Chapter 4 - The Modern Periodic Table

In the previous section, we completed the discussion on oxidation and reduction. In this section, we will discuss about the Periodic table.

A large number of elements are present in nature. We need to arrange them in order. Proper arrangement is essential in every field. Consider the example of a shop. Goods like soaps, detergents etc., will be kept at one particular place. Goods like fruits, vegetables etc., will be kept at a different place, away from soaps and detergents. In the section for soaps itself, there will be sub-sections. Bathing soaps will be kept at one place, while washing soaps will be kept at another place.

Let us now try to arrange the elements. For that, first, we will make some cards. One card for each element. A sample card (the green square) is shown in fig.4.1 below. It is the card for Iron (Fe).

Fig.4.1

The above card, and the others, that we will use for our present discussion, are taken from Wikimedia commons. Each card will show many important details about the element. For our present discussion, we need only the following details:

• The atomic number Z of the element

• Symbol of the element

• Name of the element

• Electronic configuration of the element

First we need cards upto atomic number Z = 18, that is Argon. Once we prepare those cards, we will put them side by side in sequential order. This is shown in fig.4.2 below (right click, and select 'open in new tab' for an enlarged view):

Fig.4.2

Fig.4.2 shows the cards from 1 to 18 arranged side by side. After argon, we can continue with any number of elements that we like, and arrange like this. But such an arrangement will take up a large horizontal space from left to right. Also, such an arrangement will not serve any special purpose. We want metals to be together at one place, non-metals to be together at another place, gases to be together at yet another place, etc., So let us make some modifications to the above arrangement.

Let us take out the elements after Z= 10. That is., we take out the elements from Sodium with Z = 11, upto Argon with Z = 18. We take them and put them under the first 10 elements. This is shown in fig.4.3 below(right click, and select 'open in new tab' for an enlarged view):

Fig.4.3

• Care must be taken to see that, Sodium No.11 comes under Lithium No.3. 

Now look at the electronic configurations carefully.

♦ Lithium and sodium have 1 electron in their outer most shells, and now, both are grouped together in one column

♦ Beryllium and magnesium have 2 electrons in their outer most shells, and now, both are grouped together in one column

♦ Boron and aluminium have 3 electrons in their outer most shells, and now, both are grouped together in one column

We find the above pattern for all the elements from 3 to 18. Let us write it down:

• All the elements in a column have the same number of electrons in the outer most shells

• Also, when we move from left to right, with the passing of every one column, the number of electrons in the outer most shell increase by 1

We find that hydrogen and helium are left out. Let us give them appropriate positions:

• Take out hydrogen, and put it above lithium No.3

• Take out helium, and put it above neon No.10

The modified table is shown in fig.4.4 below:

Fig.4.4

• Now, the first column has 3 elements. All of them have 1 electron in their outer most shells

• The last column also has 3 elements. But the number of electrons in the outer most shell are not the same. Helium has 2, and the others have 8. We can think about it in this way:

The elements in this column have a stable configuration. That is., maximum number of electrons in the outer most shell. These elements in the last column do not usually take part in reactions because, they are stable.

• So now our table has 3 horizontal rows. Each of these horizontal rows is called a 'period'.
• Also, the table has 7 vertical columns. Each of these vertical columns is called a 'group'.
Do we see any relation ship between the following two:

    ♦ The position of any period

    ♦ The electronic configuration of the elements in that period?

• Indeed there is a relation. Look at the 1^st period. Take a closer look at the electronic configuration of the elements in that period. The configuration has only 1 digit. That means, all the elements in the 1^st period has only 1 shell.

    ♦ In fact, 'all the elements in the world, which have only 1 shell', are included in the 1^st period. Because, after hydrogen and helium, the next element is lithium, which has 2 shells.

• Look at the 2^nd period. Take a closer look at the electronic configuration of the elements in that period. The configuration has 2 digits. That means, all the elements in the 2^nd period has two shells.

    ♦ In fact, 'all the elements in the world, which have 2 shells', are included in the 2^nd period. Because, after neon, the next element is sodium, which has 3 shells.

• Look at the 3^rd period. Take a closer look at the electronic configuration of the elements in that period. The configuration has 3 digits. That means, all the elements in the third period has 3 shells.

    ♦ In fact, 'all the elements in the world, which have 3 shells', are included in the 3^rd period. Because, after argon, the next element is potassium, which has 4 shells.

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Let us write a summary of the discussion that we had so far.

