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 2.3 - Isotopes, Isobars and Isotones

In the previous section, we saw the Bohr model electronic configuration of atoms. We also saw some solved examples. In this section, we will learn about Isotopes, Isobars and Isotones.

We have seen that the number of protons in an atom is unique. It is the atomic number Z, and it will not change. It is the identity of the element. We have also seen that, in addition to protons, the nucleus contains neutrons also. The number of neutrons may not be the same in all atoms of the same element. Consider an example:

Fig.2.15

• In fig.2.15(a), the atom has 1 proton, 1 electron, and 0 neutron. So it's mass number A = Z + No. Of neutrons = 1 + 0 = 1 
• In fig(b), the atom has 1 proton, 1 electron, and 1 neutron.  So it's mass number A = Z + No. Of neutrons = 1 + 1 = 2
• In fig(c), the atom has 1 proton, 1 electron, and 2 neutrons.  So it's mass number A = Z + No. Of neutrons = 1 + 2 = 3


In all the three cases, the number of protons is the same. So the three atoms belong to the same element. Though they belong to the same element, the mass is different. So we need a method to distinguish between the three. For this, we use the term 'isotopes'. In the above fig.2.15, the three atoms are 'isotopes of the same element'. 

In the fig., the element shown is hydrogen. Because  Z = 1. So we say: The three atoms are 'isotopes of hydrogen'. The three isotopes of hydrogen are:
• Protium Z = 1, A = 1
• Deutirium Z =1, A = 2
• Tritium Z = 1, A = 3

Representation of Isotopes

We have seen that the elements are represented by their symbols. We have also seen that the Z is written on left side bottom, and A on left side top. This same method can be used to represent isotopes also. The isotopes of hydrogen are represented as shown below:

(a) represents Protium, (b) represents Deuterium, and (c) represents Tritium  

In the same way, the isotopes of carbon can be represented as:

(a) represents Carbon-12, (b) represents Carbon-13, and (c) represents Carbon-14

Uses of isotopes
Isotopes find application in many fields. Some examples are:
• Carbon-14 is used to determine the age of fossils and prehistoric objects
• Deuterium is used in atomic reactors

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So we have seen the details about isotopes. It involves the combination of two items: (i) Atomic number Z (same), and (ii) Mass number A (different)
Other combinations are also possible. Consider the example given below:

(a) is Argon. It has Z = 18 and A = 40
(b) is Calcium. It has Z = 20 and A = 40
Note that both have A same. Such elements having same mass number but different atomic numbers are called isobars. The combination here is: (i) Atomic number Z (different), and (ii) Mass number A (same).

Another combination that is possible is, (i) Atomic number (different) and (ii) Number of neutrons (same). An example is given below:

(a) is Nitrogen. It has Z = 7 and A = 15. There fore number of neutrons = 15 -7 = 8
(b) is Carbon-14. It has Z = 6 and A = 14. There fore number of neutrons = 14 - 6 = 8 

Number of neutrons is the same. Such elements having different atomic numbers but same number of neutrons are called isotones.

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We will now see some solved examples
Solved example 2.6
Bohr models of atoms A, B, C and D (symbols are not real) are given below:

(a) Write the atomic number, mass number and electronic configuration of the atoms
(b) Among these, which are isotopes? why?
Solution:
■ (A) has 6 protons in it's nucleus. So the atomic number Z = 6
• It has 6 neutrons in the nucleus. So mass number = No. of protons + No. of neutrons = 6 + 6 = 12
• It has 2 electrons in the K-shell and 4 electrons in the L-shell. So the electronic configuration is 2,4

■ (B) has 7 protons in it's nucleus. So the atomic number Z = 7
• It has 8 neutrons in the nucleus. So mass number = No. of protons + No. of neutrons = 7 + 8 = 15
• It has 2 electrons in the K-shell and 5 electrons in the L-shell. So the electronic configuration is 2,5

■ (C) has 6 protons in it's nucleus. So the atomic number Z = 6
• It has 8 neutrons in the nucleus. So mass number = No. of protons + No. of neutrons = 6 + 8 = 14
• It has 2 electrons in the K-shell and 4 electrons in the L-shell. So the electronic configuration is 2,4

■ (D) has 8 protons in it's nucleus. So the atomic number Z = 8
• It has 8 neutrons in the nucleus. So mass number = No. of protons + No. of neutrons = 6 + 8 = 14
• It has 2 electrons in the K-shell and 6 electrons in the L-shell. So the electronic configuration is 2,6

