Chapter 9 - Hydrogen

In the previous section, we completed a discussion on redox reactions. In this chapter, we will see Hydrogen.
• We know that:
    ♦ Hydrogen atom has only one electron and one proton.
        ✰ It is the simplest atom in the universe.
    ♦ Two hydrogen atoms combine together by covalent bond to attain duplet.
        ✰ Duplet means: two electrons in the outer most shell.
    ♦ So hydrogen is usually found in nature as H₂. It is called dihydrogen.

Position of hydrogen in the periodic table

This can be explained in 14 steps:
1. We know that elements in the periodic table are arranged according to their electronic configurations.
2. Consider the electronic configuration of hydrogen. It is 1s¹
• But group 1 elements (alkali metals) also have the general outer electronic configuration of ns¹.
3. We can write:
• Alkali metals lose outer most electron to attain stability. They lose their only one outer most electron and become unipositive ions (ions with one positive charge)
For example: Li becomes Li⁺, Na becomes Na⁺
• Hydrogen also loses it’s outer most electron to attain stability. It loses it’s only one outer most electron and become unipositive ion.
That is: H becomes H⁺
4. Let us examine compounds formed by alkali metals and hydrogen.
• Alkali metals form oxides, halides and sulphides.
• Hydrogen also forms oxides, halides and sulphides.
5. Steps (1) to (4) shows the similarities between hydrogen and alkali metals. Seeing those similarities, we are tempted to include hydrogen in the group 1.
6. But we have to consider the similarity of hydrogen with group 17 elements (halogens) also.
• Halogens have the general outer electronic configuration of ns²np⁵.
    ♦ So they need one more electron to attain stability (octet).
• Hydrogen also needs one more electron to attain stability (duplet).
7. Halogens form diatomic molecules to attain stability. F₂, Cl₂ etc., are examples.
• Hydrogen also form the diatomic molecule H₂ to attain stability.
8. Let us see a comparison in terms of ionization enthalpies:
    ♦ The ionization enthalpy of hydrogen is 1312 kJ mol^-1
    ♦ The ionization enthalpy of fluorine is 1680 kJ mol^-1
• We see that:
Ionization enthalpy of hydrogen is closer to that of fluorine.
    ♦ Fluorine is a halogen.
9. Steps (6) to (8) shows the similarities between hydrogen and halogens. Seeing those similarities, we are tempted to include hydrogen in the group 17.
10. In the above steps,
    ♦ We have seen the similarities of hydrogen with alkali metals.
    ♦ We have seen the similarities of hydrogen with halogens.
• Next, we will see the differences:
11. Differences in terms of ionization enthalpies:
    ♦ The ionization enthalpy of hydrogen is 1312 kJ mol^-1
    ♦ The ionization enthalpy of lithium is 520 kJ mol^-1
• We see that,
ionization enthalpy of hydrogen is much different from that of lithium.
    ♦ Lithium is an alkali metal.
• So in terms of ionization enthalpy, we are tempted not to put hydrogen in group 1.
12. Differences in terms of reactivity:
    ♦ Halogens show high reactivity.
    ♦ But when compared to halogens, reactivity of hydrogen is very low.
• So in terms of reactivity, we are tempted not to put hydrogen in group 17.
13. We must also consider a unique behaviour of hydrogen. It can be written in 4 steps:
(i) When hydrogen loses it’s only one electron, it becomes H⁺ ion.
(ii) Since the only one electron is lost, H⁺ is a proton. It is very small in size.
Size of an H⁺ ion is 1.5 × 10^-3 pm.
• The ionic size of other atoms in general is 50 to 200 pm.
(iii) So H⁺ is extremely small when compared to other ions.
• That means, the charge is concentrated on a very small space. Consequently, the charge density is very high.
• Due to this very high charge density, the  H⁺ ion attacks any thing which comes to it’s proximity.
(iv) As a result, H⁺ is never found free in nature.
14. Due to these peculiarities, hydrogen is placed separately from the main body of the periodic table.