1. The horizontal rows are called Periods

2. The vertical columns are called Groups

3. All the elements in a period will have the same number of shells

4. All the elements in a group will have the same number of electrons in the outer most shell.
5. As we move from top to bottom in the table, with the passing of each period, 1 shell gets added

6. As we move from left to right in the table, with the passing of each new group, 1 electron gets added in the outermost shell
7. The 'name of the period' will indicate:
    ♦ the number of shells present, in each element in that period
• For example, each elements in period 3 will have 3 shells. K, L and M
8. The 'name of the group' will indicate:
    ♦ the number of electrons present in the outer most shell of each element in that group
• For example, each elements in Group V will have 5 electrons in it's outer most shell

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So we have arranged the first 18 elements. Let us now arrange the rest. Rules 7 and 8 written above will help us.

■ The next element is No.19 Potassium. 

• It has 4 shells. K, L, M and N. So, according to rule 7 above, it falls in period 4

• It has 1 electron in the outer most shell. So, according to rule 8, it will fall in group I

• Based on the above 2, potassium is the first element in the 4^th period. So it will fall just below sodium as shown in the fig.4.5 below:

Fig.4.5

■The next element is No.20 calcium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, calcium is the second element in the 4^th period. So it will fall just below magnesium. This is also shown in the fig.4.5 above.

■ The next element is No.21 scandium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, scandium is the second element in the 4^th period. So it will fall just below magnesium in the fig.4.4.

• So here we encounter a problem. We have already assigned the 'position below magnesium' to calcium. Now, scandium is also claiming the same position. Scandium has the claim because, it too has '4 shells', and '2 electrons in the outer most shell'.

■ Let us try the next element. The next element is No.22 titanium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, titanium is the second element in the 4^th period. So it will fall just below magnesium in the fig.4.4.

• Here also we encounter the same problem. We have already assigned the 'position below magnesium' to calcium. Now, scandium and titanium are also claiming the same position. Titanium has the claim because, it too has '4 shells', and '2 electrons in the outer most shell'.

■ Let us try one more element. The next element is No.23 vanadium. 

• It has 4 shells. K, L, M and N. So, according to rule 7, it falls in period 4

• It has 2 electrons in the outer most shell. So, according to rule 8, it will fall in group II

• Based on the above 2, vanadium is the second element in the 4^th period. So it will fall just below magnesium in the fig.4.4.

• Here also we encounter the same problem. We have already assigned the 'position below magnesium' to calcium. Now, scandium, titanium and vanadium are also claiming the same position. Vanadium has the claim because, it too has '4 shells', and '2 electrons in the outer most shell'.

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Why is there more than 1 claim for a single position? Let us analyse:

Consider the electron configuration of the elements that we have seen so far in Period 4:

• No.19: Potassium (K): 2,8,8,1

• No.20: Calcium (Ca): 2,8,8,2

• No.21: Scandium (Sc): 2,8,9,2

• No.22: Titanium (Ti): 2,8,10,2

• No.23: Vanadium (V): 2,8,11,2

We can see that, after No.20, the electrons get added to the second outer most shell. So there is no sequential increase in the number of electrons in the outer most shells of scandium, titanium and vanadium.

We will learn more details about such electron configuration in higher classes. At present, all we need to know is this:

■ The electrons are getting added to the second outer most shell. A phenomenon which creates more than one claim for the same position.

This special situation continues beyond No.23 vanadium. It continues up to No.30 zinc. After that normalcy is restored. That is., after No.30, the electrons get added to the outer most shell. So we have to provide a special place for the elements from 21 to 30. This special place is created in between Groups II and III. This is shown in fig.4.6 below:

Fig.4.6

How much space should be provided between Groups II and III?

• Enough space to accommodate elements from No.21 to No.30. This is shown in fig.4.7 below (right click, and select 'open in new tab' for an enlarged view):

Fig.4.7

A total of 10 elements (from No.21 to No.30) are accommodated in the newly created space. We can see that, the elements which come after No.30, that is., the elements from 31 to 36, strictly follow both the rules 7 and 8.
Now, all the elements in the world, which have 4 shells, are accommodated in the Period 4.

■ Let us move to the next period 5. In this period also, there are some problem causing elements. They have 5 shells, and thus comply with the rule 7. But they do not comply with rule 8. So they are also placed in the portion between Groups II and III. There are exactly 10 'problem causing elements', just as in Period 4

■ Let us move to the next period 6. In this period also, there are some 'problem causing elements'. They have 6 shells, and thus comply with the rule 7. But they do not comply with rule 8. So they are also placed in the portion between Groups II and III. In this period, the number of 'problem causing elements' are more. There are a total of 24 such elements. Note that there are only 10 such elements in Periods 4 and 5. If we include all 24 in between Groups II and III, then the table will become very long from left to right. We will not be able to print or draw it on a single sheet. So, 14 of them (starting from No.57 Lanthanum), are separated from the main table. These 14 are given a special name 'Lanthanides', and are kept at the bottom of the main table. This is indicated by the red arrow in the 'completed periodic table' shown in the fig.4.8 below (right click, and select 'open in new tab' for an enlarged view). This table is obtained from the Wikimedia commons, and can be seen here. All the elements in the world, which have 6 shells come under the red arrow in Period 6.

