In the previous section we saw Combination reactions, Decomposition reactions and Metal displacement reactions. The chart shown below will help us to recall those categories.
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| Fig.8.3 |
II. Hydrogen displacement reaction
• This can be explained in 7 steps:
1. Consider the general form: X + HZ → XZ + H
♦ In a hydrogen displacement reaction, a metal X replaces H in the compound HZ
2. Let us see which all metals can replace H. The following steps from (3) to (7) will enable us to make a list.
(3) All alkali metals (group I elements) can displace H even from cold water. Na is an example:
$\mathbf\small{\rm{2\,\overset{0}{Na}\,(s)+2\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)\rightarrow
2\,\overset{+1}{Na}\,\overset{-2}{O}\,\overset{-1}{H}\,(aq)
+\overset{0}{H}_2\,(g)}}$
• We see that, H gets displaced and becomes free hydrogen gas.
• Comparing with the general form in (1), we get:
♦ X is Na
♦ HZ is H2O
♦ XZ is NaOH
• So this reaction is a displacement reaction.
• By observing the oxidation numbers, we also see that, it is a redox reaction also.
♦ H is reduced
♦ Na is oxidized
• So Na is a better reductant than H. That means, Na is a better donor of electrons than H.
◼ In fact all alkali metals are better reductants than H.
(4) Some alkaline earth metals (group II elements) like Ca, Sr and Ba can also displace H even from cold water.
$\mathbf\small{\rm{\overset{0}{Ca}\,(s)+2\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)\rightarrow
2\,\overset{+1}{Ca}\,(\overset{-2}{O}\,\overset{-1}{H})_2\,(aq)+\overset{0}{H}_2\,(g)}}$
• We see that, H gets displaced and becomes free hydrogen gas.
• Comparing with the general form in (1), we get:
♦ X is Ca
♦ HZ is H2O
♦ XZ is Ca(OH)2
• So this reaction is a displacement reaction.
• By observing the oxidation numbers, we also see that, it is a redox reaction also.
♦ H is reduced
♦ Ca is oxidized
• So Ca is a better reductant than H. That means, Ca is a better donor of electrons than H.
5. Metals like Mg and Fe are less active than Ca, Sr and Ba. They will not react with cold water. They can react with water only if it is hot water. The water must be so hot that, it must be in the vapor state (steam).
Example 1:
$\mathbf\small{\rm{\overset{0}{Mg}\,(s)+2\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)\rightarrow \overset{+1}{Mg}\,(\overset{-2}{O}\,\overset{-1}{H})_2\,(aq)+\overset{0}{H}_2\,(g)}}$
• We see that, H gets displaced and becomes free hydrogen gas.
• Comparing with the general form in (1), we get:
♦ X is Mg
♦ HZ is H2O
♦ XZ is Mg(OH)2
• So this reaction is a displacement reaction.
• By observing the oxidation numbers, we also see that, it is a redox reaction also.
♦ H is reduced
♦ Mg is oxidized
Example 2:
$\mathbf\small{\rm{2\,\overset{0}{Fe}\,(s)+3\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)\rightarrow \overset{+3}{Fe}_2\,\overset{-2}{O}_3\,(aq)
+3\,\overset{0}{H}_2\,(g)}}$
• We see that, H gets displaced and becomes free hydrogen gas.
• Comparing with the general form in (1), we get:
♦ X is Fe
♦ HZ is H2O
♦ XZ is Fe2O3
• So this reaction is a displacement reaction.
• By observing the oxidation numbers, we also see that, it is a redox reaction also.
♦ H is reduced
♦ Fe is oxidized
6. We saw that, less active metals do not displace H from cold water. They need steam.
• But such 'metals which do not displace H from cold water', can easily do so from acids.
