The Periodic Table is a method of classifying elements.
(a) Identify the element which is in Group VI and Period 4.
(b) Calcium is in Group II and chlorine is in Group VII of the Periodic Table.
Explain, in terms of number of outer shell electrons and electron transfer, how calcium atoms and chlorine atoms form ions. Give the formulae of the ions formed.
(c) Group V chlorides are covalent molecules. The boiling points of some Group V chlorides are shown.
(i) Suggest the approximate boiling point of \(PCl_3\).
(ii) Explain the trend in boiling points in terms of attractive forces between particles.
(iii) Complete the dot-and-cross diagram to show the electron arrangement in a molecule of \(PCl_3\). Show outer electrons only.
(d) \(PCl_3\) reacts with chlorine, \(Cl_2\), to form \(PCl_5\). This reaction is exothermic and reaches an equilibrium.
\(PCl_3(g) + Cl_2(g) \leftrightarrow PCl_5(g)\)
(i) Describe two features of an equilibrium.
(ii) State the effect, if any, on the position of this equilibrium when the following changes are made. Explain your answers.
– Temperature is increased.
– Pressure is increased.
(iii) Explain, in terms of particles, what happens to the rate of the forward reaction when the reaction mixture is heated.
(e) \(PCl_5\) reacts with lithium fluoride, \(LiF\), to form \(LiPF_6\).
\(PCl_5 + 6LiF \rightarrow LiPF_6 + 5LiCl\)
Calculate the mass of \(LiF\) needed to form 3.04g of \(LiPF_6\) using the following steps.
- Calculate the number of moles of \(LiPF_6\) formed. [\(M_r\): \(LiPF_6\), 152]
- Deduce the number of moles of \(LiF\) needed.
- Calculate the mass of \(LiF\) needed.
(f) Lithium fluoride has ionic bonding.
(i) What is an ionic bond?
(ii) Give two physical properties of ionic compounds.
▶️ Answer/Explanation
(a) The element in Group VI and Period 4 is selenium (Se).
(b) – Calcium (Group II) has 2 outer electrons and loses them to form \(Ca^{2+}\).
– Chlorine (Group VII) has 7 outer electrons and gains 1 electron to form \(Cl^–\).
– Electron transfer occurs: \(Ca \rightarrow Ca^{2+} + 2e^–\) and \(Cl + e^– \rightarrow Cl^–\).
(c)(i) The boiling point of \(PCl_3\) is approximately 75°C (interpolating between \(NCl_3\) and \(AsCl_3\)).
(c)(ii) Boiling points increase down Group V due to stronger van der Waals forces (larger molecules have more electrons).
(c)(iii) Dot-and-cross diagram for \(PCl_3\):
– Phosphorus shares 3 electrons with chlorine atoms (3 covalent bonds).
– Each chlorine has 3 lone pairs, and phosphorus has 1 lone pair.
(d)(i) Features of equilibrium:
1. Constant concentrations of reactants/products.
2. Forward and reverse rates are equal.
(d)(ii) – Temperature increase: Shifts left (reverse reaction is endothermic).
– Pressure increase: Shifts right (fewer gas moles on product side).
(d)(iii) Heating increases the kinetic energy of particles, leading to more frequent and energetic collisions (higher reaction rate).
(e) – Moles of \(LiPF_6 = \frac{3.04}{152} = 0.02 \text{ mol}\).
– Moles of \(LiF = 0.02 \times 6 = 0.12 \text{ mol}\).
– Mass of \(LiF = 0.12 \times 26 = 3.12 \text{ g}\).
(f)(i) An ionic bond is the electrostatic attraction between oppositely charged ions.
(f)(ii) Two properties of ionic compounds:
1. High melting/boiling points (strong ionic bonds).
2. Conduct electricity when molten/dissolved (mobile ions).
Fluorine is a Group VII element. Fluorine forms compounds with metals and non-metals.
(a) Predict the physical state of fluorine at room temperature and pressure.
(b) Fluorine exists as diatomic molecules.
Complete the dot-and-cross diagram to show the electron arrangement in a molecule of fluorine. Show outer shell electrons only.
(c) Write a chemical equation for the reaction between sodium and fluorine.
(d) Explain why chlorine does not react with aqueous sodium fluoride.
(e) Tetrafluoromethane and lead(II) fluoride are fluorides of Group IV elements. Some properties of tetrafluoromethane and lead(II) fluoride are shown in the table.
(i) What is the formula of lead(II) fluoride?
(ii) What type of bonding is present between the atoms in tetrafluoromethane?
(iii) What type of structure does solid lead(II) fluoride have?
(iv) Explain, in terms of attractive forces between particles, why lead(II) fluoride has a much higher melting point than tetrafluoromethane.
In your answer refer to the types of attractive forces between particles and their relative strengths.
(f) Tetrafluoroethene is an unsaturated compound with the formula C2F4.
Tetrafluoroethene is the monomer used to make the polymer poly(tetrafluoroethene).
(i) What is meant by the term unsaturated?
(ii) Describe a test to show that tetrafluoroethene is unsaturated.
(iii) Draw the structure of a molecule of tetrafluoroethene. Show all of the atoms and all of the bonds.
(iv) Tetrafluoroethene can be polymerized to form poly(tetrafluoroethene).
Draw one repeat unit of poly(tetrafluoroethene). Show all of the atoms and all of the bonds.
(v) Deduce the empirical formula of:
tetrafluoroethene
poly(tetrafluoroethene)
▶️ Answer/Explanation
(a) Fluorine is a gas at room temperature and pressure (Group VII elements are diatomic gases).
(b) The dot-and-cross diagram for fluorine (F2) should show:
– 1 shared pair of electrons (single covalent bond).
– 6 non-bonding electrons (3 lone pairs) on each fluorine atom.
(c) The reaction between sodium and fluorine is:
\(\text{2Na} + \text{F}_2 \rightarrow \text{2NaF}\).
Sodium fluoride (NaF) is formed in this redox reaction.
(d) Chlorine does not react with aqueous sodium fluoride because chlorine is less reactive than fluorine (fluorine is higher in Group VII).
(e) (i) The formula of lead(II) fluoride is PbF2 (lead has a +2 charge).
(ii) Tetrafluoromethane has covalent bonding (shared electrons between C and F).
(iii) Lead(II) fluoride has a giant ionic lattice structure (strong electrostatic forces).
(iv) Lead(II) fluoride has a higher melting point because:
– It has strong ionic bonds between ions.
– Tetrafluoromethane has weak intermolecular forces (van der Waals).
– Ionic bonds require more energy to break than intermolecular forces.
(f) (i) Unsaturated means the compound contains double or triple bonds (not all single bonds).
(ii) Test for unsaturation:
– Bromine water (orange/brown).
– Observation: Decolorizes (turns colorless) due to addition reaction.
(iii) Structure of tetrafluoroethene:
\(\text{F}_2\text{C}=\text{CF}_2\) (double bond between carbons, single bonds to fluorine).
(iv) Repeat unit of poly(tetrafluoroethene):
\(\text{-CF}_2\text{-CF}_2\text{-}\) (extended chain with single bonds).
(v) Empirical formula:
– Tetrafluoroethene: CF2.
– Poly(tetrafluoroethene): CF2 (same as monomer).