Group 2 and the halogens Flashcards
Describe the trends in reactivity of group 2 elements down the group
All elements in group 2 have two electrons in the sub-shell of their outer shell. When group 2 elements react, these two outer electron are lost,. By transferring these two electrons to other species group 2 metals act as reducing agents
The reactivity of metals increases as we move down group 2. When group 2 elements react they lose their outer two electrons. This requires ionisation energy.
For e.g: Both the second and first ionisation energies for calcium are less than for magnesium. This is because calcium has a greater atomic radius than magnesium so the two outer electrons in calcium are further from the nucleus compared to magnesium. Calcium atoms have one more full inner electron shell than magnesium. The full inner electron shells partially shield the outer electrons from the positive charge of the nucleus. Therefore, the outer electrons in calcium are less attracted to the nucleus than the outer electron in magnesium. Calcium takes less energy to remove outer electrons. Therefore, calcium is more reactive than magnesium.
Key idea of elements in group 2
Alkaline earth metals are reactive
Group 2 elements have two electrons in their outer shell subshell. When group 2 elements react they lose they outer two electrons to form a 2+ metal ion. The group 2 metal is oxidised as it loses electrons. These are then used to reduce another chemical as a reducing agent.
The reactivity increases moving down the group as the first and second ionisation energies decrease. This means less energy is required to lose the outer two electrons as we go down the group.
Describe how group 2 elements react with oxygen
2Mg(s) + O2(g) —> 2MgO(s)
Both magnesium and oxygen have an oxidation number of zero as they are elements
During the reaction each magnesium atom is oxidised, losing two electrons. So the oxidation number of each magnesium changes from 0-+2.
Each oxygen atom is reduced, gaining two electrons. So the oxidation number of of each oxygen changes from 0– -2
The changes of oxidation number must balance so the total change of oxidation number is +4 for the two magnesium atoms and -4 for the two oxygen atoms.
Describe how group 2 elements react with water
Ca(s) + 2H2O(l) —> Ca(OH)2 (aq) + H2(g)
Reacting a group 2 metal with water produces the alkaline metal hydroxide and hydrogen gas.
At the start, the calcium has an oxidation number of zero) as it is an element.
The hydrogens in the water each have an oxidation number of +1.
During the reaction, the calcium is oxidised losing two electrons.
The oxidation number of the calcium changes to +2 in the calcium hydroxide.
The two electrons are transferred onto two hydrogen atoms, forming the hydrogen gas. These two hydrogens are reduced and their oxidation number changes to +1 to 0
The oxidation number balance. One calcium atom changes to 0 to +2. The two hydrogen atoms changes from +1 to 0.
The two hydrogen atoms in the calcium hydroxide each have an oxidation number of +1. The oxidation number of the two hydrogen atoms had not changed during the reaction.
In water they had an oxidation number of +1 and in calcium hydroxide, they still have an oxidation number of +1
Describe how group 2 elements react with dilute hydrochloric acid
Mg(s) + H2SO4(aq) —> MgSO4(aq) + H2(g)
When group 2 elements react with dilute acids, we make a metal salt and hydrogen gas. (Redox reaction)
At the start the magnesium has an oxidation number of 0 as it is an element.
During the reaction, the magnesium is oxidised,losing two outer electrons. So inthe magnesium sulfate, the magnesium has an oxidation number of +2.
In sulfuric acid the hydrogens each have an oxidation number of +1. During the reaction, these hydrogens are reduced as they each receive one electron from the magnesium. So in hydrogen gas, the oxidation number of each hydrogen is now 0 as they are elements
The changes in oxidation number must balanced. The oxidation number of magnesium has changed from 0 to +2. Each of the two hydrogens have changed from +1 to 0
The sulfate ion SO4^2- does Noto change during the reaction so we don’t need to consider it
Describe the solubility of group 2 hydroxides in water
Reactivity (solubility) of group 2 elements with water increases down the group. This explain the alkalinity of group 2 hydroxides:
When group 2 hydroxides dissolve in water, they release the metal ion and two hydroxide ions. The aqueous hydroxide ions make the solution alkaline. The concentration of the aqueous hydroxide ions determines the alkalinity of the solution. High concentration of hydroxide ions result in a high pH (alkaline solution).
