The table shows some information about elements in Group VI. | element | electronic configuration | melting point /°C | density in g/cm³ | |---|---|---|---| | oxygen | 2,6 | -218 | 0.0013 | | sulfur | | 113 | 2.1 | | selenium | 2, 8, 18, 6 | 217 | 4.8 | | tellurium | 2, 8, 18, 18, 6 | 450 | 6.3 | | polonium | 2, 8, 18, 32, 18, 6 | 254 | |
- (a)State the electronic configuration for sulfur.[1]
- (b)Predict the density of polonium.[1]
- (c)Sulfur has a boiling point of 445 °C. Predict the physical state of sulfur at 200 °C. Explain your answer. physical state explanation[1]
- (d)(i)Oxygen exists as a diatomic molecule, O2. Draw the dot-and-cross diagram for a molecule of oxygen. Show only the outer shell electrons.[1]
- (d)(ii)Explain, in terms of structure and bonding, why oxygen has a low melting point.[1]
- (e)(i)Selenium, Se, is a non-metal. Deduce the formula of selenium(IV) oxide.[1]
- (e)(ii)A small sample of selenium(IV) oxide is dissolved in water. Two drops of universal indicator are added to this aqueous solution. Predict the colour of the universal indicator in this solution. Explain your answer. colour explanation[1]
- (f)Calculate the volume, in dm³, of 30.2g of oxygen at room temperature and pressure. Give your answer to two significant figures.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [1] 2, 8, 6
- [1] any value between 6.8 to 9.4 inclusive (g/cm³)
- [1] liquid; because; melting point is below 200 °C and boiling point is above 200 °C / 200 °C is between the melting point and the boiling point
- [1] correct dot-and-cross diagram
- [1] weak intermolecular forces / intermolecular forces are easy to break or overcome
- [1] SeO2 / O2Se
- [1] red / orange / yellow / yellow-green; because; non-metal oxides (are acidic)
- [1] moles of oxygen = 0.94375 OR 30.2/32
- [1] volume = 0.94375 × 24 OR 22.65
- [1] volume = 23 (dm³)
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