■ Among the above four atoms, (A) and (C) are isotopes because they both have the same atomic number Z = 6, and different mass numbers. (A) has mass number 12, and (C) has mass number 14. It can be represented as:

Solved example 2.7
Symbols (not real symbols) of some atoms are given below

(a) Find the atomic number and mass number of these elements
(b) Which among these are isotopic pairs
(c) Draw the bohr model of atom Q 
Solution:
a. (P) has atomic number Z = 8 and mass number A = 17
(Q) has atomic number Z = 18 and mass number A = 36
(R) has atomic number Z = 8 and mass number A = 16

b. (P) and (R) are isotopic pairs. because they have the same atomic number Z = 8, and different mass numbers. P has mass number = 17 and Q has mass number = 16

c.The Bohr model of atom Q is shown below:

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In the next chapter we will see 'Chemical bonding'.

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Chapter 2.2 - Bohr model configuration - Solved examples

In the previous section, we saw the Bohr model electronic configuration of atoms. In this section, we will see some solved examples.

Solved example 2.3
Write the electronic configuration of the following elements and draw their Bohr model
(i) N with Z = 7 and A = 14  (ii) Mg with Z = 12 and A = 24  (iii) S with Z = 16 and A = 32
Solution:
(i) • Given Z = 7 and A = 14. 
• So number of protons = number of electrons = 7 
• Number of neutrons = A - Z = 14 - 7 = 7. (Number of neutrons must be marked in the nucleus when we draw the Bohr model)
• The electronic configuration can be determined by putting the arrangement of the 7 electrons in a tabular form as shown in fig. 2.10 below:

Fig.2.10

• From the table we can see that, the first 2 electrons fill up the K-shell, and the remaining 5 electrons take positions in the L-shell. So we get K = 2 and L = 5. So the electronic configuration is: 2,5
• Once we get the configuration, the Bohr model can be easily drawn. In the fig.2.10 above, the inner circle (K-shell) has 2 electrons, and the outer circle (L-shell) has 5 electrons

(ii) • Given Z = 12 and A = 24 
• So number of protons = number of electrons = 12 
• Number of neutrons = A - Z = 24 - 12 = 12. (Number of neutrons must be marked in the nucleus when we draw the Bohr model)
• The electronic configuration can be determined by putting the arrangement of the 12 electrons in a tabular form as shown in fig. 2.11 below:

Fig.2.11

• From the table we can see that, the first 2 electrons fill up the K-shell, next 8 electrons will fill up the L-shell. Maximum no. of electrons that L-shell can hold is 8. So the remaining 2 electrons take positions in the M-shell. So we get K = 2, L = 8 and M = 2. So the electronic configuration is: 2,8,2
• Once we get the configuration, the Bohr model can be easily drawn. In the fig.2.11 above, the inner circle (K-shell) has 2 electrons, the next outer circle (L-shell) has 8 electrons, and the outer most circle (M-shell) has 2 electrons.

(iii) • Given Z = 16 and A = 32 
• So number of protons = number of electrons = 16 
• Number of neutrons = A - Z = 32 - 16 = 16. (Number of neutrons must be marked in the nucleus when we draw the Bohr model)
• The electronic configuration can be determined by putting the arrangement of the 16 electrons in a tabular form as shown in fig. 2.12 below:

Fig.2.12

• From the table we can see that, the first 2 electrons fill up the K-shell, next 8 electrons will fill up the L-shell. Maximum no. of electrons that L-shell can hold is 8. So the remaining 6 electrons take positions in the M-shell. We get K = 2, L = 8 and M = 6. So the electronic configuration is: 2,8,6
• Once we get the configuration, the Bohr model can be easily drawn. In the fig.2.12 above, the inner circle (K-shell) has 2 electrons, the next outer circle (L-shell) has 8 electrons, and the outer most circle (M-shell) has 6 electrons.