The lower the solubility in water the less alkaline the solution is (lower pH)
Key idea of group 2 hydroxides
A group 2 hydroxide is created by reacting a group 2 oxide with water. (Calcium oxide and water = calcium hydroxide)
As the calcium hydroxide forms, it dissolves in water to form calcium hydroxide solution. However, calcium hydroxide is only slightly soluble in water so this solution becomes saturated. As it is continued to form this no longer dissolves but instead forms a solid calcium hydroxide.
Solutions of group 2 hydroxides become more alkaline as we move down the group
What are the uses of group 2 compounds
They are used for neutralisation reaction in agriculture and in medicine (calcium hydroxide, lime is spread on fields. This neutralizes the acids on the soil making the soil more favourable for crops.
It is also used to treat indigestion caused by excess hydrochloric acid in the stomach. A suspension of magnesium hydroxide in water is called milk of magnesia. This neutralizes hydrochloric acid to produce magnesium chloride and water. (Tablets of calcium carbonate can be eaten). This is used to neutralise hydrochloric acid producing calcium chloride, carbon dioxide and water. (We don’t use calcium hydroxide because the alkalinity of calcium hydroxide would be harmful to body issues e.g tissue lining in the throat.
State the colors of the elements in group 7 ( at room temperature and pressure)
Fluorine = pale yellow gas
Chlorine = pale green gas
Bromine = red-brown liquid
Iodine = grey-black solid
Astatine = highly radioactive and has never been observed
How does the m.p/b.p of halogens change as it moves down group 7
When the halogens are cooled, they form simple molecular lattices. The covalent bond between the two halogen atoms is strong. However, between the halogen molecules there are induced dipole-dipole interactions. (London forces) . These intermolecular forces are relatively weak and do not take much energy to break
However, the strength of London forces increases with the increased number of electrons.Therefore, London forces will be stronger on molecules which are lower down on group 7 of the periodic table.
Key idea of group 7 halogens
When halogens react they remove an electron from another species
Halogens are diatomic molecules (therefore when a redox reaction is written it is shown as one e.g 2Cl)
More reactive halogens are stronger oxidising agents (top of group 7) than less reactive halogens which are less powerful oxidising agents (bottom of group 7)
The reactivity of the halogens decreases down group 7 as they become less powerful oxidising agents
The solubility of halogens decreases as you go down the group. This is is because the number of electron increases therefore, there are more electron shells thus the atomic radius increases.
Explain why does the reactivity of the halogens decreases down group 7
When a halogen reacts as an oxidising agent the halogens stronger atom removes an electron from another species. This electron adds into the outer shell of the halogen atom, to form a full outer shell.
Why is bromine less reactive than fluorine
Bromine has a greater atomic radius than fluorine. This means that the outer electrons in bromine are further from the nucleus,
Bromine has more inner electron shells so there is greater shielding between the nucleus and the outer electrons. Therefore, there is less attraction between the nucleus and the outer electrons.
This makes it harder for a bromine atom to gain an electron from another species.Thsi makes it less reactive than fluorine and a less powerful oxidising agent.
What are the uses of chlorine
Chlorine is a widely used halogen and one of its uses is in drinking water. Small amounts of chlorine are added to drinking water to kill harmful bacteria.
By adding small amounts of of chlorine to drinking water chloric (I) acid which kills bacteria. This prevents water-borne bacteria diseases such as cholera.
Explain the reaction between chlorine and water:
Cl2(g) + H2O(l) —-> HClO(aq) + HCl(aq)
If we look at oxidation numbers we can see that this is a redox reaction:
On the left, chlorine has an oxidation number of zero as it is an element.
In hydrochloric acid, the chlorine atom has an oxidation number of zero to -1 showing that it has been reduced. However, in chloric (I) acid, the chlorine atom as an oxidation number of +1 (oxidised)
A redox reaction in which atoms of the same element are oxidised and reduced is called a disproportionation reaction
Risks of using chlorine on drinking water
Chlorine is a toxic gas and has to be handled carefully at water treatment plants
Chlorine water can react with naturally occurring hydrocarbons such as decaying material The chlorinated hydrocarbons products could increase the risk of cancer in humans (Health benefits of chlorinated drinking water outweigh the problems)
What happens during a displacement reaction
In a displacement reaction, we take a halogen in aqueous solution and we react this with an aqueous solution of a metal halide
Explain the displacement reaction of:
Cl2(aq) + 2NaBr(aq)—> 2NaCl (aq) + Br2(aq)
Cl2(aq) + 2Br(aq)^- —> 2Cl (aq)^ - + Br2(aq)
Chlorine is more reactive than bromine, in this reaction the chlorine has displaced the bromide ion. We’ve formed bromine and sodium chloride
Chlorine is a more powerful oxidising agent than bromine. (Greater ability to remove an electron).