Fig.4.8

 ■ Let us move to the next period 7. In this period also, there are 24 'problem causing elements'. They have 7 shells, and thus comply with the rule 7. But they do not comply with rule 8. 14 of them (starting from No.89 Actinium) are separated from the main table. These 14 are given a special name 'Actinides', and are kept at the bottom of the main table, below the Lanthanides. This is indicated by the blue arrow in the fig.4.8. All the elements in the world, which have 7 shells come under the blue arrow in Period 7.
• Lanthanides are also known as rare earths
• Actinides are man made artificial elements (except Thorium and Uranium)

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So we have discussed the basics about the arrangement of elements in the tabular form. The Modern Periodic table is based on the works of the Russian scientist Dmitri Ivanovich Mendeleev. Works of the English scientist Henry Moseley contributed to the modifications of the Table.

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In the next section, we will discuss more features of the Periodic table. 

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Chapter 2 - Mass of Protons, Neutrons and Electrons

In the previous section, we saw the details about Chemical equations. In this chapter, we will see Details of Atoms.

We have seen elements, compounds, molecules, atoms etc., The objects that we learn about, are becoming smaller and smaller. Compare this with the research of astronomers: They explore more and more by studying higher and higher levels. Like, upper atmosphere, outer space, deep space etc., The objects that they deal with, become larger and larger. Like planets, stars, etc.,

We are going in the opposite direction. The objects that we study, become smaller and smaller. The comparison is shown in the fig.2.1 below:

Fig.2.1

We have seen molecules and atoms. Now we go still deeper. We want to know what is inside the atoms. At the center of the atom, there is the nucleus. Different types of particles are present inside the nucleus. One such particle is the Proton. It is a positively charged particle. Each atom of a particular element will have a particular number of protons. For example,
• an atom of Sodium will have 11 protons in it’s nucleus
• an atom of Nitrogen will have 7 protons in it’s nucleus
The number of protons in the atom of a particular element is a unique value, that will never change.

Even when the nucleus carries positively charged protons, the atom as a whole, is electrically neutral. Why is that so? The answer is that, the positive charge is neutralised by another type of particle called the Electron. This electron is a negatively charged particle. And the number of electrons in an atom will be equal to the number of protons. So the charges get neutralised. The electrons revolve around the nucleus. Just like planets revolve around the sun. The electrons revolve around the nucleus in fixed orbits. These orbits are also called shells. This is shown in the fig.2.2 below:

Fig.2.2

The shells around the nucleus are given fixed numbers. The numbering is done starting from the shell which is closest to the nucleus. So the shell which is closest to the nucleus will be numbered 1. The next shell 2, and so on. Larger the number of a shell, greater is it’s distance from the nucleus. In addition to numbering, the shells are also given names. K, L, M, N etc., So the shell number 1 is called the K-shell, shell number 2 is called the L-shell, shell number 3 is called the M-shell and so on.

The above model was put forward by Danish scientist Neils Bohr, and is called the Bohr model of atom.

Mass of particles
If we take a piece of iron in our hand, we will feel some weight. This is because, the iron piece that we took has ‘mass’. If we take a piece of wood of the same size, we will not feel the same weight. This is because, iron has more mass than wood of the same size. All substances have mass, even if they are very small. Molecules, atoms, protons, electrons.., all have mass.

When scientists studied about the mass of atoms, protons, electrons etc., they noticed an important difference:
Consider the following sum:
• Total mass of all the protons in an atom + Total mass of all the electrons in that atom.
The above sum must be equal to the ‘mass of the atom as a whole’. This is because, the atom consists of protons in the central nucleus, and the electrons revolving around the nucleus.

But it was found that, the above sum was very less than the mass of the atom. This is shown schematically in the fig. 2.3 below:

Fig.2.3

Based on this difference, scientists predicted that, protons and electrons are not the only components of an atom. Some thing else is also present.

In 1932, British scientist James Chadwick discovered the unknown particle. It was named as Neutrons. They are present in the nucleus of the atom. They are neither positively charged nor negatively charged. That is., they are electrically neutral. So we can include neutrons also, in the calculations related to mass.

