Chapter 4.3 - Modern Periodic table - Solved examples

In the previous section, we saw some solved examples related to the position of elements in the periodic table. In this section we will see a few more solved examples.

Solved example 4.3

Electron configuration of elements P, Q, R and S are given below:

P – 2,2    Q – 2,8,2     R – 2,8,5     S – 2,8

(a) Which among these elements are included in the same period?

(b) Which among these elements are included in the same group?

(c) Which among these elements is a noble gas?

(d) To which group and period does the element R belong?

Solution:

(a) Elements in a same period will have the same number of shells

• In the given problem, P and S have both 2 digits in their electron configuration. So they have both 2 shells. Thus, they both belong to the Period 2

• Q and R have both 3 digits in their electron configuration. So they have both 3 shells. Thus, they both belong to the Period 3

(b) Elements in the same group will have the same number of electrons in their outermost shells

In the given problem, P and Q have both 2 electrons in their outermost shells. So they both belong to Group II

(c) S has an octet configuration in the outermost shell. So it is a noble gas

(d) • R has 5 electrons in it's outermost shell. So it belongs to Group V

• It has 3 digits in it's electron configuration. So it has 3 shells. Thus it belong to the Period 3

Solved example 4.4

An incomplete form of the periodic table is given below.

Write answers to the questions related to the position of elements in it.

(a) Which is the element with the biggest atom in group I?

(b) Which is the element with the lowest ionisation energy in group I?

(c) Which is the element with the smallest atom in period II?

(d) Which are the transition elements?

(e) Which among the two elements L and M, has the smallest electronegativity?

(f) Which among the two elements B and I, has higher metallic nature?

(g) Which are the elements that belong to the halogen family?
(h) Which type of ion will be formed from an atom of 'M'. Cation or anion?

(i) Which is the element that resembles E the most in it's chemical properties?

Solution:

(a) We have discussed the periodic trend related to the 'size of atoms', above in this section. We can write this:

• In a group, from top to bottom, size of atom increases

• In a period, from left to right, size of atom decreases

• In this problem, we have to consider the variation within a group only. The bottom most element will be the biggest. So, in group I, D is the biggest.

(b) We have discussed the periodic trend related to the 'ionisation energy', above in this section. We can write this:

• In a group, from top to bottom, the ionisation energy decreases

• In a period, from left to right, the ionisation energy increases

• In this problem, we have to consider the variation within a group only. The bottom most element will have the lowest ionisation energy. So, in group I, D will be having the lowest ionisation energy.

(c) The variation of size was written in problem (a) above. In this problem, we have to consider the variation within a period only. The element at the extreme right will be the smallest. So, in period 2, M is the smallest.

(d) The transition elements fall between groups II and III. So G and H are transition elements.

(e) We have discussed the periodic trend related to the 'electronegativity', above in this section. We can write this:

• In a group, from top to bottom, the electronegativity decreases

• In a period, from left to right, the electronegativity increases

• In this problem, we have to consider the variation within a period only. This is because, L and M fall within a period. The left most element will have the lower electronegativity. So, among L and M, the element L will be having the lower electronegativity.

(f) We have discussed the periodic trend related to the 'metallic nature'. We can write this:

• In a group, from top to bottom, the metallic nature increases

• In a period, from left to right, the metallic nature decreases

• In this problem, we have to consider the variation within a period only. This is because, B and I fall within a period. The left most element will have the higher metallic nature. So, among B and I, the element B will be having the higher metallic nature.

(g) We have learned that, halogen family is the group VII. So, among all the given elements, M and N are the halogens.
(h) • We have lerned about cations and anions here. Positive ions are Cations and negative ions are anions.
• As we move from left to right in a period, the metallic character decreases.
• That means, the tendency to lose electrons decreases. That means, tendency to gain electron increases
• As M is further to the right in it's period, it will have a tendency to gain electrons. So it will become a negative ion. That is., it will become an anion.

(i) Both B and F are close to E. But F will resemble E more because, F, like E has two electrons in the outermost shell
Solved example 4.5

Give examples for the following:

(a) Two elements that have 1 electron in their outermost shells

(b) Three elements that have 2 electrons in their outermost shells

(c) Three elements that have filled outermost shells

Solution:

(a) All elements in the group I have 1 electron in their outermost shells. We can pick any two from this group. Say lithium (Li) and sodium (Na)

(b) All elements in the group II have 2 electrons in their outermost shells. We can pick any three from this group. Say beryllium (Be), magnesium (Mg) and calcium (Ca)

(c) All elements in the group VIII have filled electrons in their outermost shells. We can pick any three from this group. Say neon (Ne), argon (Ar) and krypton (Kr)

Solved example 4.6

Lithium, sodium and potassium are all metals which react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?