Occurrence of dihydrogen

This can be written in 3 steps:
1. Dihydrogen is the most abundant element in the universe.
2. On Earth, elemental dihydrogen is found in the atmosphere. But since hydrogen molecules have only low mass, Earth’s gravity is not able to exert enough pull to keep them confined in the atmosphere.
3. However, large quantities of hydrogen are available on Earth in combined form.
• For example,
    ♦ Each molecule of water has two atoms of hydrogen.
    ♦ Carbohydrates, proteins etc., that are found in living organisms also contain hydrogen in combined form.
    ♦ Fuels like petrol, kerosene etc., which are hydrocarbons also contain hydrogen in combined form.

Isotopes of hydrogen

This can be written in 9 steps:
1. We know that, isotopes are different forms of a same substance.
• We can arrange isotopes into different groups:
    ♦ All isotopes of hydrogen in one group,
    ♦ All isotopes of carbon in another group,
    ♦ All isotopes of chlorine in yet another group so on . . .
2. All members of a group will have the same atomic number. But their mass numbers will be different.
3. Let us examine the group containing the isotopes of hydrogen.
(i) There are three members in that group.
(ii) All the three members in the group will have the same atomic number 1.
    ♦ That means, each of the three members has one proton in the nucleus.
(iii) One of the three members is our ordinary hydrogen. We know that, it has no neutrons.
• So it’s mass number = (number of electrons + number of protons) = (1 + 0 ) = 1
    ♦ This isotope is named as Protium. It’s symbol is: ₁¹H
(iv) The second member has one proton and one neutron.
• So it’s mass number = (number of electrons + number of protons) = (1 + 1) = 2
    ♦ This isotope is named as Deuterium. It’s symbol is: ₁²H
(v) The last member has one proton and two neutrons.
• So it’s mass number = (number of electrons + number of protons) = (1 + 2) = 3
    ♦ This isotope is named as Tritium. It’s symbol is: ₁³H
4. Protium is the predominant form. It is the ordinary hydrogen that we are familiar with.
5. If we take a sample of naturally occurring hydrogen, 99.98% of that sample will be protium. Only 0.0156% will be deuterium. In that sample, deuterium will be in the form of HD. That is., one H atom in the H₂ molecule is replaced by one D atom.
6. Tritium occurs only rarely. If we take 10¹⁸ atoms of protium, only one among them will be tritium.
7. Among the three isotopes of hydrogen, only tritium is radioactive.
8. The chemical properties of the three isotopes are almost same. This is because, they have the same electronic configuration.
• But the rate of reactions will differ. Because, they have different bond dissociation enthalpies.
9. The three isotopes have different physical properties because, they have different atomic masses.

Preparation of dihydrogen in the laboratory

The following two methods are usually used for preparing dihydrogen in the laboratory.
1. Reaction of granulated zinc with dilute hydrochloric acid gives dihydrogen.
Zn + 2H⁺ → Zn²⁺ + H₂
2. Reaction of zinc with aqueous alkali gives dihydrogen.
Zn + 2NaOH → Na₂ZnO₂ + H₂