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In day to day life, we require various units to make various measurements. For example, 
• We use the unit ‘meter’ to measure distances. 
• We use the unit kilogram to measure mass (or weight)
In the same way, we now need an ‘appropriate unit’ to measure very small masses. Because, electrons, protons etc., have very small masses. The unit that we use for this purpose is called the Atomic mass unit. It is abbreviated as u, just as meter is abbreviated as m and kilogram is abbreviated as kg.

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Now we must discuss an important point: 
■ One meter is the distance travelled by light in 1/299,792,458 of a second. So we have a condition: The distance travelled by light during a certain time. Such a condition puts a ‘fix’ on the meter. The advantage of having a ‘fix’ is that, different people will not take different values for ‘one meter’.
■ One gram also have a fix. It is the mass of 1 cm³ of water. From this, the mass corresponding to 1 kg can be easily calculated because, 1 kg = 1000 g
■ In the same manner, we want a fix on ‘u’ also. It is done as follows:
• One u is 1/12 of the mass of a carbon-12 atom
• One carbon-12 atom has a mass of 1.992646547 × 10 ^-27 kg
• 1/12 of that is 1.6605 × 10 ^-27 kg
• So, one u = 1.6605 × 10 ^-27

What is this ‘carbon-12 atom’ ? Why not just say ‘carbon atom’? 
Carbon-12 is a close relative of carbon. ‘12’ is strictly specified so that, there will not be any ambiguity while choosing from among close relatives who are ‘look-alikes’. We will learn about the relation later in this chapter.
Any way, it is not important for our present discussion. All we need to know at present is that 1 u = 1.6605 × 10 ^-27

So we have a 10, raised to the power of -27. let us analyse it further:
1 kg mass is known to us
• 0.1 kg = 10 ^-1 = one tenth of a kg
• 0.01 kg = 10 ^-2 = one hundredth of a kg. A smaller quantity.
• 0.001 kg = 10 ^-3 = It is a very small quantity. It comes into use when we measure the mass of precious metals like gold
• 0.0001 kg = 10 ^-4 . A still smaller quantity.

Notice that when the number of zeros (on the right side of the decimal point) increases, the quantity becomes smaller and smaller. 1.6605 × 10 ^-27 will have 26 zeros on the right side of the decimal point. It is an extremely small quantity. 

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Now we need to express the masses of electrons, protons etc., in terms of u. So that we can just write ‘u’ instead of 1.6605 × 10 ^-27. How do we do that​?

It is an ordinary case of ‘conversion from one unit to another’. Let us see a common example.
Suppose we have Rs. 2500/- with us. How many Euros is that? We can easily calculate it if we know ‘how much is one Euro in Rs.’ At the present rate, One Euro = 74.44 Rs. So Rs. 2500 = 2500/74.44 = 33.58 Euros. In this way we can convert the 'mass of a proton in kg', into 'mass of a proton in u'. The following comparison will help to understand the process better:

So we find that the procedure is to simply divide. 
• If we have Rs 2500/-, it is equivalent to having 33.58 Euros. 
• If we have 1.6605 × 10 ^-27 kg, it is equivalent to having 1.00727 u
In this way we can express the mass of neutron and electron also in terms of u. The steps are shown below:

For easy comparison, let us write the results together:
• Mass of Proton = 1.00727 u
• Mass of Neutron = 1.00866 u
• Mass of Electron = 0.005484 u

Consider the mass of proton. There are 2 zeros after the decimal point. That means, the quantity after the decimal point is small. So, for practical purposes, this portion after the decimal point is discarded. That leaves ‘1’ on the left side of the decimal point. That means, for practical purposes, the mass of one proton is taken as 1u

Next consider the mass of neutron. Here also there are 2 zeros after the decimal point. That means, the quantity after the decimal point is small. So, for practical purposes, this portion after the decimal point is discarded. That leaves ‘1’ on the left side of the decimal point. That means, for practical purposes, the mass of one neutron is taken as 1u


Now consider the mass of electron. Here also there are 2 zeros after the decimal point. That means, the quantity after the decimal point is small. So, for practical purposes, this portion after the decimal point is discarded. That leaves ‘0’ on the left side of the decimal point. That means, for practical purposes, the mass of neutron is taken as 0 u. So the neutron is given 'zero mass' in calculations.

The above calculations tallies with the experimental results. Experimentally, it is proved that most of the mass of an atom is concentrated in the nucleus. That is., total mass of the atom is due mainly to the nucleus. The electrons out side the nucleus does not contribute much to the total mass of the atom. This is shown schematically in the fig.2.4 below. 

Fig.2.4

So it is appropriate to consider electrons to have zero mass.

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In the next section we will see 'Mass number'.

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