Solved example 2.4
Bohr model of Al (Aluminium) is shown in fig.2.13 below. Analyse it and answer the following questions:

Fig.2.13

(i) Write the Atomic number and mass number of aluminium
(ii) Write the number of protons, neutrons and electrons in an aluminium atom
(iii) Write the electronic configuration of aluminium
Solution:
(i) In the Bohr model, it is written 13p. So the number of protons = 13. Also it is written 14n. So the number of neutrons = 14. Thus we get:
• Atomic number Z = No. of protons = 13
• Mass number A = No. of protons + No. of neutrons = 13 + 14 = 27
(ii) • Number of protons = 13. • No. of neutrons = 14
• No. of electrons in the k-shell = 2, No. of electrons in the L-shell = 8, and the no. of electrons in the M-shell = 3. Thus total no. of electrons = 2 + 8 + 3 = 13
(iii) We wrote the number of electrons in each shell in the previous step. From that, we get electronic configuration = 2,8,3

Solved example 2.5
The mass number of an atom = 31. It has 5 electrons in the M-shell.
(i) Write the electronic configuration of this atom
(ii) What is the atomic number of this atom ?
(iii) How many neutrons does this atom have?
(iv) Draw the Bohr model of this atom
Solution:
(i) Given that the M-shell contains 5 electrons. This is shown in fig.2.14(a) below:

Fig.2.14

• The M-shell can contain upto 2 × 3² = 18 electrons
• If the M-shell is the outermost shell, it can contain upto 8 electrons
• In our case it contains only 5 electrons. 
• So we can conclude that, in the given atom, the M-shell is the outermost shell. Because, if it were an inner shell, it would have contained more than 5 electrons
• Two other conclusions can also be made:
    ♦ The inner K-shell is completely filled up
    ♦ The inner L-shell is also completely filled up
• The above conclusions are made on the fact that, occupation of M can begin only when K and L are completely filled up. So we get: 2 electrons in K, 8 electrons in L and 5 electrons in M
• Thus the electronic configuration is 2,8,5
(ii) The total number of electrons = 2 + 8 + 5 = 15. So the number of protons is also equal to 15. Thus atomic number Z = 15
(iii) Mass number A is given as 31. So number of neutrons = A - Z = 31 - 15 = 16
(iv) The Bohr model of the given atom is shown in fig.2.14(b) above

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In the next section we will see 'Isotopes'.

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Chapter 2.1 - Bohr model Electronic configuration of Atoms

In the previous section, we saw the mass of protons, neutrons and electrons. In this section, we will learn about Mass number and Atomic number. Also later in this section, we will learn about the arrangement of electrons in the atoms.

We  have seen that the mass of a proton = 1u, mass of a neutron = 1u, and mass of an electron = 0. Consider an atom in which there are 2 protons, 2 neutrons and 2 electrons. We can calculate the mass of that atom as follows:
• Mass of 2 protons = 2 × 1 u = 2 u
• Mass of 2 neutrons = 2 × 1 u = 2 u
• Mass of 2 electrons = 2 × 0 = 0
∴ Total mass = 2 u + 2 u + 0 = 4 u
The same steps can be written in a shorter form:
• Total mass = 2 u + 2 u + 0 
• ⇒ Total mass = (2 + 2) × u = 4 u
What we have inside the parenthesis is the total number of protons and neutrons. So, to calculate the mass of the atom, all we have to do, is to add the number of protons and neutrons, and multiply the sum by u.
The steps become that easy due to the following reasons:
• Both protons and neutrons have the same '1u' as mass. So it can be taken outside the paranthesis
• The mass of electrons do not come into the calculations because it is zero

Let us verify this by taking another example:
An atom has 12 protons, 13 neutrons and 12 electrons. Find the mass of the atom:
Solution:
• Total mass = 12 × 1 u + 13 × 1 u + 12 × 0
• same as Total mass = (12 u + 13 u + 0)
• Same as Total mass = (12 + 13) × u
■ We get the same final step: Add the number of protons and neutrons, and multiply the sum by u.

This can be further simplified: 
• If we get the sum of the number of protons and number of neutrons, we can straight away calculate the mass of the atom 
• So this sum has special place in chemistry. Because, this sum gives the mass of the atom 
• Once we get the sum, all we have to do is to multiply it by 'u'. 

As the sum gives the mass directly, it is called the Mass number. So:
■ No. of protons + No. of neutrons = Mass number
■ Mass number is denoted by the letter 'A'

Significance of particles
We have seen the three particles: Protons, neutrons and electrons. Protons and neutrons are inside the nucleus. While electrons revolve around the nucleus in fixed orbits. When the atoms collide with each other, one or more electrons, especially, those in the outer orbits may be knocked out. When this happens, the number of electrons will become less than the number of protons. The ‘equality in the number of protons and electrons’ can change under other conditions also. For example, during chemical reactions, some electrons may get exchanged between atoms. That is., some atoms may lose some electrons, and those electrons will be accepted by some other atoms. So the losing atom will have a decreased number of electrons, while the accepting atom will have an increased number of electrons. 