Commercial production of dihydrogen

There are four methods available to produce hydrogen commercially.
Method 1: Electrolysis of acidified water
This can be explained in 5 steps:
1. We have already seen the basics of electrolysis in our previous classes:
    ♦ Two electrodes are partially immersed in the electrolyte.
    ♦ Those electrodes are connected to a cell.
    ♦ One electrode is connected to the positive terminal of the cell.
    ♦ The other electrode is connected to the negative terminal of the cell.
    ♦ The circuit is completed by the flow of ions through the electrolyte.
Some images can be seen here.
2. In our present case:
• Water is the electrolyte.
• But pure water contains only very few H⁺ and OH^- ions.
• So we add a small amount of acid (H₂SO₄) to the water.
• The acid dissociates into H⁺ and SO₄^2- ions.
• The newly added H⁺ ions react with H₂O molecules and produce H₃O⁺ ions and OH^- ions:
• Thus the required number of ions are produced so that, electricity can be conducted through the electrolyte.
3. Consider the negative electrode. It is the electrode connected to the negative terminal of the cell. Electrons are present in that electrode.
• The positive ions H₃O⁺ move towards this negative electrode and accepts electrons. Thus they get reduced.
2H₃O⁺  + 2e^- → H₂ + 2H₂O
• Thus we get pure dihydrogen.
• We know that, the electrode at which reduction takes place is called cathode.
◼ So we can write:
dihydrogen is obtained at the cathode.
4. Consider the positive electrode. It is the electrode connected to the positive terminal of the cell.
• The negative ions SO₄^2- and OH^- ions move towards this electrode.
• The SO₄^2- ions are stable. They do not donate electrons.
• The OH^- ions donate electrons. They get oxidized.
4OH^- → 2H₂O + O₂ + 4e^-
• Thus we get pure oxygen.
• We know that, the electrode at which oxidation takes place is called anode.
◼ So we can write:
oxygen is obtained at the anode.
5. The acid added to the water will react with the metal electrodes. This will lead to the corrosion of electrodes. To avoid such a situation, we use platinum electrodes. Platinum is a noble metal like gold. It is not corroded by acids.

Method 2: Electrolysis of warm aqueous barium hydroxide solution.
• This method is similar to method 1 except that, the electrolyte here is alkaline instead of acidic. Remember that barium hydroxide is an alkali.
• As before, the H₃O⁺ ions will get reduced at the cathode to give dihydrogen.
• The OH^- ions will get oxidized at the anode to give oxygen.
• In this method, nickel is used to make electrodes. This is because, Ni is more resistant to corrosion by alkaline solutions.

Method 3: By the electrolysis of brine solution.
This can be explained in 4 steps:
1. Brine is highly concentrated solution of NaCl in water.
• It’s electrolysis is done mainly for the production of NaOH and Cl₂
• During this process, dihydrogen is obtained as a byproduct.
2. The Cl^- ions reach the positive electrode and get oxidized.
2Cl^- → Cl₂ + 2e^-
• We know that electrode at which oxidation takes place is anode.
◼ So we can write:
Chlorine is produced at the anode.
3. The H₃O⁺ reach the this negative electrode and get reduced.
2H₃O⁺ + 2e^- → H₂ + 2H₂O
• We know that, the electrode at which reduction takes place is called cathode.
◼ So we can write:
dihydrogen is obtained at the cathode.
4. When chlorine and hydrogen are removed, what remains in the solution is Na⁺ and OH^- ions. Thus we get a highly concentrated solution of NaOH as the main product.

Method 4: Reaction of steam on hydrocarbons or coke at high temperatures in the presence of catalyst.
This can be explained in steps:
1. We know that hydrocarbons are those compounds which contain carbon and hydrogen. They have the general formula: C_nH_2n+2
2. Coke also contains carbon and hydrogen. It is produced by heating coal in the absence of oxygen.
• The coke thus produced can be used as a fuel. The advantage of using coke as a fuel over coal is that, it produces very little smoke during burning.
• Coke is also used in the manufacture of iron.
3. The reaction between hydrocarbon and steam is:
C_nH_2n+2 + nH₂O → nCO + (2n+1)H₂
For example:
CH₄ (g) + H₂O (g) → CO (g) + 3H₂ (g)
• Thus we get dihydrogen.
4. We see that, on the product side, we have a mixture of CO and H₂.
• This mixture is called water gas.
• This mixture can be used for the synthesis of methanol and a number of hydrocarbons. So it is also known as syngas.
5. We just saw that, syngas is produced from hydrocarbons and steam.
Coal can also be used instead of hydrocarbons. The equation is:
C (s) + H₂O (g) → CO + H₂ (g)
This process is called coal gasification.
6. We saw that water gas contains both CO and H₂. But what we want mainly is H₂. So we collect the CO from the water gas and react it with steam in the presence of iron chromate as catalyst. We get a mixture of carbon dioxide and H₂. This process is called water gas shift reaction.
CO (g) + H₂O → CO₂ (g) + H₂ (g)
• We can write:
The CO in the water gas is used to extract hydrogen from steam.
• We see that, on the product side, CO₂ and H₂ are present. We can remove the CO₂ by scrubbing with sodium arsenite solution.