Solution: All the three elements have one electron in the outermost shell. During chemical reactions, this single electron is readily donated by these metals

Solved example 4.7

In the modern periodic table, which are the metals among the first 10 elements?

Solution:

Lithium (Li) and Beryllium (Be).

Solved example 4.8

By considering their position in the periodic table, which one of the following elements would you expect to have the maximum metallic character?

Ga, Ge, As, Se, Br

Solution:

All the given elements belong to a same period. Among them, Ga is the one which is at the left most position. So It is the one with the maximum metallic character

Solved example 4.9

Which of the following statements is not a correct statement about the trends when going from left to right in any period

(i) The elements become less metallic in nature

(ii) The number of valence electrons increase

(iii) The atoms lose their electrons more easily

Solution:

Statement (iii) is false. As we move from left to right in any period, the metallic character decreases. That is., tendency to lose electrons decreases.

Solved example 4.10

Which element has

(i) Two shells, both of which are completely filled with electrons

(ii) Electron configuration 2,8,2

(iii) A total of 3 shells, with 4 electrons in the valence shell

(iv) A total of 2 shells, with 3 electrons in the valence shell

(v) twice as many electrons in it's second shell as in it's first shell

Solution:

(i) • Given that the element has two shells. So they are K and L

• Given that they are completely filled up. K-shell can hold a maximum of 2×1² = 2×1 = 2 electrons

• L-shell can hold a maximum of 2×2² = 2×4 = 8 electrons

• So the total number of electrons = total number of protons = 2 + 8 = 10

• Thus the atomic number Z = 10, and the element with this atomic number is Neon (Ne)

(ii) • Given that electron configuration is 2,8,2

• So number of electrons = number of protons = 2 + 8 + 2 = 12

• Thus the atomic number Z = 12, and the element with this atomic number is magnesium (Mg)

(iii) • Given that there are a total of 3 shells. So the shells are K,L and M

• Also given that there are 4 electrons in the valence shell. That means K and L are completely filled up. So number of electrons = 2 + 8 + 4 = 14

• Thus the atomic number Z = 14, and the element with this atomic number is silicon (Si)

(iv) • Given that there are a total of 2 shells. So the shells are K and L

• Also given that there are 3 electrons in the valence shell. That means K is completely filled up. So number of electrons = 2 + 3 = 5

• Thus the atomic number Z = 5, and the element with this atomic number is boron (B)

(v) • Here the total number of shells is not given. But that is not necessary to solve the problem

• Given that number of electrons in the second shell is twice that in the first. It is clear that there are some electrons in the second shell

• The second shell will begin to fill only if the first shell is completely filled up. The maximum capacity of the first shell is 2.

• So the number of electrons in the second shell = 2×2 = 4. The second shell is capable to carry more than this 4. So it follows that there are no more shells than K and L.

• So there are no more electrons than the 2 in first shell, and the 4 in the second shell.

• Thus the atomic number Z = 2 + 4 = 6, and the element with this atomic number is carbon (C)

Solved example 4.11

Nitrogen (Z=7) and Phosphorus (Z=15) belong to group V of the periodic table. Write the electron configuration of these elements. Which will be more electronegative? Why?

Solution:

• Electron configuration of nitrogen is 2,5

• Electron configuration of phosphorus is 2,8,5

• Both of them belong to the same group, with nitrogen above phosphorus. We have seen that while moving from top to bottom in a group, electropositive character increases. That is., electronegative character decreases. So, nitrogen, which is above, is more electronegative.

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We have completed the basic discussion on the Modern periodic table. We will learn more details in higher classes. In the next section, we will discuss about Non-metals. 

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Chapter 4.2 - More periodic trends and Solved examples

In the previous section, we saw the periodic trend in the size of atoms. In this section we will see more properties.

Ionisation energy

We have seen reactions in which ions are formed. Consider for example, the formation of NaCl. We have seen the details here. The sodium (Na) atom loses one electron, and becomes Na+ ion. The chlorine (Cl) atom gains one electron, and become Cl- ion.

• Consider the formation of Na⁺ ion. One electron has to be removed from the Na atom, for the formation of Na⁺. Is such a removal easy?

• The negatively charged electrons are attracted towards the central nucleus. This is because of the positively charged protons present in the nucleus.

• So the electron must over come this attractive force, in order to move away from the atom.

• To over come this attractive force, some energy is required.

