Video: GCSE Chemistry Higher Tier Pack 1 • Paper 1 • Question 3

GCSE Chemistry Higher Tier Pack 1 • Paper 1 • Question 3

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Video Transcript

Figure one shows the valence electron shell diagrams for atoms of magnesium and fluorine. One atom of magnesium reacts with two atoms of fluorine to form an ionic compound. Describe how electrons are transferred between the atoms in this reaction and give formulas for the ions produced.

The first thing you should ask yourself in this question is what information is provided in figure one that might help you answer the question. Valence electron shell diagrams depict the electrons in the outer shell of an atom or ion. Here magnesium is depicted with two outer electrons and fluorine is depicted with seven, one short of a full shell.

The question refers to an ionic compound. Ionic compounds are made up of positive ions, cations, bonded to negative ions, anions. Since we’re starting with atoms, we need them to be ionized. So magnesium and fluorine need to gain or lose electrons. You should remember the rule of thumb, which is a full outer shell is more stable and that metals lose electrons to gain full outer shells, exposing the shell beneath, and nonmetals gain electrons to fill their valence shells.

So to become more stable, magnesium will lose the two electrons in its outer shell and an atom of fluorine will gain one to fill its outer shell. Now it makes sense, one atom of magnesium reacting with two atoms of fluorine. The magnesium must give up two electrons. And one must go to one fluorine, and one must go to the other.

So the only part of the question left is what ions are produced. If magnesium loses two electrons, it becomes positively charged with a two plus charge. If fluorine gains one electron, it becomes negatively charged with a one minus charge. We now have all the information we need to write out the answer. The magnesium atom gives one electron to each of the fluorine atoms, forming an Mg²⁺ ion. The fluorine atoms gain an electron each, forming F⁻ ions.

Figure two shows a ball-and-stick model of the ionic lattice of magnesium fluoride. Identify one way in which the ball-and-stick model is not an accurate representation of the magnesium fluoride lattice.

A ball-and-stick model is a visual representation of the three-dimensional structure of a substance. Balls represent atoms, while sticks represent bonds. However, magnesium fluoride is an ionic solid, and ionic solids have ions which are attracted to one another in all directions.

The ball-and-stick model oversimplifies these bonds and misses out some of the forces of attraction between these ions. So one way that the ball-and-stick model doesn’t accurately represent magnesium fluoride is that ionic lattices arise from ionic bonds, which are not directional.

Fluorine can also form covalent bonds. Complete the dot and cross diagram to show the covalent bonding in a molecule of carbon tetrafluoride.

What are covalent bonds? The word “covalent” can be split into “co” and “valent,” “co” here meaning “shared” and “valent” referring to valency or the outer shell. So a covalent bond involves the overlapping and sharing of the outer electron shell.

In dot and cross diagrams, we express electrons from one atom using dots or crosses and electrons from another using crosses or dots. Let’s start off by drawing the diagrams for the atoms on their own.

Carbon is in group four and therefore has four outer electrons. Fluorine is in group seven and therefore has seven outer electrons. Carbon needs four electrons to make a full outer shell. Fluorine only needs one. In carbon tetrafluoride, these atoms can share electrons in order to fulfill this requirement.

Let’s start by drawing in the electrons for carbon, and now for the fluorines. And there we have it. The carbon has eight electrons in its outer shell, and each of the fluorines has eight electrons in their outer shell. This is achieved by forming a total of four carbon fluorine covalent bonds.

Calculate the relative formula mass of copper(II) tetrafluoroborate, Cu(BF₄)₂. Relative atomic masses Ar (g per mol): boron 10.8, fluorine 19.0, copper 63.5.

A relative formula mass is the sum of the relative atomic masses of the elements making up the substance. We calculate this by working out which atoms are in one unit of copper(II) tetrafluoroborate. Don’t worry about the name. You’ve been given the formula, and that’s all what matters.

In one formula unit of copper(II) tetrafluoroborate, we have one carbon atom, two boron atoms — remember to multiply three by two because of the brackets — and eight fluorine atoms — that’s four multiplied by two. What you need to write is one times the relative atomic mass of copper, which is 63.5, plus two times the relative atomic mass of boron, which is 10.8, plus eight times the relative atomic mass of fluorine, which is 19. This is equal to 237.1 grams per mol. So the relative formula mass of copper(II) tetrafluoroborate is 237.1 grams per mol.

Simple covalent materials like carbon tetrafluoride form soft solids with poor electrical conductivities in aqueous solution. Draw one line from each property to its physical explanation.

Firstly, let’s consider what we know about carbon tetrafluoride and think about which explanations we can discount. Firstly, we know that carbon tetrafluoride is a covalent solid. It is molecular. Therefore, it doesn’t have ions. It’s not ionic. Therefore, we can discount mobile ions.

What about strong intermolecular forces? Well, we know that carbon tetrafluoride is a covalent molecule. It’s not polar. There isn’t any group there that makes it strongly polar. It’s unlikely to have strong intermolecular forces.

What about delocalized electrons? In carbon tetrafluoride, there aren’t any. This leaves us with three plausible explanations for any phenomenon: the weak intermolecular forces that exist between CF₄ molecules, the strong covalent bonds that exist in CF₄ molecules, and the fact that it’s not an ionic solid and therefore has no charged particles, mobile or otherwise.

Let’s look at the first property, soft solids. This means that while there is enough attraction between the molecules, it isn’t very strong because an external force can disturb it. Therefore, the most plausible explanation for this is weak intermolecular forces. The second property, poorly conductive aqueous solutions, is down to a lack of mobile ions. The presence of strong covalent bonds in the molecule doesn’t have any impact. So the explanation for poorly conductive aqueous solutions is that there are no mobile charged particles.

Ionic materials like magnesium fluoride form hard solids with good electrical conductivities in aqueous solution. Draw one line from each property to its physical explanation.

The first thing to do is to look at the explanations and discount those that don’t apply. We can eliminate anything that refers to intermolecular forces because we are dealing with an ionic solid that doesn’t have molecules. There are also no delocalized electrons because the electrons that are there are very strongly bound to the ions they belong to. This leaves us with three explanations for the phenomenon: strong ionic bonds, no mobile charged particles, or mobile ions.

Let’s look at the first property, hard solids. Solids are hard when external forces cannot change their shape. This suggests that the forces inside are very strong. Therefore, the valid explanation here is strong ionic bonds.

Now on to the second property, highly conductive aqueous solutions. Water on its own is not conductive. Therefore, charge carriers are needed. If there were no mobile charged particles, that would not make it conductive, so we can eliminate that explanation. Therefore, the only explanation left is that mobile ions are present, and this is what you’d expect for an ionic solid like magnesium fluoride dissolving in water.

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