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◼ After the discussion about the above four methods, we can write:
Hydrogen can be produced from:
    ♦ Hydrocarbons (These are readily available from the petroleum industry)
    ♦ Coal
    ♦ Electrolysis
• 77% of the industrial dihydrogen is produced from hydrocarbons.
• 18% is produced from coal
• 4% is produced from electrolysis.
• 1% is produced from other sources.

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In the next section, we will see properties of dihydrogen.


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Chapter 2.3 - Atomic number and Mass number

In the previous section, we saw the models proposed by J.J. Thomson and Ernst Rutherford. In this section we will see details about atomic number and mass number

1. We saw that in an atom, there are electrons, protons and neutrons
• Each electron has a negative charge of -1.602176 ×10^-19 C 
• Each proton has a positive charge of +1.602176 ×10^-19 C
2. The magnitudes of both the charges are equal
• But the signs are opposite
3. So the number of protons and electrons in an atom must be equal
• Otherwise, the atom will not be electrically neutral
4. The number of protons in an atom is called the atomic number of that atom
• The symbol of atomic number is Z
Some examples:
• The number of protons in hydrogen atom is 1
    ♦ So Z of hydrogen atom is 1
• The number of protons in sodium atom is 11
    ♦ So Z of sodium atom is 11
5. We can write:
Atomic number (Z) of an atom 
= Number of protons in the nucleus of that atom
= Number of electrons in that ‘atom in neutral state’
• ‘Atom in neutral state’ has to be specifically written because, if it is not electrically neutral, the number of electrons will not be equal to the number of protons
6. Next we consider mass number
• Protons and neutrons present in the nucleus are collectively known as nucleons
• The total mass of the nucleons is equal to the mass of the atom
• The total number of nucleons in an atom is called the 'mass number' of that atom
• The symbol of mass number is A
7. We can write:
• Mass number of an atom = Number of protons (Z) + Number of neutrons (n)
• We have seen the details in our previous classes. 
8. We know that every element is represented by a unique symbol
• Z and A can be attached to that symbol
    ♦ Z is written as a subscript on the left side of the symbol 
    ♦ A is written as a superscript on the left side of the symbol 
9. Thus, if ‘X’ is the symbol of an element and Z and A the atomic and mass numbers, we can write: $\mathbf\small{\rm{^A_ZX}}$

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Isobars

• Isobars are elements with same mass number (A) but different atomic numbers (Z)
Example: $\mathbf\small{\rm{^{14}_{\;6}C\;\;\text{and}\;\;^{14}_{\;7}N}}$

Isotopes

Details about isotopes can be written in 6 steps:
1. Consider the atom of hydrogen
• It has 1 proton, 0 neutron and 1 electron
    ♦ So Z = Number of protons = 1
    ♦ Also, A = (Z+n) = (1+0) = 1
• Symbolically we can represent this as $\mathbf\small{\rm{^1_1H}}$
• It is called protium
2. Consider the atom of deuterium
• It has 1 proton, 1 neutron and 1 electron
    ♦ So Z = Number of protons = 1
    ♦ Also, A = (Z+n) = (1+1) = 2
• Symbolically we can represent this as $\mathbf\small{\rm{^2_1D}}$
3. Consider the atom of tritium
• It has 1 proton, 2 neutron and 1 electron
    ♦ So Z = Number of protons = 1
    ♦ Also, A = (Z+n) = (1+2) = 3
• Symbolically we can represent this as $\mathbf\small{\rm{^3_1T}}$
4. In all the above 3 cases, Z is the same. But A are different
• The difference in A is due to the difference in n
■ Elements with same atomic number (Z) but different mass numbers (A) are called isotopes
• $\mathbf\small{\rm{^1_1H,\;\;\,^2_1D\;\;\text{and}\;\;^3_1T}}$ are isotopes
5. If we take a sample of naturally occurring hydrogen, we get the following quantities:
(i) 99.985% of that sample will be $\mathbf\small{\rm{^1_1H}}$
(ii) The rest 0.015% will be $\mathbf\small{\rm{^2_1D}}$ 
(iii) $\mathbf\small{\rm{^3_1T}}$ is found only in traces on earth
6. Let us see some more examples of isotopes:
• Isotopes of carbon:
(i) $\mathbf\small{\rm{^{12}_{\;6}C}}$ has 6 protons and 6 neutrons
(ii) $\mathbf\small{\rm{^{13}_{\;6}C}}$ has 6 protons and 7 neutrons
(iii) $\mathbf\small{\rm{^{14}_{\;6}C}}$ has 6 protons and 8 neutrons
• Isotopes of chorine:
(i) $\mathbf\small{\rm{^{35}_{17}Cl}}$ has 17 protons and 18 neutrons
(ii)$\mathbf\small{\rm{^{37}_{17}Cl}}$ has 17 protons and 20 neutrons