• This energy is called ionisation energy

• But, for the same element, this energy may be different under different conditions. For example, the presence of some other element can cause an increase or decrease in the required energy. Also, more energy is required for an electron in an inner shell, than that in an outer shell. So the conditions are to be specified. Thus the definition of ionisation energy is as follows:

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The amount of energy required to liberate the most loosely bound electron from the outer most shell of an isolated gaseous atom of an element is called it's ionisation energy.

[Note that several conditions are specified:

• The electron must be the one which is most loosely bound

• It must be situated in the outer most shell

• The atom must be isolated. That is., there must not be any presence of any other atoms

• The atom must be in the gaseous state]

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So let us analyse the variation of ionisation energy at different parts of the periodic table:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, the size of the atom increases. We have already seen the reason here.

2. When the size increases, the distance from the nucleus to the outer most electron also increases

3. When the distance increases, the force exerted by the nucleus (on the outer most electron) decreases

4. When this force is low, low energy is sufficient to liberate the electron.

5. So we can write: The ionisation energy decreases as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, the size of the atom decreases. We have already seen the reason.

2. We have seen that the decrease in size is due to the increase in attractive force from the central nucleus

3. From (2) we can readily reach a conclusion: As the attractive force increases, more energy is required to liberate the electron.

4. So we can write: The ionisation energy increases as we move from left to right in a period

Electronegativity

We have already learned about electronegativity. It is the ability of an atom to attract the shared pairs of electrons in covalent bonds. We have seen the details here. Now we want to understand it's periodic trend.

• Electronegativity is closely related to the size of atom. Let us write an analysis:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, the size of the atom increases. We have already seen the reason here.

2. When the size increases, the distance from the nucleus to the outer most electron also increases

3. When the distance increases, the force exerted by the nucleus (on the outer most electron) decreases

4. From (3), we see that, when size increases, distance also increases, and the atom becomes lesser and lesser capable to attract 'it's own outer most electrons'.

5. Then there is nothing more to tell about the 'external shared electrons'. We can straight away say: The atom is even less capable to attract the shared pairs of electrons.

6. So we can write: The electronegativity decreases as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, the size of the atom decreases. We have already seen the reason.

2. We have seen that the decrease in size is due to the increase in attractive force from the central nucleus

3. So there is an increase in attraction on the 'outer most electrons'. This increase will be felt by the 'shared pairs of electrons' also. We can straight away say: The atom is capable to exert significant force of attraction on the shared pairs of electrons

4. So we can write: The electronegativity increases as we move from left to right in a period

Metallic and Non-metallic Nature

• Elements with a metallic nature, generally loses electrons in chemical reactions.

• Metals are called electropositive because, in chemical reactions, they lose electrons to become positive ions. (Metals are good conductors of electric current because, they readily lose electrons, and hence have lots of free electrons)

• Non-metals are called electronegative because, in chemical reactions, they gain electrons to become negative ions.

• The metallic or non-metallic nature of an element is closely related to it's ionisation energy. Why?

The reason is simple:

    ♦ If the ionisation energy is low, electrons will be easily liberated. If the electrons are easily liberated, it shows a metallic nature.

   ♦ If the ionisation energy is high, electrons will not be easily liberated. If the electrons are not easily liberated, it shows a non-metallic nature.

• So, from the periodic trend of ionisation energy, we will be able to write the periodic trend for metallic and non-metallic nature also. Thus we get:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, ionisation energy decreases

2. Decrease in ionisation energy indicates increase in metallic nature. So we can write:

3. The metallic nature increases (consequently, non-metallic nature decreases) as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, ionisation energy increases

2. Increasein ionisation energy indicates increase in non-metallic nature. So we can write:

3. The non-metallic nature increases (consequently, metallic nature decreases) as we move from left to right in a period

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Based on the results in I and II above, we can answer an interesting question:

■ Which is the element with the most metallic character in the periodic table?

Solution: 1. From II above we have: In any period, the left most element is the most metallic in nature

2. So the element that we are seeking, is some where in the first group

3. From I above we have: In any group, the bottom most element is the most metallic in nature
4. So the bottom most element in the first group (francium) is the one which is most metallic in character

• However, francium is a man made element. Among the natural elements, caesium is the one which is the most metallic in character. Look at the position of caesium. It is just above francium.

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■ In the above discussion, we derived the periodic trend of metallic and non-metallic character based on the 'trend of ionisation energy'
■ We can derive the periodic trend of metallic and non-metallic character based on the 'trend of electronegativity' also. The reader is advised to write such an analysis by him/herself.