At the start the chlorine atom has an oxidation number of zero as chlorine is an element. Each bromide ion has an oxidation number of -1. In this reaction, each chlorine atom removes an electron (oxidised) from the bromide ion. The chlorine atom are being reduced to chloride ions. At the end the bromine atom have an oxidation number of zero. The chloride ion each have an oxidation number of -1.
The changes in oxidation are balance. This happens as chlorine is a more powerful oxidising agent than bromine.
The metal ion would be a spectator ion, it does not take part in the redox reaction adn inc Luke in the redox equation
Explain the problem with halogens forming a colour change
In aqueous solution, both bromine and iodine can appear orange-brown. To solve this, we add a non-polar organic solvent, for example cyclohexane.
Cyclohexane forms an upper layer and does not take part in the reaction.However, the halogens formed dissolves in the cyclohexane. Allowing us to see the colours of the halogens more clearly.
In organic solvents, bromine appear orange and iodine appear violet.
Explain the colour change of the reaction between chlorine with sodium bromide solution
And
Chlorine with sodium iodide
And bromide with sodium iodide
At the start the chlorine solution is a pale-green colour.The reaction then forms bromine which is orange.When we add cyclohexane, the bromine dissolves in the upper layer which turns orange.
If we react chlorine with sodium iodide, then we form iodine In aqueous solution, iodine appear brown. However, in the upper cyclohexane layer, the iodine appears violet.
Reacting bromide with sodium iodide again forms iodine. This appears brown in aqueous solution but violet in the cyclohexane.
Describe the haldie test:
Ag(aq)^+ + X^- —> AgX(s)
The halide test uses a solution of silver nitrate which is soluble in water
A solution of silver nitrate contains the silver ion Ag+ as well as a nitrate ion NO3-
If we mix silver nitrate solution with a solution containing a halide ion then the silver ion reacts with the halide ion to form silver halide
Silver halide are insoluble in water therefore in this reaction a precipitate of silver halide is formed. The colour of the silver halide can be used to determine the haldie ion that we started with.
Key ideas of the halide test
Silver halides are insoluble in water
The three colours of the halide test are similar. This makes it difficult to determine which halide is present. This can be used with aqueous ammonia.
A precipitate of silver chloride will redissolve if we add dilute aqueous ammonia.
A precipitate of silver bromide will not redissolve if we use dilute aqueous ammonia but it will redissolve if we add concentrated aqueous ammonia.
A precipitate of silver iodide will not redissolve in either dilute or concentrated aqueous ammonia.
By using a solution of silver nitrate, followed by dilute or concentrated aqueous ammonia we can determine the identity of an unknown halide
Using silver nitrate solution to test for halide, has a potential problem. Silver ions will form a precipitate with both carbonate and sulfate ions. (We need to test for carbonate and sulfate ions first before testing for haldie with silver nitrate solution)
What colour precipitate does reacting silver nitrate with the different halogens form
Reacting silver nitrate with the chloride ion forms a white precipitate of silver chloride
Reacting silver nitrate with the bromdie ion forms a cream precipitate of silver bromide
Reacting silver nitrate with an iodide ion forms a pale yellow precipitate of silver iodide
How can we test for carbonate ions
We tests carbonate ion by adding dilute nitric acid to a sample . If a gas is produced, then we test the gas with lime water.
If limewater turns cloudy then the gas is carbon dioxide . This means that carbonate is present.
We keep on adding dilute nitric acid until no more gas is produced. At this point all of the carbonate ions have reacted
How can we test for sulfate ions
To test for sulfate ions, we add barium nitrate solution. A white precipitate of barium sulfate shows that sulfate ions are present.
We then add excess barium nitrate to ensure all of the sulfate ions have formed a precipitate.
At this point all, we filter off the solid barium sulfate. Because we removed all of the sulfate ions we can now test the resulting solution for halides.