Chemical properties of isotopes

1. The chemical properties of an element are controlled by the number of electrons in it’s atoms
2. The number of neutrons have very little effect on the chemical properties
3. The number of electrons is same as the atomic number (Z)
• So with regard to chemical properties, the number Z is what we consider most
4. All isotopes of an element will have the same Z
• So all isotopes of an element will show the same chemical behaviour

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Now we will see some solved examples:

Solved example 2.1
Calculate the number of protons, neutrons and electrons in $\mathbf\small{\rm{^{80}_{35}Br}}$
Solution:
1. Given : Z = 35, A = 80
2. We have: Number of protons = Number of electrons = Z = 35
3. Also we have:  Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (80-35) = 45

Solved example 2.2
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species
Solution:
1. Given: 
• Number of electrons = 18
• Number of protons = 16
    ♦ So Z = 16
    ♦ From the periodic table, we have: Z = 16 for sulphur (S)
• Number of neutrons = 16
2. We have:  A = Number of neutrons + Number of protons
⇒ A = (16+16) = 32
3. In a neutral atom, the number of electrons will be equal to that of the protons
• But in this case, number of electrons is greater by 2
• So there will be an excess of two negative charges
• So the symbol is: $\mathbf\small{\rm{^{32}_{16}S^{2-}}}$
4. Note:
• If the number of protons and electrons are not equal, the given species is an ion
• We have to determine whether it is a cation or an anion
• If Number of protons > Number of electrons
    ♦ It is a +ve ion (cation)
• If Number of protons < Number of electrons
    ♦ It is a -ve ion (anion)

Solved example 2.3
(i) Calculate the number of electrons which will together weigh one gram
(ii) Calculate the mass and charge of one mole of electrons
Solution:
Part (i)
• Mass of one electron = 9.10939 ×10^-31 kg = 9.10939 ×10^-28 g
• So number of electrons in 1 g 
= $\mathbf\small{\frac{1}{9.10939\times 10^{-28}}=0.1097768\times 10^{28}=1.098\times 10^{27}}$
Part (ii)
• One mole of electrons will contain 6.022 × 10²³ electrons
• Mass of one electron = 9.10939 ×10^-31 kg
• So mass of one mole of electrons = (6.022 × 10²³ × 9.10939 ×10^-31) = 5.48 ×10^-7 kg
• Charge of one electron = -1.602176 ×10^-19 C
• So charge of one mole of electron = (6.022 × 10²³ × -1.602176 ×10^-19) = 9.65 ×10⁴ C