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Now we will see some solved examples

Solved example 4.1

Complete the table given below:

Solution:

1. Consider lithium. It’s electron configuration is given as 2,1. So the atomic number Z = 2 + 1 = 3

2. Consider oxygen. Z is given as 8. So the electron configuration is 2,6

• From the electron configuration, we get: No. of electrons in the outer most shell = 6. So oxygen belongs to Group VI

• There are 2 digits in the electron configuration. So No. of shells = 2. Thus, oxygen belongs to the Period 2

See rules 7 and 8

3. Consider argon. Z is given as 18. So the electron configuration is 2,8,8

• From the electron configuration, we get: No. of electrons in the outer most shell = 8. So argon belongs to Group VIII

• There are 3 digits in the electron configuration. So No. of shells = 3. Thus, argon belongs to the Period 3

4. Consider calcium. It’s electron configuration is given as 2,8,8,2. So Z = 2 + 8 + 8 + 2 = 20

• From the electron configuration, we get: No. of electrons in the outer most shell = 2. So calcium belongs to Group II

• There are 4 digits in the electron configuration. So No. of shells = 4. Thus, oxygen belongs to the Period 4

The completed table is given below:

Solved example 4.2

There are 3 shells in the atom of element 'X'. 6 electrons are present in it's outermost shell.

(a) Write the electron configuration of the element

(b) What is it's atomic number Z?

(c) In which period does this element belong?

(d) In which group does this element belong?

(e) Write the name and symbol of this element

(f) To which family of elements does this element belong?

Solution:

(a) Given that the element has 3 shells. So the element has K, L, and M shells

• Given that the M shell has 6 electrons. So the L and M would be filled up to their full capacities.

• That means there are 2 electrons in K, and 8 electrons in L

• So the total number of electrons = 2 + 8 + 6 = 16

The electron configuration is 2,8,6

(b) Total number of electrons = 16 = Total number of protons

So atomic number Z = 16

(c) There are 3 shells for the atom. So element belongs to the Period 3

(d) There are 6 electrons in the outermost shell. So the element belongs to Group VI

(e) The name of the element with Z = 16 is Sulfur. It's symbol is S

(f) The family is Oxygen family

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In the next section, we will see a few more solved examples. 

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Chapter 4.1 - Families and Periodic trends in the Periodic table

We have seen how the elements are arranged in the periodic table. The arrangement is based on rules 7 and 8 that we saw in the previous section. When the elements are arranged in this way, we obtain some favourable 'side effects'. What are those side effects? Let us examine:

Look at the group I. The elements in this group are: Li, Na, K, Rb. Cs and Fr

• All of them have 1 electron in their outer most shells.

• These elements have similar properties. All of them form alkalies when reacting with water. For example, When sodium (Na) reacts with water, sodium hydroxide (NaOH) is formed.

• Because of the similarities in the properties, these elements can be regarded as 'members of a single family'. This family is given a special name: Alkali metals.

• Similarly, all elements in group II belong to the family called Alkaline Earth metals. The properties of all the elements in this family are also similar. Just like Alkali metals, this family also form alkalies. Then why the term 'earth'? 

• Because, in early days, these elements were seen only as compounds, in 'minerals obtained from earth'. They do not occur in free form in nature. Later, scientists were able to isolate them from the minerals. But the name stuck.

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The table below gives the full list of the families in the periodic table:

Group I: Alkali metals

Group II: Alkaline earth metals

Group III: Boron family

Group IV: Carbon family

Group V: Nitrogen family

Group VI: Oxygen family

Group VII: Halogens

Group VIII: Noble gases

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The elements helium, neon, argon, krypton, xenon and radon, which belong to group VIII are called Noble gases. They are monoatomic molecules. Because they do not need to combine with another atom to attain stability. (We have seen the details here also see here). They already have octet configuration. Normally, they do not combine with other elements also. Hence they are called inert gases. As they are found only in very small quantities, they are also called rare gases.

• Helium is used in weather balloons.

• Neon is used in discharge lamps to obtain orange colour

• Argon is used in electrical bulbs to prevent the evaporation of the filament

• Radon is a radioactive gas

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So we find that elements with similar properties fall together in the periodic table. This is one of the many 'favourable side effects' that we obtain, while making a systematic arrangement of elements. We will see more as we continue our discussion.

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Representative elements

The first 10 elements show a periodicity in electron filling. This is shown in the table below:

Look at the column for L-shell. The electron filling is in sequential order from 1 to 8. In the periodic table, these first 10 elements reside at the top most portions in their respective groups. They are the first elements in their families, and are called Representative elements.