So we see that the number of electrons can change. But such changes do not happen to the number of protons. Because, they are inside the nucleus. So we choose the number protons to identify an element.

■ The total number of protons in an atom is called the Atomic number of that atom. It is represented by the letter Z

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Based on A and Z, we may get problems involving the calculations of the ‘number of various particles’. An example is given below:

Solved example 2.1
For an atom, Z = 17, and A = 35. Find the number of protons, neutrons and electrons in that atom.
Solution:
• Given Z = 17 and A = 35
• For any atom, the atomic number Z is the number of protons in it. So we get:
■ No. of protons = 17   
• For any atoms (except ions), number of protons is equal to the number of electrons. So we get:
■ No. of electrons = 17
• For any atom, mass number A = No. of protons + No. of neutrons
• So 35 = 17 + no. of neutrons
■ ∴ No. of neutrons = 35 – 17 = 18

Solved example 2.2
The mass of an atom is 4u. It has 2 protons in it's nucleus. How many neutrons does it have?
Solution:
• Given mass of the atom = 4u and No. of protons = 2
• We have, Mass of atom = (No. protons + No. of neutrons) × u
• So we get 4u = (2 + No. of neutrons) × u
• ⇒ 4 = 2 + No. of neutrons
• ∴ No. of neutrons = 4 - 2 = 2

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We have learnt about 'symbol of elements'. When we write the symbol, it also represent ‘one atom’ of the element. To this symbol, we can attach the atomic number Z and mass number A. Both are written on the left side. A on left side top and Z on left side bottom. For example, Sodium (Na) has Z = 11 and Z = 23. We can write the symbol as shown in the fig.2.5(a). Fig.2.5(b) shows the general case where X is any element:

Fig.2.5

Arrangement of electrons in an atom
We have seen that the electrons orbit around the nucleus in definite orbits or shells. We have also seen that these orbits are given specific names: K, L, M, N etc., as well as specific numbers: 1, 2, 3, 4 etc.,

The electrons cannot take any orbit they like. There are strict rules by which the electrons are arranged in the orbits. Let us learn those rules:
• Rule 1: The orbits of lower energy levels gets filled first. So the K-shell (No. =1) is filled first. Because it has the lowest energy level. Then the L-shell (No. = 2) is filled. Then M-shell, and so on..
• Rule 2: The maximum number of electrons that can be accommodated in a shell is given by 2n². Where n is the shell number.
    ♦ Based on this rule, the maximum number of electrons that can be accommodated in the K-shell = 2 × 1² = 2 × 1 = 2
    ♦ Maximum number of electrons that can be accommodated in the L-shell = 2 × 2² = 2 × 4 = 8
    ♦ Maximum number of electrons that can be accommodated in the M-shell = 2 × 3² = 2 × 9 = 18
    ♦ Maximum number of electrons that can be accommodated in the N-shell = 2 × 4² = 2 × 16 = 32
• Rule 3: The outer most shell in any atom can accommodate upto a maximum of 8 electrons only

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We will now arrange the electrons in various atoms. While doing the arrangement, we will see the application of the three rules also. First we take the element hydrogen with atomic number 1. It has only one electron. That electron must occupy the shell with the lowest energy level, which is the K-shell. This is shown in the fig.2.6 below: A tabular form of the arrangement is also shown at the side.

Fig.2.6

In the tabular form, the electron is indicated by a small red square. It comes under the K-shell.
Next consider Helium, which has an atomic number 2. The fig. 2.7 below shows the arrangement:

Fig.2.7

The K-shell can accommodate 2 electrons. So the 2 electrons of Helium will occupy the K-shell.
Next consider Lithium, which has an atomic number 3. The fig. 2.8 below shows the arrangement:

Fig.2.8

The K-shell can accommodate a maximum of only 2 electrons. So the third electron of Lithium must go to the L-shell.
In this way, we can write the configuration for various elements. What happens when the L-shell gets filled up ? The next electron will go to the M-shell. To demonstrate this, we look at the configuration of Neon and Sodium together. The fig. 2.9 below shows the details:

Fig.2.9

Neon has 10 electrons. 2 electrons go to the K-shell. There are 8 electrons remaining. They can occupy the L-shell. But then, the L-shell is filled up. Because 8 is the maximum no. of electrons that the L-shell can accommodate. It does not cause a problem for Neon because, it does not have more than 10 electrons. But when we come to the next element Sodium, the number of electrons is 11. So the eleventh electron must occupy the M-shell.

The configuration of elements up to Argon can be seen here. 

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In the next section we will see some solved examples.

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