Example 1:
$\mathbf\small{\rm{\overset{0}{Zn}\,(s)+2\,\overset{+1}{H}\,\overset{-1}{Cl}\,(aq)\rightarrow \overset{+2}{Zn}\,\overset{-1}{Cl}_2\,(aq)+\overset{0}{H}_2\,(g)}}$
Example 2:
$\mathbf\small{\rm{\overset{0}{Mg}\,(s)+2\,\overset{+1}{H}\,\overset{-1}{Cl}\,(aq)\rightarrow \overset{+2}{Mg}\,\overset{-1}{Cl}_2\,(aq)+\overset{0}{H}_2\,(g)}}$
Example 3:
$\mathbf\small{\rm{\overset{0}{Fe}\,(s)+2\,\overset{+1}{H}\,\overset{-1}{Cl}\,(aq)\rightarrow
\overset{+2}{Fe}\,\overset{-1}{Cl}_2\,(aq)+\overset{0}{H}_2\,(g)}}$
• In each of the above three examples,
♦ the metal is oxidized
♦ the hydrogen is reduced.
• The above three reactions are used to produce H2 gas in the laboratory.
• The ‘rate of H2 production’ (amount of H2 produced in unit time) varies for the three metals.
♦ The rate is highest for Mg, which is the most reactive among the three.
♦ The rate is lowest for Fe, which is the least reactive among the three.
7. Metals like silver (Ag) and gold (Au) are very less active. They occur free in nature. They do not react even with HCl.
III. Halogen displacement reaction
• This can be explained in 7 steps:
1. We know that halogens are electronegative. So they pull electron from other atoms. This results in the oxidation of those other atoms.
2. The main halogens are: F, Cl, Br and I
• The ‘electron pulling capacity’ decreases in the order: F > Cl > Br > I
3. F has so much pulling capacity that, it will pull electrons even from water.
♦ So we will discuss F later separately.
♦ At present, we will discuss Cl, Br and I
4. Let us see a reaction which demonstrates the competition between Cl and Br. It can be written in 6 steps:
(i) Consider KBr. It is an ionic compound. K+ and Br- stick together due to the electrostatic force of attraction.
(ii) In an aqueous solution of KBr, they exist as independent ions K+ and Br-
(iii) Suppose that, to this aqueous solution, Cl2 is added.
♦ The Cl atoms will pull the electron away from each of the Br- ions.
♦ The Br- ions will thus become Br atoms
(iv) But Br atoms are electron deficient. They are unstable. So the only option available for the Br atoms is to form covalent bonds among themselves.
♦ Thus free Br2 molecules will be formed in the solution.
• Br2 molecules are colored. So the formation of Br2 atoms can be easily observed in the solution.
(v) The balanced equation of the reaction is:
$\mathbf\small{\rm{\overset{0}{Cl}_2\,(g)+2\,\overset{+1}{K}\,\overset{-1}{Br}\,(aq)\rightarrow 2\,\overset{+1}{K}\,\overset{-1}{Cl}\,(aq)
+\overset{0}{Br}_2\,(l)}}$
• We see that, Cl displaces Br from KBr. So it is a displacement reaction.
• From the oxidation numbers, we also see that,
♦ Cl is reduced.
♦ Br is oxidized.
(vi) This reaction can be written in ionic form as:
$\mathbf\small{\rm{\overset{0}{Cl}_2\,(g)+2\,\overset{-1}{Br^-}\,(aq)
\rightarrow 2\,\overset{-1}{Cl^-}\,(aq)+\overset{0}{Br}_2\,(l)}}$
• We see that, Cl pulls away the electron from Br- ions
• This equation gives us an idea to detect the presence of Br- ions:
♦ If Br- ions are present in the solution, colored Br2 molecules will be formed,
♦ when gaseous chlorine is passed through that solution
• This test is known as Layer test.
5. Let us see a reaction which demonstrates the competition between Cl and I. It can be written in 6 steps:
(i) Consider KI. It is an ionic compound. K+ and I- stick together due to the electrostatic force of attraction.
(ii) In an aqueous solution of KI, they exist as independent ions K+ and I-
(iii) Suppose that, to this aqueous solution, Cl2 is added.
♦ The Cl atoms will pull the electron away from each of the I- ions.
♦ The I- ions will thus become I atoms
(iv)
But I atoms are electron deficient. They are unstable. So the only
option available for the I atoms is to form covalent bonds among
themselves.
