, an adduct with the formula $\text{H}_3\text{N} \rightarrow \text{BF}_3$H3​N→BF3​. In this product a new bond joins the nitrogen atom to the boron atom.

In $\text{BF}_3$BF3​ the boron atom has only 6 electrons in its outer shell (three bonding pairs and no lone pair); in $\text{NH}_3$NH3​ the nitrogen atom has three bonding pairs and one lone pair.

(a) Identify the type of bond formed between the nitrogen and boron atoms, and explain, in terms of the electrons supplied, how it forms. (2 marks)

(b) State and explain the shape and bond angle around the boron atom before the reaction (in $\text{BF}_3$BF3​) and after the reaction (in the adduct). (3 marks)

Phosphorus pentafluoride, $\text{PF}_5$PF5​, is a colourless gas. Its molecule contains five bonding pairs of electrons around the central phosphorus atom and no lone pairs.

(a) Predict the shape of a $\text{PF}_5$PF5​ molecule and state the two different bond angles present. (3 marks)

(b) State why all the P–F bonds are polar. (1 mark)

This question is about electron configuration, acids and oxidation numbers.

(a) Write the full electron configuration of a sulfur atom (proton number 16), showing all the sub-shells. (1 mark)

(b) State the definition of a Brønsted-Lowry acid. (1 mark)

(c) Deduce the oxidation number of sulfur in the sulfate ion, $\text{SO}_4^{2-}$SO42−​. (1 mark)