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• The position of hydrogen is still under debate. Some of it's properties makes it eligible to be a member of the Alkali metal family (group I)

• Some of it's other properties makes it eligible to be a member of the halogen family (group VII). Let us examine:
1. Hydrogen loses one electron in some chemical reactions. Alkali metals also lose one electron in chemical reactions

2(a) Hydrogen is a non metal. Halogens are also non metals

(b) Hydrogen is diatomic. Halogens are also diatomic

(c) Hydrogen is not a solid. Halogens are also not solid. (Alkali metals are solid)

(d) Hydrogen is not a metal. Halogens are also not metals. (Alkali 'metals' are metals)

(e) Hydrogen has high ionisation energy. Halogens also have high ionisation energy. (Alkali metals have low ionisation energy). We will learn about ionisation energy later in the next section.

(g) Hydrogen gains one electron in some chemical reactions. Halogens also gain one electron in chemical reactions

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Transition elements

Consider the groups I and II. The elements in these two groups are metals. Consider the groups III to VIII. They are non metals. [However, some elements (inside green squares coming under groups III, IV, V and VI in fig.4.8) show properties of both metals and non metals. They are classified as Metalloids] The elements in the space between group II and group III are called Transition elements. Because, they form a transition from metals to non metals. The following are some of the properties of transition elements:

• They are metals

• They form coloured compounds

• They show similarities in chemical properties in groups as well as in periods. That is:

    ♦ If we take any single vertical group from within the transition elements, all the elements in that group will have similar chemical properties

    ♦ If we take any single horizontal period from within the transition elements, all the elements in that period will have similar chemical properties

• In compounds, they exhibit different oxidation states. For example: Fe²⁺ and Fe³⁺

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Periodic trends in the Periodic table

The word 'period' means an interval of time. An example for the usage of the word 'periodic' is:

Periodic checkup of a car. It means, that, the car has to undergo regular checkup at the ends of definite periods. So the event 'checkup of the car' occurs at the end of regular periods. In other word, it occur periodically.

In the same way, some of the properties of elements occur periodically. Let us see an example:

■ Point your finger on Lithium No.3

• It has 1 electron in it's outer most shell.

• Now add 8 to it's number. We get 3 + 8 = 11. Element no. 11 is sodium.

• It also has 1 electron in it's outer most shell

■ Point your finger on Fluorine No.9

• It has 7 electrons in it's outer most shell.

• Now add 8 to it's number. We get 9 + 8 = 17. Element no. 17 is Chlorine.

• It also has 7 electrons in it's outer most shell

■ So 8 is the period. This period '8' works for elements upto magnesium No.12. 
Let us see the period for the elements after magnesium:

■ Point your finger on Potassium No. 19.

• It has 1 electron in it's outer most shell.

• Now add 18 to it's number. We get 19 + 18 = 37. Element no. 37 is Rubidium.

• It also has 1 electron in it's outer most shell

■ Point your finger on germanium No. 32.

• It has 4 electrons in it's outer most shell.

• Now add 18 to it's number. We get 32 + 18 = 50. Element no. 50 is Tin.

• It also has 4 electron in it's outer most shell

■ So 18 is the period for the elements after 12.

What is the difference between the two periods '8' and '18'?

The difference is 18 – 8 = 10. Note that 10 is the 'number of groups' of the transition elements
It is not just the 'number of electrons in the outer most shell'. Chemical properties of the element obtained by adding the appropriate period 8 or 18 will also be similar.

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• We have not checked the period for any transition elements.

• For the transition elements, we may not get the same number of electrons in the outer most shell after the period 18. 
• But still, the element that we get by adding 18, will have similar chemical properties.

• Within the transition elements, after barium No.56, the period will become '32' instead of '18'. This is because of the presence of 14 Lanthanides and Actinides. [18 + 14 = 32] 

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Now let us see some important properties that vary periodically:

Size of atoms in a group

When we move down any group, the size of atom increases. This is because, the number of shells increases.

Example: In group II, Ca has 4 shells. Sr, which is just below Ca, has 5 shells. Naturally, the atom with 5 shells will be larger than that with 4 shells. Some Bohr models showing the details of shells can be seen here.

Size of atoms in a period

We want to know how the size of atoms vary when we move from left to right in any period. Let us write an analysis:

1. In any period, the 'number of shells' is same for all the elements in that period

2. In any period, when we move from left to right, the number of electrons get increased by '1'

3. The number of electrons is same as the number of protons in the nucleus. So the number of protons also increase as we move from left to right.

4. Greater number of protons and electrons means that there will be greater force of attraction between the positively charged protons and the negatively charged electrons.

5. The greater force of attraction pulls the shells more and more towards the nucleus.

6. As a result, the overall size of the atom decreases as we move from left to right in a period.

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In the next section, we will discuss more Periodic properties. 

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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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