Solved example 2.4
(i) Calculate the total number of electrons present in one mole of methane
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of $\mathbf\small{\rm{^{14}C}}$
(Assume that mass of a neutron = 1.675 × 10^-27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed ?
Solution:
Part (i)
1. The molecular formula of methane is CH₄
• So in one molecule of methane, there are:
    ♦ 1 atom of C
    ♦ 4 atoms of H
2. In one atom of C there are 6 electrons
• In one atom of H there is 1 electron
• So in 1 molecule of methane, there are (1×6 + 4×1) = 10 electrons
3. One mole of methane contains 6.022 × 10²³ molecules of methane
• So in one mole of methane, there will be 6.022 × 10²³ × 10 = 6.022 × 10²⁴  electrons
Part (ii)
1. Molar mass of $\mathbf\small{\rm{^{14}C}}$ is 14 g
⇒ In 14 g of $\mathbf\small{\rm{^{14}C}}$, there will be 6.022 × 10²³ atoms
⇒ In 1 g, there will be (6.022 × 10²³ / 14) atoms
⇒ In 1 mg there will be (6.022 × 10²³ / 14000) atoms 
⇒ In 7 mg there will be (6.022 × 10²³ / 2000) = 3.011 × 10²⁰ atoms
2. Number of neutrons in 1 $\mathbf\small{\rm{^{14}C}}$ atom = (A-Z) = (14-6) = 8
• So total number of neutrons in 7 mg = 8 × 3.011 × 10²⁰ = 24.088 × 10²⁰
3. Mass of this much neutrons = (24.088 × 10²⁰ × 1.675 × 10^-27) 
= 40.35 × 10^-7 
= 4.035 × 10^-6 kg
Part (iii)
1. Molar mass of NH₃ = (14+3) = 17 g
⇒ 17 g of NH₃ at STP will contain 6.022 × 10²³ molecules
⇒ 1 g of NH₃ at STP will contain (6.022 × 10²³ / 17) molecules
⇒ 1 mg of NH₃ at STP will contain (6.022 × 10²³ / 17000) molecules
⇒ 34 mg of NH₃ at STP will contain (6.022 × 2 × 10²⁰) molecules
2. one molecule of NH₃ will contain (7+3) = 10 protons 
• So number of protons in (6.022 × 2 × 10²⁰) molecules 
= 6.022 × 2 × 10 × 10²⁰ 
= 12.044 × 10²¹
= 1.2044 × 10²²
3. Mass of this much protons 
= (1.2044 × 10²² × mass of one proton)
= (1.2044 × 10²² × 1.67262 × 10^-27) 
= 2.015 × 10^-5 kg

Solved example 2.5
How many neutrons and protons are there in the following nuclei ?
$\mathbf\small{\rm{^{13}_{\;6}C,\;\;^{16}_{\;8}O,\;\;^{24}_{12}Mg,\;\;^{56}_{26}Fe,\;\;^{88}_{38}Sr}}$
Solution:
Part (i)
1. Given : Z = 6, A = 13
2. We have: Number of protons = Z = 6
3. We have:  Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (13-6) = 7
Part (ii)
1. Given : Z = 8, A = 16
2. We have: Number of protons = Z = 8
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (16-8) = 8
Part (iii)
1. Given : Z = 12, A = 24
2. We have: Number of protons = Z = 12
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (24-12) = 12
Part (iv)
1. Given : Z = 26, A = 56
2. We have: Number of protons = Z = 26
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (56-26) = 30
Part (v)
1. Given : Z = 38, A = 88
2. We have: Number of protons = Z = 38
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (88-38) = 50

Solved example 2.6

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35
(ii) Z = 92 , A = 233
(iii) Z = 4 , A = 9
Solution:
Part (i)
• Given: Z = 17, A = 35
    ♦ From the periodic table, we have: Z = 17 for chlorine (Cl)
• So symbol is $\mathbf\small{\rm{^{35}_{17}Cl}}$
Part (ii)
• Given: Z = 92, A = 233
    ♦ From the periodic table, we have: Z = 92 for uranium (U)
• So symbol is $\mathbf\small{\rm{^{233}_{92}U}}$
Part (iii)
• Given: Z = 4, A = 9
    ♦ From the periodic table, we have: Z = 4 for beryllium (Be)
• So symbol is $\mathbf\small{\rm{^{9}_{4}Be}}$

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In the next section we will see some more solved examples. We will also see the drawbacks of Rutherford's model

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