♦ Thus free I2 molecules will be formed in the solution.
• I2 molecules are colored. So the formation of I2 molecules can be easily observed in the solution.
(v) The balanced equation of the reaction is:
$\mathbf\small{\rm{\overset{0}{Cl}_2\,(g)+2\,\overset{+1}{K}\,\overset{-1}{I}\,(aq)\rightarrow
2\,\overset{+1}{K}\,\overset{-1}{Cl}\,(aq)
+\overset{0}{I}_2\,(l)}}$
• We see that, Cl displaces I from KI. So it is a displacement reaction.
• From the oxidation numbers, we also see that,
♦ Cl is reduced.
♦ I is oxidized.
(vi) This reaction can be written in ionic form as:
$\mathbf\small{\rm{\overset{0}{Cl}_2\,(g)+2\,\overset{-1}{I^-}\,(aq)
\rightarrow 2\,\overset{-1}{Cl^-}\,(aq)+\overset{0}{I}_2\,(s)}}$
• We see that, Cl pulls away the electron from I- ions
• This equation gives us an idea to detect the presence of I- ions:
♦ If I- ions are present in the solution, colored I2 molecules will be formed,
♦ when gaseous chlorine is passed through that solution
• As mentioned above, this test is known as Layer test.
6. Consider the series that we wrote in (2): F > Cl > Br > I
• We see that, I is to the right of Br. So we would expect Br to have greater electron pulling capacity than I.
• Indeed, it is the case. Just as in (4) and (5), we can show reactions where addition of Br leads to the formation of I2 molecules.
• The equation is:
$\mathbf\small{\rm{\overset{0}{Br}_2\,(g)+2\,\overset{-1}{I^-}\,(aq)
\rightarrow 2\,\overset{-1}{Br^-}\,(aq)+\overset{0}{I}_2\,(s)}}$
7. Now we can write about F. It can be written in steps:
(i) Consider the series that we wrote in (2): F > Cl > Br > I
• We see that, F is in the left most position. So it will have the greatest electron pulling capacity.
(ii) The capacity of F is so high that, it can pull away electrons even from the O atom in H2O.
• The equation is:
$\mathbf\small{\rm{2\,\overset{0}{F}_2\,(g)+2\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)\rightarrow 4\,\overset{+1}{H}\,\overset{-1}{F}\,(aq)
+\overset{0}{O}_2\,(g)}}$
(iii) We saw that:
♦ Cl can be used to obtain Br2 molecules
♦ Cl can be used to obtain I2 molecules
♦ Br can be used to obtain I2 molecules
• Likewise, if we want to use F to obtain Cl2, Br2 or I2, the reaction should not be carried out in an aqueous solution. Because, F will attack the water.
8. The above seven steps give us an idea about halogen displacement reactions. Those reactions have direct industrial application. This can be easily explained in 4 steps:
(i) Imagine that, we have some quantity of KI. We want to obtain pure halogen molecules (I2 molecules)
(ii) We make an aqueous solution of KI and pass chlorine gas through it. The I2 molecules will precipitate.
(iii) The general formula is: 2X- → X2 + 2e-
♦ Here X denotes a halogen atom.
(iv) Remember that, there are no elements to the left of F in the series:
F > Cl > Br > I
• So to obtain F2 molecules, we need to use electricity. We will see those details in later sections.
We have completed a discussion on the three types of displacement reactions. We can write a simple comparison between the three types. It can be written in 3 steps:
1. Metal displacement reactions:
• Here the competition is between various metals.
• Some metals have greater capacity to donate electrons. The other metals will be forced to accept electrons.
2. Hydrogen displacement reactions:
• Here the competition is between metals and hydrogen.
• Some metals can force hydrogen to accept electrons even in cold water.
• Some other metals can force hydrogen only if it is steam.
• Most metals can force hydrogen in acids.
3. Halogen displacement reactions:
• Here the competition is between various halogens.
• Some halogens have greater capacity to pull electrons away. The other halogens will be forced to donate electrons.
In the next section, we will see disproportionation reactions.
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